Question

9. The first step in industrial nitric acid production is the catalyzed oxidation of ammonia. Without a catalyst, a different reaction predominates: 4NH3(g) + 3O2(g) ⇔ 2N2(g) + 6H2O(g) When 0.0120 mol gaseous NH3 and 0.0170 mol gaseous O2 are placed in a 1.00 L container at a certain temperature, the N2 concentration at equilibrium is 2.20×10-3 M. Calculate Keq for the reaction at this temperature.

Answer 1

Answer: The value of $K_{eq}$ is $4.66\times 10^{-5}$

Explanation:

We are given:

Initial moles of ammonia = 0.0120 moles

Initial moles of oxygen gas = 0.0170 moles

Volume of the container = 1.00 L

Concentration of a substance is calculated by:

$\text{Concentration}=\frac{\text{Number of moles}}{\text{Volume}}$

So, concentration of ammonia = $\frac{0.0120}{1.00}=0.0120M$

Concentration of oxygen gas = $\frac{0.0170}{1.00}=0.0170M$

The given chemical equation follows:

                   $4NH_3(g)+3O_2(g)\rightleftharpoons 2N_2(g)+6H_2O(g)$

Initial:          0.0120    0.0170

At eqllm:     0.0120-4x   0.0170-3x       2x       6x

We are given:

Equilibrium concentration of nitrogen gas = $2.20\times 10^{-3}M=0.00220M$

Evaluating the value of 'x', we get:

$\Rightarrow 2x=0.00220\\\\\Rightarrow x=\frac{0.00220}{2}=0.00110M$

Now, equilibrium concentration of ammonia = $0.0120-4x=[0.0120-(4\times 0.00110)]=0.00760M$

Equilibrium concentration of oxygen gas = $0.0170-3x=[0.0170-(3\times 0.00110)]=0.0137M$

Equilibrium concentration of water = $6x=[6\times 0.00110]=0.00660M$

The expression of $K_{eq}$ for the above reaction follows:

$K_{eq}=\frac{[N_2]^2\times [H_2O]^6}{[NH_3]^4\times [O_2]^3}$

Putting values in above expression, we get:

$K_{eq}=\frac{(0.00220)^2\times (0.00660)^6}{(0.00760)^4\times (0.0137)^3}\\\\K_{eq}=4.66\times 10^{-5}$

Hence, the value of $K_{eq}$ is $4.66\times 10^{-5}$