Energetics, Rates of Reaction and Chemical Equilibrium Alıştırma Soruları — JAMB UTME

Soru 1Soru →

Consider the reversible industrial synthesis of ammonia gas: $N_{2(g)} + 3H_{2(g)} \rightleftharpoons 2NH_{3(g)} \quad \Delta H = -92\text{ kJ mol}^{-1}$ . A chemical engineer introduces a finely divided iron catalyst into the reaction vessel while maintaining constant temperature and pressure. Which of the following best describes the effect of adding the catalyst on the system?

It selectively lowers the activation energy of the forward reaction, shifting the equilibrium position to the right and increasing the yield of

NH_3

.

It increases the average kinetic energy of the reactant molecules, causing an increase in the equilibrium constant (

K_c

) and a higher yield of

NH_3

.

It lowers the activation energy for both the forward and reverse reactions equally, increasing their rates and reducing the time taken to reach equilibrium without altering the final yield of

NH_3

.

It shifts the position of equilibrium to the left because the forward reaction is exothermic, thereby favoring the decomposition of

NH_3

.

Cevabı ve açıklamayı göster

Cevap: It lowers the activation energy for both the forward and reverse reactions equally, increasing their rates and reducing the time taken to reach equilibrium without altering the final yield of $NH_3$ .

Cevap

Adding a catalyst lowers the activation energy for both forward and reverse reactions by equal amounts, increasing their rates and reducing the time needed to reach dynamic equilibrium without altering the equilibrium yield of ammonia.

A catalyst functions by offering an alternative pathway with a lower activation energy barrier. In a reversible system, this reduction in activation energy applies equally to both the forward and reverse directions. Consequently, the rates of both reactions increase by the same proportion, enabling the system to reach dynamic equilibrium faster without changing the equilibrium composition or the yield of products.

Adım Adım Çözüm

1

Analyze the function of a catalyst in chemical kinetics.

A catalyst provides an alternative reaction pathway with a lower activation energy (

E_a

).

Lowering

E_a

allows a greater fraction of reactant molecules to possess sufficient energy to undergo effective collisions per unit time.

2

Evaluate the symmetry of activation energy reduction in reversible reactions.

The catalyst lowers the activation energy barrier for both the forward reaction and the reverse reaction by the exact same amount (

Δ E_a

).

Since the initial reactants and final products remain in the same energy states, the difference between the forward and reverse activation energy barriers (

Δ H

) is unchanged.

3

Determine the impact of the catalyst on chemical equilibrium and product yield.

The rates of both forward and reverse reactions increase by the same factor, so the equilibrium position and equilibrium constant (

K_c

) remain unchanged; only the time to reach equilibrium decreases.

Equal acceleration of both directions leaves the relative equilibrium concentrations of products and reactants unaffected.

Anahtar Kavram

Effect of Catalysts on Reaction Kinetics and Chemical Equilibrium

Tahmini Süre:1m 30s

Soru 2Soru →

For a reversible gas-phase reaction $X_{(g)} + Y_{(g)} \rightleftharpoons Z_{(g)}$ , the total enthalpy of the reactants is $+120\text{ kJ mol}^{-1}$ . The reaction is exothermic with a standard enthalpy change ( $\Delta H$ ) of $-45\text{ kJ mol}^{-1}$ . In the absence of a catalyst, the activation energy for the reverse reaction ( $E_{a,\text{rev}}$ ) is $+185\text{ kJ mol}^{-1}$ . If a catalyst is introduced that lowers the activation energy barrier by $30\text{ kJ mol}^{-1}$ , what is the activation energy for the catalyzed forward reaction?

+110\text{ kJ mol}^{-1}

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

+155\text{ kJ mol}^{-1}

+200\text{ kJ mol}^{-1}

Cevabı ve açıklamayı göster

Cevap: $+110\text{ kJ mol}^{-1}$

Cevap

+110\text{ kJ mol}^{-1}

The correct answer is

+110\text{ kJ mol}^{-1}

. For an exothermic reaction with

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

and reverse activation energy

E_{a,\text{rev}} = +185\text{ kJ mol}^{-1}

, the uncatalyzed forward activation energy is

E_{a,\text{fwd}} = E_{a,\text{rev}} + \Delta H = 185 - 45 = +140\text{ kJ mol}^{-1}

. Adding a catalyst lowers this activation energy barrier by

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

, yielding

140 - 30 = +110\text{ kJ mol}^{-1}

.

Adım Adım Çözüm

1

Calculate the uncatalyzed forward activation energy (

E_{a,\text{fwd, uncatalyzed}}

)

E_{a,\text{fwd, uncatalyzed}} = E_{a,\text{rev}} + \Delta H = +185\text{ kJ mol}^{-1} + (-45\text{ kJ mol}^{-1}) = +140\text{ kJ mol}^{-1}

For any chemical system, the relationship between forward activation energy, reverse activation energy, and enthalpy change is

\Delta H = E_{a,\text{fwd}} - E_{a,\text{rev}}

.

2

Apply the catalyst reduction to the forward activation energy

E_{a,\text{fwd, catalyzed}} = E_{a,\text{fwd, uncatalyzed}} - 30\text{ kJ mol}^{-1} = 140 - 30 = +110\text{ kJ mol}^{-1}

A catalyst lowers both forward and reverse activation energies by equal amounts, reducing the energy barrier height.

Anahtar Kavram

Relationship between forward activation energy, reverse activation energy, enthalpy change, and catalyst effect in energy profile diagrams

Alternatif Yöntem

Calculate the energy levels directly: Reactants =

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

. Products =

120 + (-45) = +75\text{ kJ mol}^{-1}

. Uncatalyzed Transition State = Products +

E_{a,\text{rev}} = 75 + 185 = +260\text{ kJ mol}^{-1}

. Catalyzed Transition State =

260 - 30 = +230\text{ kJ mol}^{-1}

. Catalyzed

E_{a,\text{fwd}} = \text{Catalyzed Transition State} - \text{Reactants} = 230 - 120 = +110\text{ kJ mol}^{-1}

.

