Chapter 14, Problem 1SAQ

Interpretation Introduction

Introduction: For a reaction, the equilibrium constant is given by the law of mass action. It is defined as the ratio of concentrations of products to that of the reactants, each raised to the power of stoichiometric coefficients.

Answer to Problem 1SAQ

Correct answer: Option $\left(d\right)$ is correct.

Explanation of Solution

Justify reasons for correct statement:

The given reaction is as follows:

$\text{C}\left(\text{s}\right){\text{+2H}}_2\left(\text{g}\right)\rightleftharpoons {\text{CH}}_4\left(\text{g}\right)$

Here, the product is ${\text{CH}}_4\left(\text{g}\right)$ and reactants are $\text{C}\left(\text{s}\right){\text{and H}}_2\left(\text{g}\right)$. Thus, equilibrium constant must be expressed in terms of these components as follows:

$K_{\text{c}}=\frac{\left[{\text{CH}}_4\right]}{{\left[{\text{H}}_2\right]}^2}$

It includes the concentrations of only gaseous terms; all other terms are excluded. In option $\left(d\right)$, all these conditions are fulfilled; There fore, option $\left(d\right)$ is correct.

Justify reasons for incorrect statements:

In options $\left(b\right)$ and (c), the concentrations of solid carbon are taken as zero in writing the equilibrium constant. In option $\left(a\right)$, the concentration of carbon is excluded, but the concentration of hydrogen is not raised to the correct power of stoichiometry. There fore, options $\left(b\right)$, $\left(c\right)$, and $\left(a\right)$ are incorrect.

Conclusion

Equilibrium constant is given as $K_{\text{c}}=\frac{\left[{\text{CH}}_4\right]}{{\left[{\text{H}}_2\right]}^2}$

There fore, option $\left(d\right)$ is correct.