All practice questions

Question 8221Question →

What mass of calcium hydroxide, $\text{Ca(OH)}_2$ , must be dissolved in distilled water to make $250\text{ cm}^3$ of an aqueous solution with a pH of $12.00$ at $25^\circ\text{C}$ ? $[\text{Molar mass of Ca(OH)}_2 = 74.0\text{ g mol}^{-1}]$

0.0925\text{ g}

0.185\text{ g}

0.370\text{ g}

0.740\text{ g}

Show answer & explanation

Answer: $0.0925\text{ g}$

Answer

0.0925\text{ g}

At

25^\circ\text{C}

,

\text{pOH} = 14 - 12 = 2

, giving

[\text{OH}^-] = 10^{-2} = 0.01\text{ mol dm}^{-3}

. Because

\text{Ca(OH)}_2

produces two

\text{OH}^-

ions per formula unit upon dissociation, the molar concentration of

\text{Ca(OH)}_2

is

0.005\text{ mol dm}^{-3}

. In

250\text{ cm}^3

(

0.25\text{ dm}^3

), the number of moles is

0.005 \times 0.25 = 0.00125\text{ mol}

. Multiplying by the molar mass (

74.0\text{ g mol}^{-1}

) gives

0.0925\text{ g}

.

Step-by-Step Solution

1

Calculate the pOH and hydroxide ion concentration

[\text{OH}^-]

from the given pH.

\text{pOH} = 14.00 - 12.00 = 2.00 \implies [\text{OH}^-] = 10^{-2.00} = 0.01\text{ mol dm}^{-3}

.

Water ion product relationship

\text{pH} + \text{pOH} = 14.00

at

25^\circ\text{C}

connects pH to hydroxide ion concentration.

2

Determine the molar concentration of

\text{Ca(OH)}_2

.

[\text{Ca(OH)}_2] = \frac{[\text{OH}^-]}{2} = \frac{0.01\text{ mol dm}^{-3}}{2} = 0.005\text{ mol dm}^{-3}

.

Calcium hydroxide dissociates completely according to

\text{Ca(OH)}_2(aq) \rightarrow \text{Ca}^{2+}(aq) + 2\text{OH}^-(aq)

, yielding two moles of

\text{OH}^-

per mole of base.

3

Calculate the number of moles of

\text{Ca(OH)}_2

required for

250\text{ cm}^3

of solution.

\text{Moles} = 0.005\text{ mol dm}^{-3} \times 0.25\text{ dm}^3 = 0.00125\text{ mol}.

Volume must be converted from

\text{cm}^3

to

\text{dm}^3

by dividing by 1000.

4

Calculate the required mass of

\text{Ca(OH)}_2

.

\text{Mass} = 0.00125\text{ mol} \times 74.0\text{ g mol}^{-1} = 0.0925\text{ g}.

Mass equals number of moles multiplied by molar mass.

Key Concept

Calculating required solute mass from pH for a dibasic base by converting pH to pOH, adjusting for stoichiometry, and scaling by volume and molar mass.

Question 8222Question →

For the thermal decomposition of a compound, the standard enthalpy change ( $\Delta H^\circ$ ) is $+117.0\text{ kJ mol}^{-1}$ and the standard entropy change ( $\Delta S^\circ$ ) is $+180.0\text{ J K}^{-1}\text{ mol}^{-1}$ . What is the minimum temperature, in kelvin ( $\text{K}$ ), above which the reaction becomes spontaneous?

Show answer & explanation

Answer: 650

Answer

The minimum temperature above which the reaction becomes spontaneous is 650 K.

For a reaction with positive

\Delta H^\circ

and positive

\Delta S^\circ

, spontaneity depends on temperature. Spontaneity occurs when

\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ < 0

, which simplifies to

T > \frac{\Delta H^\circ}{\Delta S^\circ}

. Converting

\Delta H^\circ

to joules gives

117000\text{ J mol}^{-1}

, so

T = \frac{117000}{180.0} = 650\text{ K}

.

Step-by-Step Solution

1

Convert enthalpy change units from kilojoules per mole to joules per mole

\Delta H^\circ = 117.0\text{ kJ mol}^{-1} \times 1000\text{ J/kJ} = 117000\text{ J mol}^{-1}

Enthalpy and entropy must be in consistent energy units (joules) before performing thermodynamic calculations.

2

Apply the condition for reaction spontaneity threshold

\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ = 0

A reaction is spontaneous when

\Delta G^\circ < 0

. The threshold temperature occurs exactly when

\Delta G^\circ = 0

.

3

Calculate the threshold temperature T

T = \frac{\Delta H^\circ}{\Delta S^\circ} = \frac{117000\text{ J mol}^{-1}}{180.0\text{ J K}^{-1}\text{ mol}^{-1}} = 650\text{ K}

Solving the threshold condition for

T

yields the absolute temperature in kelvin above which

T\Delta S^\circ > \Delta H^\circ

, making

\Delta G^\circ

negative.

Key Concept

Relationship between Gibbs free energy, enthalpy, entropy, and reaction spontaneity threshold

Question 8223Question →

In a chemical reaction, the potential energy of the reactants is $45\text{ kJ mol}^{-1}$ , while the potential energy of the activated complex at the peak of the energy profile diagram is $125\text{ kJ mol}^{-1}$ . What is the activation energy for the forward reaction in $\text{kJ mol}^{-1}$ ?

Show answer & explanation

Answer: 80

Answer

The activation energy for the forward reaction is 80 kJ mol^{-1}.

The activation energy (

E_a

) for a forward chemical reaction is defined as the difference in energy between the activated complex (peak of the energy profile) and the reactants. Subtracting the reactant potential energy (

45\text{ kJ mol}^{-1}

) from the activated complex potential energy (

125\text{ kJ mol}^{-1}

) gives an activation energy of

80\text{ kJ mol}^{-1}

.

