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For the reaction $A(g) \rightleftharpoons B(g)$ at 495 K ....

For the reaction $A(g) \rightleftharpoons B(g)$ at 495 K, $\Delta_r G^\circ = -9.478 kJ.mol^{-1} $. If we start the reaction in a closed container at 495 K with 22 millimoles of A, the amount of B in the equilibrium mixture is _ _ _ _ millimoles. (Round off to the nearest integer)

[$R=8.314 J.mol^{-1} .K^{-1} $; ln 10 = 2.303 ]

Solution

For the equilibrium, $\mathop {A(g)}\limits_{a-x} \rightleftharpoons \mathop {B(g)}\limits_x $,

${K_C} = \frac{{\frac{x}{V}}}{{\frac{{a - x}}{V}}} = \frac{x}{{a - x}} = {K_P}{(RT)^{ - 0}} = {K_P}$

$\Delta G^\circ =-2.303 RT log K_P $

$\therefore \log {K_P} = \frac{{ - 9.478 \times 1000}}{{ - 2.303 \times 8.314 \times 495}} = 1$

So, ${K_P} = 10 = \frac{x}{{a - x}}$

$ \Rightarrow 0.1 = \frac{a}{x} - 1$

$ \Rightarrow \frac{a}{x} = 1.1$

$\therefore x = \frac{{22}}{{1.1}} = 20$