Question

Be sure to answer all parts. The first step in HNO3 production is the catalyzed oxidation of NH3. Without a catalyst, a different reaction predominates: 4 NH3(g) +3 O2 (g) ⇌ 2 N2(g) + 6 H2O(g) When 0.0150 mol of NH3(g) and 0.0150 mol of O2(g) are placed in a 1.00−L container at a certain temperature, the N2 concentration at equilibrium is 1.96 × 10−3 M. Calculate Kc.

Answer 1

Answer: The value of $K_c$ for the reaction is $6.005\times 10^{-6}$

Explanation:

We are given:

Initial moles of $NH_3=0.0150mol$

Initial moles of $O_2=0.0150mol$

Volume of the container = 1.00 L

Molarity of the solution = $\frac{\text{Number of moles}}{\text{Volume of container}}$

$[NH_3]_i=\frac{0.0150}{1.00}=0.0150M$

$[O_2]_i=\frac{0.0150}{1.00}=0.0150M$

The given chemical equation follows:

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

Initial:          0.0150        0.0150

At eqllm:   0.0150-4x     0.0150-3x    2x       6x

The expression of $K_c$ for above equation follows:

$K_c=\frac{[N_2]^2[H_2O]^6}{[NH_3]^4[O_2]^3}$         .......(1)

We are given:

Equilibrium concentration of $N_2=1.96\times 10^{-3}$

Equating the equilibrium concentrations of nitrogen, we get:

$2x=1.96\times 10^{-3}\\\\x=0.98\times 10^{-3}M$

Calculating the equilibrium concentrations:

Concentration of $NH_3=(0.0150-4x)=0.0150-4(0.00098)=0.01108M$

Concentration of $O_2=(0.0150-3x)=0.0150-3(0.00098)=0.01206M$

Concentration of $N_2=2x=2(0.00098)=0.00196M$

Concentration of $H_2O=6x=6(0.00098)=0.00588M$

Putting values in expression 1, we get:

$K_c=\frac{(0.00196)^2\times (0.00588)^6}{(0.01108)^4\times (0.01206)^3}\\\\K_c=6.005\times 10^{-6}$

Hence, the value of $K_c$ for the reaction is $6.005\times 10^{-6}$