Step-by-Step Solution

1

Identify the potential energy levels of the reactants and the activated complex from the given data.

Energy of reactants = 45 kJ mol^{-1}; Energy of activated complex = 125 kJ mol^{-1}

Forward activation energy depends directly on the difference between these two energy states.

2

Subtract the potential energy of the reactants from the potential energy of the activated complex.

Ea = 125 kJ mol^{-1} - 45 kJ mol^{-1} = 80 kJ mol^{-1}

The activation energy represents the minimum energy barrier that reacting particles must overcome to reach the transition state.

Key Concept

Forward Activation Energy Calculation from Energy Profile Data

Question 8224Question →

In a municipal water treatment plant, raw river water undergoes several sequential processing stages to ensure it is safe for domestic consumption. Arrange the following key stages of municipal water purification in the correct chronological order from first to last.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence of municipal water treatment stages from start to finish is: Screening to remove large floating debris, followed by Coagulation using alum to clump fine particles, then Sand filtration to remove tiny remaining suspended solids, and finally Chlorination to kill disease-causing germs.

The correct sequence follows the standard municipal waterworks workflow: physical removal of large debris (Screening)

\rightarrow

chemical aggregation of clay suspensions (Coagulation)

\rightarrow

mechanical straining of micro-solids (Filtration)

\rightarrow

chemical destruction of disease-causing bacteria (Chlorination).

Step-by-Step Solution

1

Identify the primary intake step.

Screening is the initial physical process to filter out large objects like leaves and sticks.

Large debris must be removed first to protect pump machinery and piping.

2

Identify the chemical clumping step.

Coagulation involves adding chemical coagulants such as alum to bind fine suspended clay particles into larger flocs.

Fine particles will not settle or filter easily without prior coagulation.

3

Identify the physical clarification step.

Filtration passes water through sand and gravel layers to remove remaining micro-suspended matter.

Water must be visually clear before chemical disinfection so pathogens are fully exposed to chlorine.

4

Identify the final disinfection step.

Chlorination is carried out last to kill bacteria and ensure biological safety for drinking.

Adding chlorine last ensures residual disinfection in the municipal piping network.

Key Concept

Sequential Stages of Municipal Water Purification

Question 8225Question →

Match each industrial chemical process or biotechnology application listed on the left with its corresponding catalyst or biological agent on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Haber process matches Finely divided iron (

Fe

); Contact process matches Vanadium(V) oxide (

V_2O_5

); Fermentation of glucose matches Zymase enzyme; Hydrogenation of vegetable oils matches Finely divided nickel (

Ni

).

Each process relies on a specific catalyst or biological agent: the Haber process utilizes finely divided iron, the Contact process uses vanadium(V) oxide, glucose fermentation relies on the biological enzyme zymase, and vegetable oil hydrogenation uses finely divided nickel.

Step-by-Step Solution

1

Identify the catalyst used in ammonia production via the Haber process.

Finely divided iron accelerates the reversible reaction between

N_2

and

H_2

.

Iron lowers the activation energy required to break the strong triple bond in nitrogen.

2

Identify the catalyst in the Contact process stage converting

SO_2

to

SO_3

.

Vanadium(V) oxide (

V_2O_5

) is the modern industrial catalyst employed.

V_2O_5

provides high efficiency and is resistant to catalytic poisoning compared to platinum.

3

Identify the biocatalyst in ethanol fermentation.

Zymase, an enzyme complex produced by yeast cells, catalyzes glucose breakdown.

Fermentation is an anaerobic biotechnological process relying on enzymatic activity.

4

Identify the catalyst used in margarine synthesis.

Finely divided nickel is used during unsaturated oil hydrogenation.

Nickel adsorbs hydrogen gas and unsaturated hydrocarbon chains onto its surface to facilitate addition.

Key Concept

Industrial catalysts and biological enzymes in commercial chemical processes

Question 8226Question →

Match each type of organic isomerism on the left with its corresponding example pair of compounds on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Chain isomerism pairs with butane and

2

-methylpropane; Positional isomerism pairs with drop-in pair butan-

1

-ol and butan-

2

-ol; Functional group isomerism pairs with ethanoic acid and methyl methanoate; Geometric isomerism pairs with

cis

-but-

2

-ene and

trans

-but-

2

-ene.

Each type of isomerism is matched to its definitive structural characteristic: chain isomerism involves skeleton branching changes (butane and

2

-methylpropane), positional isomerism involves relocations of the same functional group (butan-

1

-ol and butan-

2

-ol), functional group isomerism involves different organic families sharing a formula (ethanoic acid and methyl methanoate), and geometric isomerism involves spatial orientations across a double bond (

cis

-but-

2

-ene and

trans

-but-

2

-ene).

Step-by-Step Solution

1

Analyze the carbon skeletons of butane and

2

-methylpropane.

Both have formula

C_4H_{10}

, but one is unbranched while the other is branched, defining chain isomerism.

Chain isomers possess identical molecular formulas but different carbon chain arrangements.

2

Examine the functional group positions in butan-

1

-ol and butan-

2

-ol.

The

-\text{OH}

group is on carbon-

1

in butan-

1

-ol and carbon-

2

in butan-

2

-ol, defining positional isomerism.

Positional isomers have the same functional group located on different carbon atoms along the same parent chain.

3

Compare the functional groups of ethanoic acid and methyl methanoate.

Ethanoic acid is a carboxylic acid (

CH_3COOH

) while methyl methanoate is an ester (

HCOOCH_3

). Both have the formula

C_2H_4O_2

, defining functional group isomerism.

