Chemical EquilibriumHard
Question
An amount of 1 mole of PCl3(g) and 1 mole of PCl5(g) is taken in a vessel of 10 L capacity maintained at 400 K. At equilibrium, the moles of Cl2 is found to be 0.004.
Options
A.KC for the reaction: PCl5(g) $\rightleftharpoons$PCl3(g) + Cl2(g) is 0.0004 M.
B.KP for the reaction: PCl3(g) + Cl2(g) $\rightleftharpoons$PCl5(g) is 0.0004 × (0.082 × 400) atm.
C.If PCl3(g) is added to the equilibrium mixture, ΔG at the new equilibrium becomes greater than the ΔG at old equilibrium.
D.After equilibrium is achieved, the moles of PCl3 are doubled and moles of Cl2 are halved simultaneously, then the partial pressure of PCl5 remains unchanged.
Solution
PCl5(g) $\rightleftharpoons$ PCl3(g) + Cl2(g)
Initial moles 1 1 0
Equ. moles 1 – x 1 + x x
≈1 ≈1 = 0.004
$\therefore K_{C} = \frac{1 \times 0.004}{1} \times \left( \frac{1}{10} \right) = 0.0004\text{ M}$
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