Chemical EquilibriumHard

Question

A gaseous substance AB2(g) converts to AB(g) in the presence of solid A as AB2(g) + A(s)$\rightleftharpoons$ 2AB(g).

The initial pressure and equilibrium pressure are 0.7 and 0.95 bar, respectively. Now, the equilibrium mixture is expanded reversibly and isothermally till the gas pressure falls to 0.4 bar. Then, which of the following statements is correct?

Options

A.The volume percent of AB(g) at final equilibrium is nearly 27%.
B.The volume percent of AB2(g) at final equilibrium is nearly 32.5%.
C.KP for the equilibrium is 5/9.
D.The ratio of equilibrium pressure of AB2(g) at initial and final equilibrium is more than 2 but less than 3.

Solution

AB2 (g) + A(s)$\rightleftharpoons$ 2AB(g)

Initial partial pressure 0.7 bar 0

Equ. partial pressure (0.7 – x) bar 2x bar

Second equ. partial pressure y bar (0.4 – y) bar

From question, (0.7 – x) + 2x = 0.95 ⇒ x = 0.25

$\therefore K_{P} = \frac{(2x)^{2}}{(0.7 - x)} = \frac{(0.5)^{2}}{0.45} = \frac{5}{9} $$$\text{Now, }\frac{5}{9} = \frac{(0.4 - y)^{2}}{y}$$

⇒ y = 0.13

∴ At second equilibrium, the volume per cent of $AB_{2} = \frac{0.13}{0.4} \times 100 = 32.5\%$

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