Functional group isomers share a molecular formula but belong to different organic homologous families.

4

Evaluate the spatial arrangement in

cis

-but-

2

-ene and

trans

-but-

2

-ene.

The double bond restricts rotation, placing methyl groups on the same side (

cis

) or opposite sides (

trans

), defining geometric isomerism.

Geometric isomerism is a stereoisomerism type arising from restricted double-bond rotation.

Key Concept

Classification of Structural Isomerism and Stereoisomerism

Question 8227Question →

During the fractional distillation of crude oil (petroleum), different fractions are separated based on their boiling points. Arrange the following petroleum fractions in order of their distillation from first to last (lowest boiling point to highest boiling point).

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct order of petroleum fractions from first to last distilled is: Petroleum gas, Petrol / Gasoline, Kerosene / Paraffin, and Diesel oil / Gas oil.

Fractional distillation separates liquids based on differences in their boiling points. The fraction with the lowest boiling point (petroleum gas) is the most volatile and vaporizes first, exiting at the top of the fractionating column. As the temperature of the mixture increases, fractions with progressively higher boiling points (petrol, then kerosene, then diesel oil) vaporize and distil off sequentially.

Step-by-Step Solution

1

Identify the separation principle of fractional distillation.

Components with lower boiling points vaporize first and distil off before components with higher boiling points.

Lower boiling point liquids are more volatile and require less thermal energy to enter the vapor phase.

2

Compare the boiling points of the given petroleum fractions.

Petroleum gas (<

20^\circ\text{C}

) < Petrol (

40^\circ\text{C}-170^\circ\text{C}

) < Kerosene (

170^\circ\text{C}-250^\circ\text{C}

) < Diesel oil (

250^\circ\text{C}-350^\circ\text{C}

).

Hydrocarbon chain length increases from gas to diesel, increasing intermolecular forces and boiling points.

3

Arrange the items from lowest boiling point to highest boiling point.

Order: Petroleum gas → Petrol / Gasoline → Kerosene / Paraffin → Diesel oil / Gas oil.

This matches the sequence of collection during fractional distillation.

Key Concept

Fractional Distillation of Petroleum

Question 8228Question →

For a reversible gaseous reaction $A + B \rightleftharpoons C + D$ , the potential energy of the reactants is $120\text{ kJ mol}^{-1}$ and the potential energy of the products is $45\text{ kJ mol}^{-1}$ . If the activation energy for the reverse reaction is $140\text{ kJ mol}^{-1}$ , calculate the activation energy for the forward reaction in $\text{kJ mol}^{-1}$ .

Show answer & explanation

Answer: 65

Answer

The activation energy for the forward reaction is

65\text{ kJ mol}^{-1}

.

The forward activation energy (

E_{a,\text{forward}}

) is the energy barrier measured from the energy level of the reactants (

120\text{ kJ mol}^{-1}

) to the peak of the activated complex. Since the products lie at

45\text{ kJ mol}^{-1}

and require

140\text{ kJ mol}^{-1}

to reach the activated complex, the peak energy is

45 + 140 = 185\text{ kJ mol}^{-1}

. Subtracting the reactant energy gives

185 - 120 = 65\text{ kJ mol}^{-1}

. Alternatively, using

\Delta H = E_{a,\text{forward}} - E_{a,\text{reverse}}

, where

\Delta H = 45 - 120 = -75\text{ kJ mol}^{-1}

, gives

-75 = E_{a,\text{forward}} - 140 \implies E_{a,\text{forward}} = 65\text{ kJ mol}^{-1}

.

Step-by-Step Solution

1

Determine the potential energy of the activated complex (transition state).

Transition state energy

E_{\text{peak}} = H_{\text{products}} + E_{a,\text{reverse}} = 45 + 140 = 185\text{ kJ mol}^{-1}

.

The reverse activation energy is the energy required to overcome the barrier going from products up to the peak.

2

Calculate the forward activation energy.

E_{a,\text{forward}} = E_{\text{peak}} - H_{\text{reactants}} = 185 - 120 = 65\text{ kJ mol}^{-1}

.

The forward activation energy is the energy difference between the peak (activated complex) and the energy level of the reactants.

Key Concept

Activation energy and potential energy relationships in energy profile diagrams

Estimated Time:1m 30s

Question 8229Question →

A water sample containing $0.010\text{ mol}$ of dissolved calcium hydrogentrioxocarbonate(IV), $\text{Ca(HCO}_3)_2$ , responsible for temporary hardness, is boiled until complete precipitation of calcium trioxocarbonate(IV) occurs. The resulting solid residue is isolated and treated with an excess of dilute hydrochloric acid. What volume of carbon(IV) oxide gas, measured at room temperature and pressure (RTP, molar volume of gas $= 24.0\text{ dm}^3\text{ mol}^{-1}$ ), is liberated during the acid treatment?

0.240\text{ dm}^3

0.224\text{ dm}^3

0.480\text{ dm}^3

0.120\text{ dm}^3

Show answer & explanation

Answer: $0.240\text{ dm}^3$

Answer

The volume of carbon(IV) oxide gas liberated during the acid treatment is

0.240\text{ dm}^3

.

Boiling

0.010\text{ mol}

of

\text{Ca(HCO}_3)_2

yields

0.010\text{ mol}

of

\text{CaCO}_3

precipitate. Reacting this precipitate with excess dilute acid yields

0.010\text{ mol}

of

\text{CO}_2

gas. At RTP,

0.010\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 0.240\text{ dm}^3

.

Step-by-Step Solution

1

Determine moles of

\text{CaCO}_3

precipitate formed during boiling.