Tahmini Süre:2m 0s

Soru 3Soru →

According to collision theory, which of the following conditions must be satisfied for a collision between reacting particles to result in a chemical reaction?

The colliding particles must possess kinetic energy equal to or greater than the activation energy and have proper spatial orientation.

A catalyst must be present to shift the position of equilibrium towards product formation.

The overall enthalpy change (

\Delta H

) of the reaction must be negative regardless of activation energy.

The concentration of reactants must be decreased to reduce acid strength and pH.

Cevabı ve açıklamayı göster

Cevap: The colliding particles must possess kinetic energy equal to or greater than the activation energy and have proper spatial orientation.

Cevap

The colliding particles must possess kinetic energy equal to or greater than the activation energy and have proper spatial orientation.

Collision theory establishes that a reaction occurs only when colliding particles possess energy at least equal to the activation energy (

E_a

) and are oriented properly during impact.

Adım Adım Çözüm

1

Identify the basic postulates of Collision Theory.

Not all collisions lead to a chemical reaction; only 'effective' collisions produce products.

Collision theory requires reacting species to meet specific energy and alignment criteria.

2

Analyze the energy requirement.

Colliding molecules must have kinetic energy

E \ge E_a

(activation energy).

Energy is required to overcome repulsive forces and break existing chemical bonds.

3

Analyze the orientation requirement.

Particles must collide with proper spatial alignment.

Correct alignment ensures reactive sites interact to form the activated complex.

Anahtar Kavram

Conditions for Effective Collisions

Tahmini Süre:45s

Soru 4Soru →

An endothermic reaction has a forward activation energy of $75\text{ kJ mol}^{-1}$ and an enthalpy change ( $\Delta H$ ) of $+30\text{ kJ mol}^{-1}$ . What is the activation energy of the reverse reaction?

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

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

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

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

Cevabı ve açıklamayı göster

Cevap: $45\text{ kJ mol}^{-1}$

Cevap

The activation energy of the reverse reaction is

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

.

For an endothermic reaction, the products have higher potential energy than the reactants by an amount equal to

\Delta H

. The energy barrier to go from products to the transition state (reverse activation energy) is therefore smaller than the barrier from reactants to the transition state (forward activation energy). Using

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

, we get

30 = 75 - E_{a(\text{reverse})}

, giving

E_{a(\text{reverse})} = 45\text{ kJ mol}^{-1}

.

Adım Adım Çözüm

1

Identify the given thermodynamic parameters for the endothermic reaction.

Forward activation energy

E_{a(\text{forward})} = 75\text{ kJ mol}^{-1}

and enthalpy change

\Delta H = +30\text{ kJ mol}^{-1}

.

Establishing the known energy levels from the problem stem.

2

Apply the relationship between forward activation energy, reverse activation energy, and enthalpy change.

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

.

The difference between the energy barrier of the forward reaction and reverse reaction determines the net heat content change.

3

Substitute the values and solve for

E_{a(\text{reverse})}

.

E_{a(\text{reverse})} = 75\text{ kJ mol}^{-1} - 30\text{ kJ mol}^{-1} = 45\text{ kJ mol}^{-1}

.

Rearranging the equation yields the activation energy needed for the reverse process.

Anahtar Kavram

Relationship between forward activation energy, reverse activation energy, and enthalpy change in energy profile diagrams.

Tahmini Süre:45s

Soru 5Soru →

A sample of

0.13\text{ g}

of zinc granules reacts completely with an excess of dilute hydrochloric acid according to the reaction equation:

\text{Zn}(s) + 2\text{HCl}(aq) \rightarrow \text{ZnCl}_2(aq) + \text{H}_2(g)

If the reaction takes exactly

40\text{ seconds}

to reach completion, what is the average rate of consumption of hydrochloric acid in

\text{mol s}^{-1}

? (Molar mass of

\text{Zn} = 65\text{ g mol}^{-1}

)

Cevabı ve açıklamayı göster

Cevap: 0.0001

Cevap

The average rate of consumption of hydrochloric acid is

0.0001\text{ mol s}^{-1}

(or

1.0 \times 10^{-4}\text{ mol s}^{-1}

).

To determine the average rate of consumption of hydrochloric acid, first convert the mass of zinc to moles (

0.13\text{ g} / 65\text{ g mol}^{-1} = 0.002\text{ mol}

). According to the stoichiometric coefficients in the balanced equation

\text{Zn} + 2\text{HCl} \rightarrow \text{ZnCl}_2 + \text{H}_2

,

2\text{ moles}

of

\text{HCl}

react for every

1\text{ mole}

of

\text{Zn}

. Therefore,

0.004\text{ mol}

of

\text{HCl}

is consumed. Dividing this quantity by the reaction time (

40\text{ seconds}

) gives an average rate of

0.0001\text{ mol s}^{-1}

.

Adım Adım Çözüm

1

Calculate the amount in moles of zinc reacted

\text{Moles of Zn} = \frac{0.13\text{ g}}{65\text{ g mol}^{-1}} = 0.002\text{ mol}

Mass divided by molar mass yields the quantity in moles.

2

Determine the moles of hydrochloric acid consumed using the mole ratio

\text{Moles of HCl} = 2 \times 0.002\text{ mol} = 0.004\text{ mol}

The balanced chemical equation shows a

1:2

stoichiometric ratio between

\text{Zn}

and

\text{HCl}

.

3

Calculate the average rate of consumption of HCl per unit time

\text{Rate of HCl consumption} = \frac{0.004\text{ mol}}{40\text{ s}} = 0.0001\text{ mol s}^{-1}

Rate of reaction is defined as the change in moles of reactant divided by elapsed time.

