Chemical and Ionic Equilibrium 2 Question 54

54. One mole of nitrogen is mixed with three moles of hydrogen in a four litre container. If 0.25 per cent of nitrogen is converted to ammonia by the following reaction

$$ \mathrm{N}{2}(g)+3 \mathrm{H}{2}(g) \rightleftharpoons 2 \mathrm{NH}_{3}(g) \text {, then } $$

calculate the equilibrium constant, $K_{c}$ in concentration units. What will be the value of $K_{c}$ for the following equilibrium?

$$ \frac{1}{2} \mathrm{~N}{2}(g)+\frac{3}{2} \mathrm{H}{2}(g) \rightleftharpoons \mathrm{NH}_{3}(g) $$

$(1981,4 M)$

Passage Based Questions

Thermal decomposition of gaseous $X_{2}$ to gaseous $X$ at $298 \mathrm{~K}$ takes place according to the following equation:

$$ X_{2}(g) \rightleftharpoons 2 X(g) $$

The standard reaction Gibbs energy, $\Delta_{r} G^{\circ}$, of this reaction is positive. At the start of the reaction, there is one mole of $X_{2}$ and no $X$. As the reaction proceeds, the number of moles of $X$ formed is given by $\beta$. Thus, $\beta_{\text {equilibrium }}$ is the number of moles of $X$ formed at equilibrium. The reaction is carried out at a constant total pressure of 2 bar. Consider the gases to behave ideally.

(Given, $R=0.083 \mathrm{~L} \mathrm{bar} \mathrm{K}^{-1} \mathrm{~mol}^{-1}$ )

Show Answer

Solution:

  1. (i) $\mathrm{CH}{3} \mathrm{COOH} \rightleftharpoons \mathrm{CH}{3} \mathrm{COO}^{-}+\mathrm{H}^{+}$

$$ C(1-\alpha) \quad C \alpha \quad C \alpha $$

If no $\mathrm{HCl}$ is present,

$$ \begin{aligned} {[\mathrm{HCl}] } & =\frac{0.2}{2}=0.10 \mathrm{M} \ {\left[\mathrm{CH}_{3} \mathrm{COOH}\right] } & =0.10 \mathrm{M} \end{aligned} $$

The major contributor of $\mathrm{H}^{+}$in solution is $\mathrm{HCl}$.

$$ \begin{aligned} K_{a} & =\frac{C \alpha(0.1)}{C(1-\alpha)}=1.75 \times 10^{-5} \ \alpha & =1.75 \times 10^{-4} \end{aligned} $$

(ii) $\mathrm{mmol}$ of $\mathrm{NaOH}$ added $=\frac{6}{40} \times 1000=150$

$$ \mathrm{mmol} \text { of } \mathrm{HCl}=500 \times 0.2=100 $$

$\mathrm{mmol}$ of $\mathrm{CH}_{3} \mathrm{COOH}=500 \times 0.2=100$

After neutralisation, $\mathrm{mmol}$ of $\mathrm{CH}_{3} \mathrm{COOH}=50$

$\mathrm{mmol}$ of $\mathrm{CH}_{3} \mathrm{COONa}=50$

$$ \mathrm{pH}=\mathrm{p} K_{a}=4.75 $$