The balanced thermal decomposition equation is

\text{Ca(HCO}_3)_2(aq) \rightarrow \text{CaCO}_3(s) + \text{H}_2\text{O}(l) + \text{CO}_2(g)

. Thus,

0.010\text{ mol}

of

\text{Ca(HCO}_3)_2

produces

0.010\text{ mol}

of

\text{CaCO}_3

.

Temporary hardness is removed by boiling, which decomposes hydrogentrioxocarbonate(IV) into insoluble trioxocarbonate(IV).

2

Determine moles of

\text{CO}_2

gas evolved from the reaction of

\text{CaCO}_3

with dilute

\text{HCl}

.

The balanced chemical equation is

\text{CaCO}_3(s) + 2\text{HCl}(aq) \rightarrow \text{CaCl}_2(aq) + \text{H}_2\text{O}(l) + \text{CO}_2(g)

.

0.010\text{ mol}

of

\text{CaCO}_3

reacts to release

0.010\text{ mol}

of

\text{CO}_2

.

Trioxocarbonate(IV) salts react with acids in a 1:1 mole ratio with respect to carbon(IV) oxide gas produced.

3

Calculate the volume of

\text{CO}_2

gas evolved at RTP.

\text{Volume} = 0.010\text{ mol} \times 24.0\text{ dm}^3\text{ mol}^{-1} = 0.240\text{ dm}^3$.

At room temperature and pressure, one mole of any gas occupies

24.0\text{ dm}^3

.

Key Concept

Thermal decomposition of temporary water hardness salts and acid reactions of trioxocarbonate(IV) salts

Question 8230Question →

Match each redox reaction mixture (left) with its corresponding characteristic laboratory observation (right).

Click a left item, then click its matching right item

Show answer & explanation

Answer

Acidified

KMnO_4

with

SO_2

changes from purple to colorless (

Mn^{2+}

); Acidified

K_2Cr_2O_7

with

FeSO_4

changes from orange to green (

Cr^{3+}

); Chlorine gas with

KI

forms brown iodine (

I_2

); Hydrogen sulfide gas with

FeCl_3

reduces

Fe^{3+}

to

Fe^{2+}

(pale green) with a yellow sulfur precipitate.

Each test pair matches an oxidizing or reducing reagent with its definitive qualitative laboratory test observation: permanganate decolorizes upon reduction, dichromate turns green upon reduction, iodide oxidizes to brown iodine, and iron(III) reduces to pale green iron(II) alongside yellow sulfur precipitation.

Step-by-Step Solution

1

Determine the redox behavior of

SO_2

with acidified

KMnO_4

SO_2

is oxidized while reducing purple

MnO_4^-

to colorless

Mn^{2+}

.

Potassium permanganate test for reducing agents involves reduction of manganese from oxidation state +7 to +2.

2

Determine the redox behavior of

FeSO_4

with acidified

K_2Cr_2O_7

Fe^{2+}

ions oxidize to

Fe^{3+}

while

Cr_2O_7^{2-}

(orange) is reduced to

Cr^{3+}

(green).

Dichromate(VI) is a standard oxidizing agent whose reduced form contains green chromium(III) ions.

3

Determine the reaction of

Cl_2

with

KI

Cl_2

oxidizes colorless

I^-

ions to elemental iodine (

I_2

), turning the solution brown.

Chlorine is a stronger oxidizing agent than iodine and displaces iodide from solution.

4

Determine the reaction of

H_2S

with

FeCl_3

H_2S

reduces yellow-brown

Fe^{3+}

to pale green

Fe^{2+}

while forming a insoluble yellow deposit of sulfur.

Hydrogen sulfide acts as a reducing agent and deposits elemental sulfur upon oxidation.

Key Concept

Laboratory tests and characteristic color changes for oxidizing and reducing agents.

Question 8231Question →

Arrange the following sequential electrochemical and physical steps occurring during the reduction of alumina in the Hall-Héroult cell to extract molten aluminium metal, from initial electrolyte preparation to final anode gas emission.

Drag items to arrange them in the correct order

Show answer & explanation

Answer

The correct sequence of steps in the Hall-Héroult cell is: (1) Dissolving alumina in molten cryolite, (2) Dissociation of alumina into

Al^{3+}

and

O^{2-}

ions, (3) Migration of

Al^{3+}

ions to the cathode, (4) Reduction of

Al^{3+}

to form molten aluminium, and (5) Oxidation of

O^{2-}

at the graphite anodes yielding carbon dioxide gas.

The Hall-Héroult process operates sequentially by first dissolving alumina (

Al_2O_3

) in molten cryolite (

Na_3AlF_6

) at around

950^\circ\text{C}

to create a conducting medium. Upon melting, alumina dissociates into mobile

Al^{3+}

and

O^{2-}

ions. Applying an electric current causes

Al^{3+}

cations to migrate to the carbon cathode at the bottom, where they are reduced to liquid aluminium metal (

Al^{3+} + 3e^- \rightarrow Al_{(l)}

). Concurrently,

O^{2-}

anions migrate to the top graphite anodes and undergo oxidation to oxygen gas (

2O^{2-} \rightarrow O_{2(g)} + 4e^-

), which reacts with the hot carbon anodes to form carbon dioxide gas (

C + O_2 \rightarrow CO_2

).

Step-by-Step Solution

1

Identify the electrolyte preparation step in the Hall-Héroult cell.

Alumina is dissolved in molten cryolite at about

950^\circ\text{C}

.

Cryolite acts as a solvent and flux to lower the high melting point of pure alumina and enhance conductivity.

2

Determine the ionization behavior of the dissolved alumina.

Alumina dissociates into mobile

Al^{3+}

cations and

O^{2-}

anions.

Liquid state ionic dissociation is necessary for current transport through the electrolyte.

3

Trace the movement of cations under the applied electric field.

Al^{3+}

cations migrate to the negatively charged carbon cathode lining at the cell floor.