Anahtar Kavram

Stoichiometric determination of reaction rate from reactant consumption

Soru 6Soru →

In a $1.0\text{ dm}^3$ rigid reaction vessel, $4.0\text{ moles}$ of gas $A$ and $3.0\text{ moles}$ of gas $B$ are mixed and allowed to reach equilibrium at a constant temperature according to the equation:

2A_{(g)} + B_{(g)} \rightleftharpoons C_{(g)}

If analysis shows that $1.0\text{ mole}$ of gas $C$ is present at equilibrium, what is the value of the equilibrium constant, $K_c$ , for this reaction?

0.125\text{ dm}^6\text{mol}^{-2}

0.250\text{ dm}^6\text{mol}^{-2}

0.056\text{ dm}^6\text{mol}^{-2}

8.000\text{ dm}^6\text{mol}^{-2}

Cevabı ve açıklamayı göster

Cevap: $0.125\text{ dm}^6\text{mol}^{-2}$

Cevap

The equilibrium constant

K_c

is

0.125\text{ dm}^6\text{mol}^{-2}

.

To find

K_c

, set up an equilibrium concentration table. Producing

1.0\text{ mole}

of product

C

consumes

2.0\text{ moles}

of

A

and

1.0\text{ mole}

of

B

per

\text{dm}^3

. The equilibrium concentrations are

[A] = 2.0\text{ mol dm}^{-3}

,

[B] = 2.0\text{ mol dm}^{-3}

, and

[C] = 1.0\text{ mol dm}^{-3}

. Substituting these values into

K_c = \frac{[C]}{[A]^2[B]}

yields

\frac{1.0}{2.0^2 \times 2.0} = \frac{1.0}{8.0} = 0.125\text{ dm}^6\text{mol}^{-2}

.

Adım Adım Çözüm

1

Determine initial molar concentrations

Since container volume is

1.0\text{ dm}^3

, initial concentrations are

[A]_0 = 4.0\text{ mol dm}^{-3}

,

[B]_0 = 3.0\text{ mol dm}^{-3}

, and

[C]_0 = 0\text{ mol dm}^{-3}

.

Concentration is moles divided by volume in

\text{dm}^3

.

2

Calculate equilibrium concentrations using stoichiometry (ICE Table)

At equilibrium,

[C]_{eq} = 1.0\text{ mol dm}^{-3}

.
According to

2A + B \rightleftharpoons C

:
Moles of

A

consumed =

2 \times 1.0 = 2.0\text{ mol dm}^{-3} \implies [A]_{eq} = 4.0 - 2.0 = 2.0\text{ mol dm}^{-3}

.
Moles of

B

consumed =

1 \times 1.0 = 1.0\text{ mol dm}^{-3} \implies [B]_{eq} = 3.0 - 1.0 = 2.0\text{ mol dm}^{-3}

.

Stoichiometric coefficients govern the mole ratio of reactants consumed to products formed.

3

Write the

K_c

expression and substitute equilibrium concentrations

K_c = \frac{[C]}{[A]^2[B]} = \frac{1.0}{(2.0)^2 \times 2.0} = \frac{1.0}{4.0 \times 2.0} = \frac{1.0}{8.0} = 0.125\text{ dm}^6\text{mol}^{-2}

The equilibrium constant expression raises each concentration to the power of its stoichiometric coefficient.

Anahtar Kavram

Equilibrium constant

K_c

expression and stoichiometric calculations

Tahmini Süre:2m 0s

Soru 7Soru →

Given the following thermochemical equations:

\text{I. } 2\text{Fe}(s) + \frac{3}{2}\text{O}_2(g) \rightarrow \text{Fe}_2\text{O}_3(s) \quad \Delta H^\circ = -824.2\text{ kJ mol}^{-1}

\text{II. } \text{CO}(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H^\circ = -283.0\text{ kJ mol}^{-1}

What is the standard enthalpy change,

\Delta H^\circ

, for the reduction of iron(III) oxide by carbon monoxide according to the following reaction?

\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(s) + 3\text{CO}_2(g)

-24.8\text{ kJ mol}^{-1}

-1673.2\text{ kJ mol}^{-1}

+1673.2\text{ kJ mol}^{-1}

+541.2\text{ kJ mol}^{-1}

Cevabı ve açıklamayı göster

Cevap: $-24.8\text{ kJ mol}^{-1}$

Cevap

-24.8\text{ kJ mol}^{-1}

According to Hess's Law, the total enthalpy change for a reaction is the sum of the enthalpy changes for individual intermediate steps. Reversing the formation equation of

\text{Fe}_2\text{O}_3(s)

flips its enthalpy sign from

-824.2\text{ kJ mol}^{-1}

to

+824.2\text{ kJ mol}^{-1}

. Multiplying the oxidation reaction of

\text{CO}(g)

by 3 scales its enthalpy from

-283.0\text{ kJ mol}^{-1}

to

-849.0\text{ kJ mol}^{-1}

. Adding both modified values yields

+824.2\text{ kJ mol}^{-1} - 849.0\text{ kJ mol}^{-1} = -24.8\text{ kJ mol}^{-1}

.

Adım Adım Çözüm

1

Reverse Equation I so that

\text{Fe}_2\text{O}_3(s)

becomes a reactant.

\text{Fe}_2\text{O}_3(s) \rightarrow 2\text{Fe}(s) + \frac{3}{2}\text{O}_2(g) \quad \Delta H_1^\circ = +824.2\text{ kJ mol}^{-1}

Reversing a thermochemical equation changes the sign of its standard enthalpy change,

\Delta H^\circ

.

2

Multiply Equation II by 3 so that the stoichiometric coefficient of

\text{CO}(g)

matches the target reaction.

