In a 1.0\text{ dm}^3 rigid reaction vessel, 4.0\text{ moles} of gas A and 3.0\text{ moles}… — JAMB UTME

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?

$0.125\text{ dm}^6\text{mol}^{-2}$ Answer
B
$0.250\text{ dm}^6\text{mol}^{-2}$
C
$0.056\text{ dm}^6\text{mol}^{-2}$
D
$8.000\text{ dm}^6\text{mol}^{-2}$

Answer

The equilibrium constant

K_c

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}

A

and

1.0\text{ mole}

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}

Step-by-Step Solution

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

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.

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.

Key Concept

Equilibrium constant

K_c

expression and stoichiometric calculations

Estimated Time:2m 0s

Question

Answer

Step-by-Step Solution

Key Concept

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