Electrostatic attraction draws positive ions toward the negative electrode.

4

Determine the chemical reaction occurring at the cathode.

Al^{3+}

ions gain electrons to form molten aluminium metal (

Al^{3+} + 3e^- \rightarrow Al_{(l)}

).

Cation gain of electrons at the cathode represents the reduction process that isolates elemental aluminium.

5

Determine the chemical reaction occurring at the anode and the fate of the anode material.

O^{2-}

ions lose electrons to produce oxygen gas, which reacts with graphite anodes to produce

CO_2

gas.

Anode oxidation releases oxygen gas at high temperature, causing carbon anodes to burn away continuously.

Key Concept

Electrolytic reduction of alumina in the Hall-Héroult process

Estimated Time:1m 30s

Question 8232Question →

A chemical reaction has a standard enthalpy change ( $\Delta H^\circ$ ) of $+75.0\text{ kJ mol}^{-1}$ and a standard entropy change ( $\Delta S^\circ$ ) of $+250\text{ J K}^{-1}\text{mol}^{-1}$ . Calculate the minimum temperature in Kelvin ( $\text{K}$ ) at which this reaction becomes spontaneous.

Show answer & explanation

Answer: 300

Answer

The minimum temperature at which the reaction becomes spontaneous is 300 K.

A chemical process becomes spontaneous when the Gibbs free energy change (

\Delta G^\circ

) is negative (

\Delta G^\circ < 0

). Setting

\Delta G^\circ = 0

defines the threshold temperature for spontaneity. From

\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ

, re-arranging gives

T = \frac{\Delta H^\circ}{\Delta S^\circ}

. Expressing

\Delta H^\circ

as

75000\text{ J mol}^{-1}

and

\Delta S^\circ

as

250\text{ J K}^{-1}\text{mol}^{-1}

yields

T = \frac{75000}{250} = 300\text{ K}

.

Step-by-Step Solution

1

Convert standard enthalpy change

\Delta H^\circ

from kilojoules per mole to Joules per mole.

\Delta H^\circ = 75.0 \times 10^3\text{ J mol}^{-1} = 75000\text{ J mol}^{-1}

.

Units of enthalpy and entropy must be compatible (Joules) before performing calculation.

2

Apply the Gibbs free energy relationship

\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ

at the spontaneity boundary condition

\Delta G^\circ = 0

.

0 = \Delta H^\circ - T\Delta S^\circ \implies T = \frac{\Delta H^\circ}{\Delta S^\circ}

.

A reaction is spontaneous when

\Delta G^\circ < 0

, making

\Delta G^\circ = 0

the exact minimum temperature threshold.

3

Divide the enthalpy change in Joules per mole by the entropy change in Joules per Kelvin-mole.

T = \frac{75000\text{ J mol}^{-1}}{250\text{ J K}^{-1}\text{mol}^{-1}} = 300\text{ K}

.

Dividing Joules by Joules per Kelvin yields the temperature in Kelvin.

Key Concept

Gibbs free energy and temperature dependence of reaction spontaneity

Estimated Time:1m 15s

Question 8233Question →

Match each redox reaction scenario involving an oxidizing or reducing agent on the left with its corresponding characteristic laboratory observation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct pairings match each redox reagent with its specific electron-transfer observation:

SO_2

with acidified

KMnO_4

produces a purple to colorless change;

Cl_2

with aqueous

KI

turns colorless solution brown; conc.

HNO_3

with

FeSO_4

converts pale green solution to brown; and

H_2S

with

FeCl_3

converts yellow/brown solution to pale green with yellow sulfur deposit.

Each pair correctly links the specific chemical species undergoing oxidation or reduction to its empirical qualitative test result. Sulfur(IV) oxide decolorizes acidified potassium tetraoxomanganate(VII); chlorine oxidizes iodide ions to brown iodine; concentrated trioxonitrate(V) acid converts green iron(II) to brown iron(III); and hydrogen sulfide reduces brown iron(III) to green iron(II) with yellow sulfur precipitation.

Step-by-Step Solution

1

Analyze the redox roles of the reagents in each left item.

SO_2

and

H_2S

act as reducing agents;

Cl_2

and conc.

HNO_3

act as oxidizing agents.

Identifying whether a species donates or accepts electrons determines the expected chemical transformation of the target solution.

2

Determine the oxidation state change and color change for

SO_2

+ acidified

KMnO_4

.

MnO_4^-

(oxidation state +7, purple) is reduced to

Mn^{2+}

(oxidation state +2, colorless).

Manganate(VII) reduction is the standard test for reducing agents like

SO_2

.

3

Determine the oxidation state change and color change for

Cl_2

+ aqueous

KI

.

I^-

(oxidation state -1, colorless) is oxidized to

I_2

(oxidation state 0, brown).

Halogen displacement shows chlorine's higher electronegativity and oxidizing strength compared to iodine.

4

Determine the oxidation state change for conc.

HNO_3

+

FeSO_4

and

H_2S

+

FeCl_3

.

Conc.

HNO_3

oxidizes pale green

Fe^{2+}

to brown

Fe^{3+}

.

H_2S

reduces yellow/brown

Fe^{3+}

to pale green

Fe^{2+}

with precipitate of sulfur.

Iron transitions between +2 (pale green) and +3 (yellow/brown) depending on whether an oxidant or reductant is introduced.

Key Concept

Laboratory identification of oxidizing and reducing agents via characteristic color changes and oxidation state transitions

Estimated Time:2m 0s

Question 8234Question →

Consider the unbalanced redox reaction taking place in an acidic medium:

\text{a MnO}_4^-(\text{aq}) + \text{b SO}_3^{2-}(\text{aq}) + \text{c H}^+(\text{aq}) \rightarrow \text{d Mn}^{2+}(\text{aq}) + \text{e SO}_4^{2-}(\text{aq}) + \text{f H}_2\text{O}(\text{l})

When this chemical equation is balanced using the smallest set of whole-number coefficients, what is the value of the stoichiometric coefficient

c

for

\text{H}^+(\text{aq})

?