3\text{CO}(g) + \frac{3}{2}\text{O}_2(g) \rightarrow 3\text{CO}_2(g) \quad \Delta H_2^\circ = 3 \times (-283.0\text{ kJ mol}^{-1}) = -849.0\text{ kJ mol}^{-1}

Enthalpy change is an extensive property, so multiplying reaction coefficients requires multiplying

\Delta H^\circ

by the same factor.

3

Sum the modified thermochemical equations according to Hess's Law.

\text{Fe}_2\text{O}_3(s) + 3\text{CO}(g) \rightarrow 2\text{Fe}(s) + 3\text{CO}_2(g)

\Delta H^\circ = \Delta H_1^\circ + \Delta H_2^\circ = +824.2\text{ kJ mol}^{-1} + (-849.0\text{ kJ mol}^{-1}) = -24.8\text{ kJ mol}^{-1}

Hess's Law states that the overall enthalpy change for a chemical reaction is independent of the pathway taken.

Anahtar Kavram

Hess's Law of Constant Heat Summation

Tahmini Süre:2m 0s

Soru 8Soru →

When the temperature of a reacting system is increased, the rate of reaction increases. According to collision theory, which of the following best explains this increase?

The fraction of colliding particles with kinetic energy equal to or greater than the activation energy increases significantly.

The activation energy of the chemical reaction is lowered by the increase in thermal energy.

The enthalpy change (

\Delta H

) of the reaction increases, making the reaction more exothermic.

The total number of reactant molecules increases, raising the molar concentration of the system.

Cevabı ve açıklamayı göster

Cevap: The fraction of colliding particles with kinetic energy equal to or greater than the activation energy increases significantly.

Cevap

The rate of reaction increases because higher temperature increases the fraction of colliding particles possessing kinetic energy equal to or exceeding the activation energy.

According to collision theory, increasing temperature increases the average kinetic energy of the molecules. This results in a much larger fraction of colliding molecules having energy equal to or greater than the activation energy (

E_a

), thereby dramatically increasing the rate of successful collisions.

Adım Adım Çözüm

1

Recall the two main requirements for an effective collision according to collision theory.

Effective collisions require particles to collide with correct orientation and with energy greater than or equal to the activation energy (

E_a

).

Collision theory states that not all collisions produce products.

2

Analyze the impact of increasing temperature on molecular kinetic energy.

Higher temperature increases the average kinetic energy of reactant particles, broadening the Maxwell-Boltzmann energy distribution curve.

Temperature is a measure of the average kinetic energy of particles.

3

Determine why this leads to a faster reaction rate.

A significantly larger fraction of collisions now possess energy

E \ge E_a

, leading to a higher frequency of successful (effective) collisions.

This exponential increase in effective collisions is the primary reason for the increased rate of reaction.

Anahtar Kavram

Effect of Temperature on Kinetic Energy and Activation Energy in Collision Theory

Soru 9Soru →

The hydration of ethene to produce liquid ethanol is represented by the chemical equation:

\text{C}_2\text{H}_4(g) + \text{H}_2\text{O}(g) \rightarrow \text{C}_2\text{H}_5\text{OH}(l)

Given the following thermochemical equations:
1. $\text{C}_2\text{H}_5\text{OH}(l) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 3\text{H}_2\text{O}(l) \quad \Delta H^\circ = -1367\text{ kJ mol}^{-1}$
2. $\text{C}_2\text{H}_4(g) + 3\text{O}_2(g) \rightarrow 2\text{CO}_2(g) + 2\text{H}_2\text{O}(l) \quad \Delta H^\circ = -1411\text{ kJ mol}^{-1}$
3. $\text{H}_2\text{O}(g) \rightarrow \text{H}_2\text{O}(l) \quad \Delta H^\circ = -44\text{ kJ mol}^{-1}$

Calculate the standard enthalpy change, $\Delta H^\circ$ , for the hydration reaction in $\text{kJ mol}^{-1}$ .

Cevabı ve açıklamayı göster

Cevap: -88

Cevap

The standard enthalpy change for the hydration reaction is -88 kJ/mol.

According to Hess's law, the standard enthalpy change of a net reaction can be determined by algebraically combining component reactions and their enthalpy values. Adding Equation 2 as written, Equation 3 as written, and the reverse of Equation 1 cancels out intermediate species (carbon dioxide, oxygen, and liquid water), leaving the net hydration reaction. Summing their respective enthalpy values gives -1411 kJ/mol + (-44 kJ/mol) + 1367 kJ/mol = -88 kJ/mol.

Adım Adım Çözüm

1

Identify the required target thermochemical equation

Target:

\text{C}_2\text{H}_4(g) + \text{H}_2\text{O}(g) \rightarrow \text{C}_2\text{H}_5\text{OH}(l)

This establishes the stoichiometry and physical states required for reactants and products.

2

Apply Hess's law to reverse and combine the given thermochemical equations

Keep Equation 2:

\Delta H_2 = -1411\text{ kJ mol}^{-1}

Keep Equation 3:

\Delta H_3 = -44\text{ kJ mol}^{-1}

Reverse Equation 1:

\Delta H_1' = +1367\text{ kJ mol}^{-1}

Reversing Equation 1 places liquid ethanol on the product side, requiring the sign of its enthalpy change to be inverted.

3

Sum the enthalpy changes of the modified reaction steps

\Delta H^\circ = (-1411) + (-44) + (+1367) = -88\text{ kJ mol}^{-1}

According to Hess's law, the total enthalpy change of an overall reaction equals the sum of the enthalpy changes for individual component steps.

Anahtar Kavram

Hess's Law of Constant Heat Summation

Soru 10Soru →

Match each chemical process on the left with its corresponding thermodynamic energy classification and enthalpy characteristics on the right.

