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, t

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If 0.3 moles of hydrogen gas and 2.0 moles of sulphur solid are heated to 87^o 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): K_C = 0.08

Accepted Answer

H2(g) + S(s) $ightleftharpoons$ H2S(g)0.3 mole 2 mole 0Equilibrium (0.3 – x) mole (2 – x) mole x mole$H_{C} = \frac{x/2}{(0.3 - x)/2} = 0.08 \Rightarrow 0.022 $$$	ext{Now, }P_{H_{2}S} = \frac{x 	imes 0.08 	imes 360}{2} = 0.32	ext{ atm}$$Le-Chatelier’s Principle

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