Show answer & explanation

Answer: 6

Answer

The value of the stoichiometric coefficient c for H+(aq) is 6.

Balancing the reduction half-reaction (

2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}

) and oxidation half-reaction (

5\text{SO}_3^{2-} + 5\text{H}_2\text{O} \rightarrow 5\text{SO}_4^{2-} + 10\text{H}^+ + 10e^-

) gives a combined total of

16\text{H}^+

on the reactant side and

10\text{H}^+

on the product side. Subtracting

10\text{H}^+

from both sides leaves a net coefficient of 6 for

\text{H}^+(\text{aq})

on the reactant side.

Step-by-Step Solution

1

Write the balanced reduction half-reaction for permanganate ion in acidic medium.

\text{MnO}_4^-(\text{aq}) + 8\text{H}^+(\text{aq}) + 5e^- \rightarrow \text{Mn}^{2+}(\text{aq}) + 4\text{H}_2\text{O}(\text{l})

Manganese goes from oxidation state +7 to +2, requiring 5 electrons, 8 H+ ions to balance oxygen atoms, forming 4 H2O molecules.

2

Write the balanced oxidation half-reaction for sulfite ion to sulfate ion.

\text{SO}_3^{2-}(\text{aq}) + \text{H}_2\text{O}(\text{l}) \rightarrow \text{SO}_4^{2-}(\text{aq}) + 2\text{H}^+(\text{aq}) + 2e^-

Sulfur goes from oxidation state +4 to +6, releasing 2 electrons and 2 H+ ions while consuming 1 H2O molecule.

3

Equalize the number of transferred electrons by multiplying the reduction half-reaction by 2 and the oxidation half-reaction by 5.

2\text{MnO}_4^- + 16\text{H}^+ + 10e^- \rightarrow 2\text{Mn}^{2+} + 8\text{H}_2\text{O}

5\text{SO}_3^{2-} + 5\text{H}_2\text{O} \rightarrow 5\text{SO}_4^{2-} + 10\text{H}^+ + 10e^-

The least common multiple of 5 and 2 transferred electrons is 10.

4

Combine the half-reactions and subtract common species (

10e^-

,

10\text{H}^+

, and

5\text{H}_2\text{O}

) from both sides.

2\text{MnO}_4^-(\text{aq}) + 5\text{SO}_3^{2-}(\text{aq}) + 6\text{H}^+(\text{aq}) \rightarrow 2\text{Mn}^{2+}(\text{aq}) + 5\text{SO}_4^{2-}(\text{aq}) + 3\text{H}_2\text{O}(\text{l})

Subtracting

10\text{H}^+

from

16\text{H}^+

leaves

6\text{H}^+

on the reactant side, giving

c = 6

.

Key Concept

Balancing Redox Equations using the Ion-Electron Method in Acidic Medium

Question 8235Question →

Match each observed physical property or phenomenon of ionic (electrovalent) compounds on the left with its correct underlying thermodynamic or structural explanation on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

1 matches with the explanation of charge product dependence on lattice energy; 2 matches with the explanation of Fajans' rules of polarization; 3 matches with the explanation of hydration enthalpy exceeding lattice enthalpy; 4 matches with the explanation of ion mobility in molten versus solid states.

Each physical property directly corresponds to its underlying quantum mechanical or thermodynamic principle: lattice energy scales with charge product (

MgO

vs

NaCl

), polarization of anion electron clouds by small high-charge cations creates covalent character (

AlI_3

), exothermic dissolution occurs when hydration energy exceeds lattice energy (

NaOH

), and electrical conduction requires mobile ions that are locked in solids but liberated upon melting (

CaF_2

).

Step-by-Step Solution

1

Analyze the high melting point of

MgO

versus

NaCl

Lattice energy is governed by Coulomb's law:

E \propto \frac{|q_1 q_2|}{r}

.

MgO

consists of

Mg^{2+}

and

O^{2-}

(product = 4), while

NaCl

consists of

Na^+

and

Cl^-

(product = 1). Higher charge product leads to stronger lattice attraction and a higher melting point.

Identify the primary thermodynamic factor controlling lattice strength in ionic crystals.

2

Analyze the anomalous covalent behavior of

AlI_3

Apply Fajans' rules: Covalency increases with high cation charge density and large anion size.

Al^{3+}

has high charge density and

I^-

is large and easily polarized, leading to electron cloud sharing (covalent character).

Explain deviations from purely electrovalent behavior using polarization principles.

3

Analyze the thermochemistry of dissolution of

NaOH

Dissolution enthalpy

\Delta H_{soln} = \Delta H_{lat} + \Delta H_{hyd}

. If hydration enthalpy released is greater in magnitude than the lattice enthalpy required to separate ions, the net process is exothermic.

Relate lattice energy and hydration energy to dissolution energetics.

4

Analyze electrical conductivity in solid versus molten

CaF_2

Solid ionic compounds contain ions held rigidly in a lattice structure. When melted, thermal energy breaks the lattice, producing free-moving ions capable of carrying electrical current.

Distinguish between mobile charge carriers (molten state) and immobile lattice positions (solid state).

Key Concept

Thermodynamic and structural factors governing ionic lattice stability, polarization (Fajans' rules), solution energetics, and state-dependent conductivity.