Soldaki öğeye tıklayın, sonra eşleşen sağdaki öğeye tıklayın

Cevabı ve açıklamayı göster

Cevap

Dissolution of ammonium chloride matches Endothermic dissolution (absorbs heat causing solution temperature drop); Combustion of methane matches Exothermic combustion (releases heat energy into surroundings); Neutralization of acid and base matches Exothermic neutralization (releases heat during water formation); Thermal decomposition of calcium carbonate matches Endothermic thermal decomposition (requires continuous external heat input).

The pairs are assigned based on thermodynamic principles: reactions that absorb energy from their surroundings (ammonium chloride dissolution and calcium carbonate breakdown) are endothermic (

\Delta H > 0

), whereas reactions that release heat to their surroundings (methane combustion and acid-base neutralization) are exothermic (

\Delta H < 0

).

Adım Adım Çözüm

1

Identify whether each reaction absorbs or releases energy from its surroundings.

Dissolution of

NH_4Cl

and decomposition of

CaCO_3

absorb heat (

\Delta H > 0

). Combustion of

CH_4

and neutralization of

HCl/NaOH

evolve heat (

\Delta H < 0

).

Endothermic processes absorb heat from surroundings (

\Delta H > 0

), while exothermic processes release heat to surroundings (

\Delta H < 0

).

2

Pair dissolution of ammonium chloride with its thermodynamic behavior.

Matches 'Endothermic dissolution (

\Delta H > 0

), where heat is absorbed from the surroundings, resulting in a temperature drop of the solution.'

Lattice dissociation energy exceeds hydration energy in

NH_4Cl

dissolution, causing a cooling effect in the solution.

3

Pair combustion of methane gas with its thermodynamic behavior.

Matches 'Exothermic combustion (

\Delta H < 0

), where bond formation in products releases substantial thermal energy to the surroundings.'

Energy released upon forming

C=O

and

O-H

bonds is greater than energy required to break

C-H

and

O=O

bonds.

4

Pair acid-base neutralization with its thermodynamic behavior.

Matches 'Exothermic neutralization (

\Delta H < 0

), releasing heat as hydrogen ions react with hydroxide ions to form liquid water.'

The reaction

H^+_{(aq)} + OH^-_{(aq)} \rightarrow H_2O_{(l)}

has a standard enthalpy change of approximately

-57.3\text{ kJ mol}^{-1}

.

5

Pair thermal decomposition of calcium carbonate with its thermodynamic behavior.

Matches 'Endothermic thermal decomposition (

\Delta H > 0

), requiring continuous thermal energy input to break chemical bonds in the reactant.'

Thermal breakdown of limestone into calcium oxide and carbon dioxide requires continuous high-temperature heat input.

Anahtar Kavram

Classification and Enthalpy Changes of Exothermic and Endothermic Reactions

Soru 11Soru →

An exothermic reversible reaction $P_{(g)} + Q_{(g)} \rightleftharpoons R_{(g)}$ has an enthalpy change ( $\Delta H$ ) of $-40\text{ kJ mol}^{-1}$ and an uncatalyzed forward activation energy of $65\text{ kJ mol}^{-1}$ . In the presence of a catalyst, the activation energy for the forward reaction is lowered by $25\text{ kJ mol}^{-1}$ . What is the activation energy for the reverse catalyzed reaction?

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

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

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

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

Cevabı ve açıklamayı göster

Cevap: $80\text{ kJ mol}^{-1}$

Cevap

The activation energy for the reverse catalyzed reaction is

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

.

In an exothermic reaction with

\Delta H = -40\text{ kJ mol}^{-1}

, the products reside at a lower energy level than the reactants. The catalyzed forward activation energy is

65 - 25 = 40\text{ kJ mol}^{-1}

. To go from products back to the activated complex, reactants must gain the

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

lost during the forward reaction plus the

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

required to reach the transition state. Thus, the reverse catalyzed activation energy is

40 - (-40) = 80\text{ kJ mol}^{-1}

.

Adım Adım Çözüm

1

Calculate the forward catalyzed activation energy

E_{a,\text{f(cat)}} = 65\text{ kJ mol}^{-1} - 25\text{ kJ mol}^{-1} = 40\text{ kJ mol}^{-1}

A catalyst lowers the forward activation energy by the given reduction amount.

2

Relate reverse activation energy to forward activation energy and reaction enthalpy

E_{a,\text{r}} = E_{a,\text{f}} - \Delta H

For any reaction step, the energy barrier in the reverse direction equals the forward energy barrier minus the enthalpy change.

3

Calculate the reverse catalyzed activation energy using the catalyzed forward value and enthalpy change

E_{a,\text{r(cat)}} = 40\text{ kJ mol}^{-1} - (-40\text{ kJ mol}^{-1}) = 80\text{ kJ mol}^{-1}

Substituting

E_{a,\text{f(cat)}} = 40\text{ kJ mol}^{-1}

and

\Delta H = -40\text{ kJ mol}^{-1}

yields the energy required to convert products back to the transition state under catalysis.

Anahtar Kavram

Relationship between forward activation energy, reverse activation energy, enthalpy change, and catalytic lowering in reaction profile diagrams

Soru 12Soru →

During a chemical reaction between dilute hydrochloric acid and calcium carbonate, $60\text{ cm}^3$ of carbon dioxide gas is collected over a period of $30\text{ seconds}$ . What is the average rate of evolution of the gas in $\text{cm}^3\text{ s}^{-1}$ ?

Cevabı ve açıklamayı göster

Cevap: 2

Cevap

The average rate of gas evolution is

2.0\text{ cm}^3\text{ s}^{-1}

.

The average rate of a reaction yielding a gas is given by dividing the total volume of gas produced by the total time taken. Dividing

60\text{ cm}^3

by

30\text{ seconds}

gives an average rate of

2.0\text{ cm}^3\text{ s}^{-1}

.

