Chemical EquilibriumHard

Question

If 0.3 moles of hydrogen gas and 2.0 moles of sulphur solid are heated to 87o C in a 2.0 L vessel, then what will be the partial pressure of H2S gas at equilibrium? (Given: R = 0.08 L-atm/K-mol)

H2(g) + S(s) $\rightleftharpoons$H2S(g): KC = 0.08

Options

A.0.32 atm
B.0.43 atm
C.0.62 atm
D.0.48 atm

Solution

H2(g) + S(s) $\rightleftharpoons$ H2S(g)

0.3 mole 2 mole 0

Equilibrium (0.3 – x) mole (2 – x) mole x mole

$H_{C} = \frac{x/2}{(0.3 - x)/2} = 0.08 \Rightarrow 0.022 $$$\text{Now, }P_{H_{2}S} = \frac{x \times 0.08 \times 360}{2} = 0.32\text{ atm}$$

Le-Chatelier’s Principle

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