Question 8236Question →

Match each organic reagent setup or chemical process in Column A with its corresponding chemical reaction product or characteristic diagnostic observation in Column B.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Bubbling

but-1-yne

into ammoniacal silver nitrate forms a white precipitate of silver but-1-ynide due to acidic acetylenic hydrogen replacement; passing

propene

into cold dilute alkaline

KMnO_4

decolourizes purple

MnO_4^-

forming a diol and brown

MnO_2

; heating ethanol with excess concentrated

H_2SO_4

at

170^\circ\text{C}

produces ethene via intramolecular dehydration; adding water to

CaC_2

yields ethyne gas via hydrolysis.

The matches correctly pair each chemical reaction with its distinctive mechanism or diagnostic test outcome: terminal alkyne acidity forming silver salts, alkene hydroxylation via Baeyer's reagent, alcohol elimination yielding ethene at elevated temperature, and carbide hydrolysis generating ethyne.

Step-by-Step Solution

1

Analyze the reaction of terminal alkynes with ammoniacal silver nitrate

but-1-yne

is a 1-alkyne containing a hydrogen atom bonded to an

sp

-hybridized carbon (

\text{CH}_3\text{CH}_2\text{C}\equiv\text{CH}

). This hydrogen is slightly acidic and is readily displaced by

Ag^+

to form a insoluble white precipitate of silver but-1-ynide.

To distinguish terminal alkynes from internal alkynes and alkenes.

2

Analyze the reaction of alkenes with Baeyer's reagent

propene

(

\text{CH}_3\text{CH}=\text{CH}_2

) undergoes syn-hydroxylation with cold, dilute alkaline

KMnO_4

to form

propane-1,2-diol

. The purple trioxomanganate(VII) is reduced to brown manganese(IV) oxide (

MnO_2

).

To identify mild oxidation of double bonds in unsaturation testing.

3

Analyze the acid-catalyzed dehydration condition of ethanol

Heating ethanol with concentrated

H_2SO_4

at a high temperature (

170^\circ\text{C}

) favours intramolecular elimination of water yielding ethene (

C_2H_4

).

Lower temperatures (

140^\circ\text{C}

) yield ethoxyethane instead, so temperature controls product selectivity.

4

Analyze the laboratory preparation of ethyne

Calcium dicarbide (

CaC_2

) reacts directly with water to yield ethyne gas (

C_2H_2

) and calcium hydroxide (

Ca(OH)_2

).

This is the primary laboratory synthesis method for ethyne.

Key Concept

Chemical reactions, laboratory preparation methods, and diagnostic unsaturation tests for alkenes and alkynes.

Question 8237Question →

Match each chemical process or application involving iron and its compounds on the left with its corresponding chemical reaction characteristic or product on the right.

Click a left item, then click its matching right item

Show answer & explanation

Answer

The correct matches pair thermal decomposition of siderite with production of

\text{FeO}

and

\text{CO}_2

; reaction of iron with hot concentrated

\text{H}_2\text{SO}_4

with formation of iron(III) sulfate, water, and

\text{SO}_2

; galvanization with zinc acting as a sacrificial anode; and fluxing action of

\text{CaO}

with molten slag (

\text{CaSiO}_3

) formation.

Each process matches its corresponding chemical principle: siderite decomposes into

\text{FeO}

and

\text{CO}_2

; hot concentrated

\text{H}_2\text{SO}_4

oxidizes iron to iron(III) sulfate releasing

\text{SO}_2

; galvanization uses zinc as a sacrificial anode; and calcium oxide combines with silica to form slag.

Step-by-Step Solution

1

Analyze the thermal decomposition of siderite

Siderite (

\text{FeCO}_3

) decomposes upon heating to yield iron(II) oxide (

\text{FeO}

) and carbon(IV) oxide (

\text{CO}_2

).

Transition metal carbonates decompose thermally into the corresponding metal oxide and carbon dioxide.

2

Examine the reaction of iron with hot concentrated oxidizing acid

Hot concentrated

\text{H}_2\text{SO}_4

oxidizes iron metal to the iron(III) oxidation state, producing

\text{Fe}_2(\text{SO}_4)_3

,

\text{SO}_2

gas, and

\text{H}_2\text{O}

.

Hot concentrated acid acts as a strong oxidizing agent rather than liberating hydrogen gas as dilute acid would.

3

Evaluate galvanization for rusting prevention

Galvanization coats iron with zinc metal, which undergoes preferential cathodic protection as a sacrificial anode.

Zinc is higher in the electrochemical series than iron, so it corrodes preferentially even if scratched.

4

Identify the fluxing action of basic oxides in the blast furnace

Calcium oxide (

\text{CaO}

) reacts with acidic sandy impurities (

\text{SiO}_2

) to form molten slag (

\text{CaSiO}_3

).

Slag floats on top of molten iron, preventing re-oxidation of the extracted metal while removing silicon impurities.

Key Concept

Extraction, Reaction Properties, and Corrosion Prevention of Iron

Question 8238Question →

Match each redox reagent mixture or reaction system with its corresponding characteristic diagnostic observation.

Click a left item, then click its matching right item

Show answer & explanation

Answer

Acidified potassium heptaoxodichromate(VI) with sulfur(IV) oxide matches the orange to green color change. Freshly prepared iron(II) tetraoxosulfate(VI) with concentrated trioxonitrate(V) acid matches the pale green to reddish-brown change. Aqueous potassium iodide with chlorine gas and starch matches the blue-black complex formation. Acidified potassium tetraoxomanganate(VII) with hydrogen peroxide matches decolorization from purple to colorless with oxygen gas effervescence.

Each test reagent matches its specific chemical observation based on electron transfer and oxidation state changes:

K_2Cr_2O_7

turns green when reduced by

SO_2

;

Fe^{2+}

turns brown when oxidized by conc.

