Q9 / 160 · AP EAMCET 2019 (21 Apr Shift 1) · Chemistry

The equilibrium constant at $850 \mathrm{~K}$ for the reaction $\mathrm{N}_2(g)+\mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{NO}(g)$ is 0.5625 . The equilibrium concentration of $\mathrm{NO}(g)$ is $3.0 \times 10^{-3} \mathrm{M}$. If the equilibrium concentrations of $\mathrm{N}_2(g)$ and $\mathrm{O}_2(g)$ are equal, the concentrations of $\mathrm{N}_2(g)$ in $M$ is

  • Correct answer
    A
    $4.0 \times 10^{-3}$
  • B
    $4.0 \times 10^{-2}$
  • C
    $1.6 \times 10^{-3}$
  • D
    $3.0 \times 10^{-3}$

Solution

Given, $\mathrm{N}_2(g)+\mathrm{O}_2(g) \rightleftharpoons 2 \mathrm{NO}(g)$ i.e. equilibrium constant, $K_C=0.5625$ Equilibrium concentration of $[\mathrm{NO}]=3 \times 10^{-3} \mathrm{M}$ $\begin{array}{rlrl} & & K_C & =\frac{\left[\mathrm{NO}^2\right.}{\left[\mathrm{N}_2\right]\left[\mathrm{O}_2\right]} \\ & \text {But, } & {\left[\mathrm{N}_2\right]} & =\left[\mathrm{O}_2\right] \quad \text{(Given)} \\ & \therefore & K_C & =\frac{[\mathrm{NO}]^2}{\left[\mathrm{~N}_2\right]^2} \\ & & & \\ \text {or, } & 0.5625 & =\frac{\left[3 \times 10^{-3}\right]^2}{\left[\mathrm{~N}_2\right]^2} \\ & \text {or, } & {\left[\mathrm{N}_2\right]^2} & =\frac{\left[3 \times 10^{-3}\right]^2}{0.5625}=\frac{9 \times 10^{-6}}{0.5625} \\ & \text {or, } & {\left[\mathrm{N}_2\right]^2} & =16 \times 10^{-6} \\ \text {or, } & {\left[\mathrm{N}_2\right]} & =4 \times 10^{-3} \end{array}$ Hence, option (a) is the correct answer.

Hence, the correct answer is option: 'A'.

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