Question

(ii) Explain the effect of change of the pressure and temperature in the Haber's process:
$2S{O}_{2\left(g\right)}+O_{2\left(g\right)}\rightleftharpoons 2S{O}_{3\left(g\right)}{\Delta }_r{H}^{\circ }=-193.2KL.$
(iii) Bromine monochloride (BrCl) decomposes into bromine and chlorine and reaches the equilibrium :
$2BrC{l}_{\left(g\right)}\rightleftharpoons B{r}_{2\left(g\right)}+C{l}_{2\left(g\right)}$.
For which $kc=32$ at 500 K. If initially pure BrCl is present at a concentration of $3.30\times {10}^{-3}mol{L}^{-1}$ what is its molar concentration in the mixture at equilibrium?

Answer 1

$\underset{g}{\underset{}{2S{O}_4}}+\underset{g}{\underset{}{O_2}}\rightleftharpoons \underset{g}{\underset{}{2S{o}_3}}\Delta H=-193.2KJ$

when there is an increase in pressure the equilibrium will shifts towards the side of the reaction with fewer moles of gas. When there is a decrease in pressure, the equilibrium will shift towards the side of reaction with more moles of gas.

Therefore in above reaction, the equilibrium shifts backward.

The above reaction is exothermic (since $\Delta H$ is -ve). In case of exothermic reaction with increase in temperature, the equilibrium shifts backwards and with the decrease in temperature it shifts forward.

$2BrCl\rightleftharpoons \underset{B{r}_2}{\underset{}{g}}+\underset{C{l}_2}{\underset{}{g}}$

At $t=03.3\times {10}^{-3}$

At $t=t_{eq}3.3\times {10}^{-3}-2xxx$

$K_c=\frac{x\times x}{3.3\times {10}^{-3}-2x}=32$

$x^2=105.6\times {10}^{-3}-64x$

$x^2+64x-10.56\times {10}^{-2}=0$

$x=\frac{-{64}_{-}^{+}\sqrt{(64{)}^2+4\times 1\times 10.56\times {10}^{-2}}}{2\times 1}$

$x=\frac{-{64}_{-}^{+}\sqrt{4096+0.4224}}{2\times 1}$

$x=\frac{-64+64.00330}{2}$

$x=\frac{0.003}{2}$

$x=0.0015=1.5\times {10}^3$

$\therefore$ At $t=t_{eq}$

$2Brcl\rightleftharpoons B{r}_2+C{l}_2$

$=3.3\times {10}^{-3}-2\times 1.5\times {10}^{-3}1.5\times {10}^{-3}1.5\times {10}^{-3}$

$=0.3\times {10}^{-3}1.5\times {10}^{-3}1.5\times {10}^{-3}$

$\left[BrCl\right]=0.3\times {10}^{-3}$

$\left[C{l}_2\right]=1.5\times {10}^{-3}$

$\left[B{r}_2\right]=1.5\times {10}^{-3}$