Adım Adım Çözüm

1

Identify the volume of gas evolved and the time duration.

Volume =

60\text{ cm}^3

, Time =

30\text{ seconds}

.

These are the measured parameters required to calculate the average rate of reaction.

2

Calculate the average rate of reaction.

\text{Rate} = \frac{60\text{ cm}^3}{30\text{ s}} = 2.0\text{ cm}^3\text{ s}^{-1}

.

The reaction rate measures the change in concentration or amount of a reactant or product per unit time.

Anahtar Kavram

Rate of Reaction Calculation

Soru 13Soru →

Consider the thermochemical equations below:

\text{I. } \text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H = -393.5\text{ kJ mol}^{-1}

\text{II. } \text{CO}(g) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H = -283.0\text{ kJ mol}^{-1}

What is the standard enthalpy of formation of carbon(II) oxide, $\text{CO}(g)$ ?

-110.5\text{ kJ mol}^{-1}

+110.5\text{ kJ mol}^{-1}

-676.5\text{ kJ mol}^{-1}

+676.5\text{ kJ mol}^{-1}

Cevabı ve açıklamayı göster

Cevap: $-110.5\text{ kJ mol}^{-1}$

Cevap

-110.5\text{ kJ mol}^{-1}

To find the enthalpy of formation of

\text{CO}(g)

from

\text{C}(s)

and

\text{O}_2(g)

, Equation I is kept as written (

\Delta H_1 = -393.5\text{ kJ mol}^{-1}

) while Equation II is reversed so that

\text{CO}(g)

appears on the product side (changing

\Delta H_2

from

-283.0\text{ kJ mol}^{-1}

to

+283.0\text{ kJ mol}^{-1}

). Summing both steps gives

-393.5 + 283.0 = -110.5\text{ kJ mol}^{-1}

.

Adım Adım Çözüm

1

Write the target equation for the standard enthalpy of formation of

\text{CO}(g)

\text{C}(s) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g) \quad \Delta H_f^\circ = ?

The standard enthalpy of formation is the heat change when 1 mole of a substance is formed from its constituent elements in their standard states.

2

Manipulate the given equations so their sum yields the target equation

Keep Equation I as written:

\text{C}(s) + \text{O}_2(g) \rightarrow \text{CO}_2(g) \quad \Delta H_1 = -393.5\text{ kJ mol}^{-1}

Reverse Equation II:

\text{CO}_2(g) \rightarrow \text{CO}(g) + \frac{1}{2}\text{O}_2(g) \quad \Delta H_2' = +283.0\text{ kJ mol}^{-1}

Reversing a reaction changes the sign of its enthalpy change according to Hess's Law.

3

Sum the manipulated equations and their corresponding

\Delta H

values

\text{C}(s) + \frac{1}{2}\text{O}_2(g) \rightarrow \text{CO}(g)

\Delta H_f^\circ = -393.5\text{ kJ mol}^{-1} + 283.0\text{ kJ mol}^{-1} = -110.5\text{ kJ mol}^{-1}

Hess's Law states that the overall enthalpy change of a reaction is equal to the sum of the enthalpy changes for each step.

Anahtar Kavram

Hess's Law and Standard Enthalpy of Formation

Tahmini Süre:1m 30s

Soru 14Soru →

Which of the following statements correctly describes the effect of adding a positive catalyst to a chemical reaction?

It lowers the activation energy barrier by providing an alternative reaction pathway.

It shifts the equilibrium position to favor product formation.

It increases the overall enthalpy change (

\Delta H

) of the reaction.

It increases the potential energy level of the initial reactants.

Cevabı ve açıklamayı göster

Cevap: It lowers the activation energy barrier by providing an alternative reaction pathway.

Cevap

A catalyst lowers the activation energy barrier by providing an alternative reaction pathway.

A positive catalyst increases the rate of reaction by providing an alternative pathway with a lower activation energy barrier, allowing more colliding particles to possess energy equal to or exceeding the activation energy.

Adım Adım Çözüm

1

Identify the role of a catalyst in chemical kinetics.

A catalyst speeds up a reaction without being consumed.

By introducing an alternative mechanism, the activation energy (

E_a

) peak on the energy profile diagram is lowered.

2

Distinguish between kinetic effects and thermodynamic properties.

Thermodynamic properties like enthalpy change (

\Delta H

) and equilibrium position remain unaffected.

The initial energy of reactants and final energy of products remain constant regardless of whether a catalyst is present.

Anahtar Kavram

Effect of a Catalyst on Activation Energy and Energy Profile Diagrams

Soru 15Soru →

Calculate the standard enthalpy of combustion of liquid carbon disulfide (

\text{CS}_2(l)

) in

\text{kJ mol}^{-1}

, given the following standard enthalpies of formation:

\Delta H_f^\circ[\text{CS}_2(l)] = +88\text{ kJ mol}^{-1}

\Delta H_f^\circ[\text{CO}_2(g)] = -394\text{ kJ mol}^{-1}

\Delta H_f^\circ[\text{SO}_2(g)] = -297\text{ kJ mol}^{-1}

The balanced chemical equation for the combustion process is:

\text{CS}_2(l) + 3\text{O}_2(g) \rightarrow \text{CO}_2(g) + 2\text{SO}_2(g)

What is the standard enthalpy change of combustion in $\text{kJ mol}^{-1}$ ?

Cevabı ve açıklamayı göster

Cevap: -1076

Cevap

The standard enthalpy of combustion of liquid carbon disulfide is -1076 kJ mol^-1.

Applying Hess's law using standard enthalpies of formation gives

\Delta H^\circ = \sum \Delta H_f^\circ(\text{products}) - \sum \Delta H_f^\circ(\text{reactants})

. For the combustion of carbon disulfide, this evaluates to

[(-394) + 2(-297)] - (+88) = -988 - 88 = -1076\text{ kJ mol}^{-1}

.