HNO_3

;

KI

yields free

I_2

which gives a blue-black color with starch when oxidized by

Cl_2

; and acidified

KMnO_4

is decolorized with

O_2

evolution when reduced by

H_2O_2

.

Step-by-Step Solution

1

Analyze the reduction of acidified

K_2Cr_2O_7

by

SO_2

Chromium decreases in oxidation number from

+6

in

Cr_2O_7^{2-}

to

+3

in

Cr^{3+}

, causing a distinct color shift from orange to green.

SO_2

acts as a reductant and

K_2Cr_2O_7

as an oxidant.

2

Analyze the oxidation of

FeSO_4

by concentrated

HNO_3

Iron increases in oxidation state from

+2

(

Fe^{2+}

, pale green) to

+3

(

Fe^{3+}

, brown/yellowish-brown).

Concentrated

HNO_3

is a strong oxidizing agent.

3

Analyze halogen displacement of

KI

by

Cl_2

Chlorine oxidizes

I^-

to

I_2

. Liberated

I_2

forms a blue-black starch-iodine inclusion complex.

Chlorine has a higher standard reduction potential than iodine.

4

Analyze the redox reaction between acidified

KMnO_4

and

H_2O_2

Manganese is reduced from

+7

(

MnO_4^-

, purple) to

+2

(

Mn^{2+}

, colorless), while peroxide oxygen is oxidized from

-1

to

0

(

O_2

gas bubbles).

In the presence of a stronger oxidant (

KMnO_4

),

H_2O_2

behaves as a reducing agent.

Key Concept

Oxidizing and Reducing Agents and Diagnostic Chemical Tests

Estimated Time:1m 30s

Question 8239Question →

A gas sealed inside a rigid container of fixed volume is connected to a pressure gauge that displays a gauge pressure of $1.50\text{ atm}$ at an initial temperature of $27^\circ\text{C}$ . If the ambient atmospheric pressure is $1.00\text{ atm}$ , calculate the temperature in degrees Celsius ( $^\circ\text{C}$ ) to which the gas must be heated for its total absolute pressure to double.

Show answer & explanation

Answer: 327

Answer

The final temperature required is

327^\circ\text{C}

.

To find the required temperature, first calculate the initial absolute pressure (

1.50\text{ atm} + 1.00\text{ atm} = 2.50\text{ atm}

) and convert the initial temperature to Kelvin (

27^\circ\text{C} + 273 = 300\text{ K}

). Doubling the absolute pressure to

5.00\text{ atm}

requires doubling the absolute temperature according to the Pressure Law (

T_2 = 600\text{ K}

). Converting

600\text{ K}

back to Celsius yields

600 - 273 = 327^\circ\text{C}

.

Step-by-Step Solution

1

Determine the initial absolute pressure of the gas.

P_1 = 2.50\text{ atm}

Absolute pressure is the sum of gauge pressure and ambient atmospheric pressure (

P_{\text{abs}} = P_{\text{gauge}} + P_{\text{atm}}

).

2

Convert the initial temperature from Celsius to Kelvin.

T_1 = 300\text{ K}

Gas law calculations strictly require thermodynamic temperature expressed in Kelvin (

T_{\text{K}} = t_{^\circ\text{C}} + 273

).

3

Calculate the required final absolute pressure.

P_2 = 5.00\text{ atm}

The total absolute pressure doubles, so

P_2 = 2 \times 2.50\text{ atm} = 5.00\text{ atm}

.

4

Apply Gay-Lussac's Pressure Law to calculate the final absolute temperature.

T_2 = 600\text{ K}

At constant volume, pressure is directly proportional to absolute temperature (

\frac{P_1}{T_1} = \frac{P_2}{T_2}

), yielding

T_2 = T_1 \times \frac{P_2}{P_1} = 300 \times 2 = 600\text{ K}

.

5

Convert the final absolute temperature back to degrees Celsius.

t_2 = 327^\circ\text{C}

Subtract 273 from the Kelvin temperature (

t_{^\circ\text{C}} = 600 - 273 = 327^\circ\text{C}

).

Key Concept

Pressure Law (Gay-Lussac's Law) states that the pressure of a fixed mass of gas at constant volume is directly proportional to its absolute temperature (

P \propto T

). Calculations must strictly use absolute pressure and Kelvin temperatures.

Question 8240Question →

During the fractional distillation of crude oil in a refining column, components separate based on differences in their boiling point ranges. Which of the following petroleum fractions is collected at the very top of the fractionating column?

Refinery gas

Bitumen

Kerosene

Diesel oil

Show answer & explanation

Answer: Refinery gas

Answer

Refinery gas is collected at the very top of the fractionating column.

Refinery gas comprises short-chain alkanes containing 1 to 4 carbon atoms. Because these small molecules experience weak intermolecular forces, they possess the lowest boiling points among all petroleum fractions and rise to the coolest region at the very top of the fractionating tower.

Step-by-Step Solution

1

Analyze how fractional distillation separates petroleum fractions.

Fractions separate according to their boiling point ranges, which depend on hydrocarbon chain length and molar mass.

Smaller alkane molecules have weaker van der Waals forces and lower boiling points.

2

Determine the temperature gradient in a fractionating column.

The bottom of the column is the hottest, while the top is the coolest.

Vapors rise through the column and condense when the temperature drops below their respective boiling points.

3

Identify the fraction with the lowest boiling point.

Refinery gas (

C_1

–

C_4

alkanes like methane, ethane, propane, and butane) has the lowest boiling point range (

< 20^\circ\text{C}

) and remains gaseous at the top.

Components with the lowest boiling points travel to the coolest section at the top before exiting.

Key Concept

Fractional Distillation of Crude Oil and Boiling Point Trends

Estimated Time:45s

Answer

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