Chemical and Ionic Equilibrium - Result Question 110
####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
$$ N _2(g)+3 H _2(g) \rightleftharpoons 2 NH _3(g) \text {, then } $$
calculate the equilibrium constant, $K _c$ in concentration units. What will be the value of $K _c$ for the following equilibrium?
$$ \frac{1}{2} N _2(g)+\frac{3}{2} H _2(g) \rightleftharpoons NH _3(g) $$
$(1981,4 M)$
Passage Based Questions
Thermal decomposition of gaseous $X _2$ to gaseous $X$ at $298 K$ 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, $R=0.083 L bar K^{-1} mol^{-1}$ )
Show Answer
Solution:
- (i) $CH _3 COOH \rightleftharpoons CH _3 COO^{-}+H^{+}$
$$ C(1-\alpha) \quad C \alpha \quad C \alpha $$
If no $HCl$ is present,
$$ \begin{aligned} {[HCl] } & =\frac{0.2}{2}=0.10 M \ {\left[CH _3 COOH\right] } & =0.10 M \end{aligned} $$
The major contributor of $H^{+}$in solution is $HCl$.
$$ \begin{aligned} K _a & =\frac{C \alpha(0.1)}{C(1-\alpha)}=1.75 \times 10^{-5} \ \alpha & =1.75 \times 10^{-4} \end{aligned} $$
(ii) $mmol$ of $NaOH$ added $=\frac{6}{40} \times 1000=150$
$$ mmol \text { of } HCl=500 \times 0.2=100 $$
$mmol$ of $CH _3 COOH=500 \times 0.2=100$
After neutralisation, $mmol$ of $CH _3 COOH=50$
$mmol$ of $CH _3 COONa=50$
$$ pH=p K _a=4.75 $$