Adım Adım Çözüm

1

Formulate the standard enthalpy of reaction equation using standard enthalpies of formation.

\Delta H^\circ = \sum n\Delta H_f^\circ(\text{products}) - \sum m\Delta H_f^\circ(\text{reactants})

According to Hess's Law, the net standard enthalpy change for a chemical process equals the sum of standard formation enthalpies of products minus reactants.

2

Substitute the provided standard enthalpy of formation values into the expression, taking into account stoichiometry.

\Delta H^\circ = [-394 + 2(-297)] - [88] = -1076\text{ kJ mol}^{-1}

Sulfur dioxide is formed with a mole ratio of 2, so its formation enthalpy must be doubled; oxygen gas has a formation enthalpy of zero.

Anahtar Kavram

Standard Enthalpy Changes and Hess's Law

Soru 16Soru →

Consider the reversible industrial reaction represented by the thermochemical equation below:

2SO_2(g) + O_2(g) \rightleftharpoons 2SO_3(g) \quad \Delta H = -197\text{ kJ mol}^{-1}

If finely divided vanadium(V) oxide ( $V_2O_5$ ) catalyst is introduced into the reaction vessel at constant volume while the temperature of the system is lowered, which of the following best describes the combined effect on the equilibrium yield of $SO_3(g)$ and the rate of reaching equilibrium?

The yield of

SO_3(g)

increases because the exothermic forward reaction is favored by cooling, and the equilibrium state is reached more rapidly due to the catalyst lowering activation energy.

The yield of

SO_3(g)

increases further because the catalyst preferentially accelerates the forward reaction, while the lower temperature slows down the overall rate.

The yield of

SO_3(g)

decreases because a negative enthalpy change ($\Delta H < 0$) signifies an endothermic process that is disfavored when temperature is reduced.

The yield of

SO_3(g)

remains unchanged because the catalytic effect counteracts the equilibrium shift induced by the temperature reduction.

Cevabı ve açıklamayı göster

Cevap: The yield of $SO_3(g)$ increases because the exothermic forward reaction is favored by cooling, and the equilibrium state is reached more rapidly due to the catalyst lowering activation energy.

Cevap

The yield of

SO_3(g)

increases because the exothermic forward reaction is favored by cooling, and the equilibrium state is reached more rapidly due to the catalyst lowering activation energy.

Lowering the temperature favors the heat-producing (exothermic) forward reaction because $\Delta H < 0$, resulting in an increased equilibrium yield of

SO_3(g)

. Simultaneously, adding vanadium(V) oxide provides an alternative reaction pathway with lower activation energy, speeding up the attainment of dynamic equilibrium without altering the equilibrium position.

Adım Adım Çözüm

1

Analyze the thermochemical equation for enthalpy change ($\Delta H$).

$\Delta H = -197\text{ kJ mol}^{-1}$, which confirms the forward reaction is exothermic (releases heat).

Determining whether the forward reaction is exothermic or endothermic is necessary to predict the impact of temperature changes according to Le Chatelier's principle.

2

Apply Le Chatelier's principle to the temperature decrease.

Lowering the temperature shifts the equilibrium in the heat-producing (exothermic forward) direction, increasing the yield of

SO_3(g)

.

The system counteracts the loss of thermal energy by shifting toward the side that produces heat.

3

Evaluate the effect of adding a catalyst (

V_2O_5

).

The catalyst lowers activation energy for both forward and reverse pathways, accelerating the rate at which equilibrium is attained without affecting the position of equilibrium or yield.

Catalysts increase reaction rates equally in both directions and have zero effect on thermodynamic equilibrium position.

Anahtar Kavram

Effect of temperature and catalysts on dynamic equilibrium (Le Chatelier's Principle)

Tahmini Süre:1m 30s

Soru 17Soru →

A chemical reaction has an enthalpy change ( $\Delta H$ ) of $+40.0\text{ kJ mol}^{-1}$ and an entropy change ( $\Delta S$ ) of $+100\text{ J K}^{-1}\text{ mol}^{-1}$ . Above what minimum temperature, in degrees Celsius ( $^\circ\text{C}$ ), will the reaction become spontaneous?

127^\circ\text{C}

400^\circ\text{C}

673^\circ\text{C}

-272.6^\circ\text{C}

Cevabı ve açıklamayı göster

Cevap: $127^\circ\text{C}$

Cevap

The minimum temperature above which the reaction becomes spontaneous is

127^\circ\text{C}

.

According to the Gibbs free energy relationship

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

, a reaction is spontaneous when

\Delta G < 0

. At the transition temperature between spontaneous and non-spontaneous states,

\Delta G = 0

. Substituting

\Delta H = 40,000\text{ J mol}^{-1}

and

\Delta S = 100\text{ J K}^{-1}\text{ mol}^{-1}

gives

T = \frac{40000}{100} = 400\text{ K}

. Converting to degrees Celsius by subtracting

273

yields

127^\circ\text{C}

. Therefore, above

127^\circ\text{C}

, the reaction becomes spontaneous.

Adım Adım Çözüm

1

Convert the enthalpy change from kilojoules to joules to ensure consistent units with entropy change.

\Delta H = +40.0\text{ kJ mol}^{-1} = +40,000\text{ J mol}^{-1}

\Delta S

is given in

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

, so

\Delta H

must be expressed in Joules.

2

Determine the threshold temperature (

T

) at equilibrium where

\Delta G = 0

using the Gibbs free energy equation

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

.

0 = \Delta H - T\Delta S \implies T = \frac{\Delta H}{\Delta S} = \frac{40000\text{ J mol}^{-1}}{100\text{ J K}^{-1}\text{ mol}^{-1}} = 400\text{ K}

A reaction is spontaneous when

\Delta G < 0

, which occurs when temperature exceeds the threshold temperature

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

for endothermic reactions with positive entropy change.

3

Convert the temperature from Kelvin (

\text{K}

) to degrees Celsius (

^\circ\text{C}

).

T(^\circ\text{C}) = 400\text{ K} - 273 = 127^\circ\text{C}

The question specifically requests the temperature in degrees Celsius.

Anahtar Kavram

Gibbs Free Energy Equation and Temperature Dependence of Spontaneity

Soru 18Soru →

What is the correct equilibrium constant expression, $K_c$ , for the reaction $2\text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g)$ ?

K_c = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2}

K_c = \frac{[\text{NO}_2]^2}{[\text{N}_2\text{O}_4]}

K_c = \frac{[\text{N}_2\text{O}_4]}{2[\text{NO}_2]}

K_c = \frac{2[\text{NO}_2]}{[\text{N}_2\text{O}_4]}

Cevabı ve açıklamayı göster

Cevap: $K_c = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2}$

Cevap

The correct equilibrium constant expression is

K_c = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2}

.

According to the Law of Mass Action, the equilibrium constant expression

K_c

for a general reversible reaction

aA + bB \rightleftharpoons cC + dD

is given by

K_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}

. For the reaction

2\text{NO}_2(g) \rightleftharpoons \text{N}_2\text{O}_4(g)

, the concentration of the product

\text{N}_2\text{O}_4

is raised to the power of 1 in the numerator, and the concentration of the reactant

\text{NO}_2

is raised to the power of 2 in the denominator, yielding

K_c = \frac{[\text{N}_2\text{O}_4]}{[\text{NO}_2]^2}

.

Adım Adım Çözüm

1

Identify the products and reactants along with their stoichiometric coefficients from the balanced equation.

Reactant:

\text{NO}_2

with coefficient 2; Product:

\text{N}_2\text{O}_4

with coefficient 1.

Equilibrium constant expressions require product concentrations in the numerator and reactant concentrations in the denominator, each raised to the power of its stoichiometric coefficient.

2

Formulate the equilibrium constant ratio

K_c = \frac{[\text{Products}]^{\text{coefficients}}}{[\text{Reactants}]^{\text{coefficients}}}

.

K_c = \frac{[\text{N}_2\text{O}_4]^1}{[\text{NO}_2]^2}

.

Placing

[\text{N}_2\text{O}_4]

in the numerator and

[\text{NO}_2]^2

in the denominator satisfies the law of mass action.

Anahtar Kavram

Equilibrium Constant Expression (

K_c

)

Soru 19Soru →

In an experiment to measure the rate of a chemical reaction, $0.50\text{ g}$ of calcium carbonate reacts completely with excess dilute hydrochloric acid in $25\text{ seconds}$ . What is the average rate of reaction with respect to the loss of mass of calcium carbonate in $\text{g s}^{-1}$ ?

Cevabı ve açıklamayı göster

Cevap: 0.02

Cevap

The average rate of reaction with respect to the mass of calcium carbonate consumed is

0.02\text{ g s}^{-1}

.

The average rate of a reaction is calculated as the ratio of the change in amount of reactant or product to the time taken. Substituting the given values gives

\text{Rate} = \frac{0.50\text{ g}}{25\text{ s}} = 0.02\text{ g s}^{-1}

.

Adım Adım Çözüm

1

Extract the given values from the problem statement.

Mass of

\text{CaCO}_3 = 0.50\text{ g}

, Time

= 25\text{ s}

.

These parameters define the total change in quantity and the time interval for the reaction.

2

Calculate the average rate of reaction by dividing the change in mass by the time elapsed.

\text{Rate} = \frac{0.50\text{ g}}{25\text{ s}} = 0.02\text{ g s}^{-1}

.

The rate of a chemical reaction measures how rapidly a reactant is consumed per unit time.

Anahtar Kavram

Rate of Reaction Calculation

Tahmini Süre:1m 0s

Soru 20Soru →

According to collision theory, which of the following best explains why adding a positive catalyst increases the rate of a chemical reaction?

It provides an alternative reaction pathway with a lower activation energy, increasing the proportion of effective collisions.

It shifts the position of dynamic equilibrium to favor the formation of products.

It increases the average kinetic energy of the reacting particles in the system.

It increases the total enthalpy change (

\Delta H

) of the overall reaction.

Cevabı ve açıklamayı göster

Cevap: It provides an alternative reaction pathway with a lower activation energy, increasing the proportion of effective collisions.

Cevap

A catalyst increases the reaction rate by providing an alternative pathway with a lower activation energy, thereby increasing the fraction of colliding particles with energy

E \ge E_a

.

The correct option explains that a positive catalyst lowers the activation energy (

E_a

) by providing an alternative mechanism. Consequently, a greater percentage of molecular collisions possess the requisite energy to overcome the energy barrier, resulting in a higher rate of effective collisions.

Adım Adım Çözüm

1

Define collision theory criteria for effective collisions

For a collision to result in a chemical reaction, colliding particles must possess minimum activation energy (

E_a

) and correct molecular orientation.

Establishing the essential requirements for a successful chemical transformation.

2

Analyze the action of a positive catalyst

A positive catalyst offers an alternative reaction mechanism featuring an activated complex of lower energy.

Determining how the energy barrier is modified in the presence of a catalyst.

3

Evaluate the effect on reaction rate and equilibrium

Lowering

E_a

means a higher fraction of reactant particles have kinetic energy

E \ge E_a

, increasing collision frequency successfully without shifting equilibrium or changing overall

\Delta H

.

Connecting activation energy reduction directly to rate increase.

Anahtar Kavram

Role of Catalysts in Collision Theory

Tahmini Süre:1m 0s

Energetics, Rates of Reaction and Chemical Equilibrium

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