At the following equilibrium is established +\boldsymbol{S}(s) \rightleftharpoons \boldsymbol{H}_{2} \boldsymbol{S}(\boldsymbol{g}) \boldsymbol{K}_{\boldsymbol{p}}=\mathbf{6 . 8} \times \) If 0.2 mol of hydrogen and 1.0 mol of sulphur are heated to in a 1.0 litre vessel, what will be the partial pressure of at equilibrium?

Chemistry · Equilibrium involving chemical processes: Law of chemical equilibrium, equilibrium constants (Kp and Kc) and their significance, the significance of Delta G and Delta G° in chemical equilibrium

At the following equilibrium is established +\boldsymbol{S}(s) \rightleftharpoon

At \( 90^{\circ} C, \) the following equilibrium is established \( \boldsymbol{H}_{2}(\boldsymbol{g})+\boldsymbol{S}(s) \rightleftharpoons \boldsymbol{H}_{2} \boldsymbol{S}(\boldsymbol{g}) \boldsymbol{K}_{\boldsymbol{p}}=\mathbf{6 . 8} \times \) \( \mathbf{1 0}^{\mathbf{2}} \) If 0.2 mol of hydrogen and 1.0 mol of sulphur are heated to \( 90^{\circ} \mathrm{C} \) in a 1.0 litre vessel, what will be the partial pressure of \( \boldsymbol{H}_{2} \boldsymbol{S} \) at equilibrium?

A. 0.19 atm
B. 0.379 atm
C. 0.75 atm
D. None of these

Step-by-step solution

The equilibrium constant expression is Kp = P_H2S / P_H2. Since the number of moles of gas is constant, partial pressures are proportional to moles. Let x be the moles of H2 reacted. Then at equilibrium: n_H2 = 0.2 - x, n_H2S = x. Using Kp = 6.8 × 10^2, solving x/(0.2 - x) = 680 gives x ≈ 0.1997, yielding P_H2S ≈ 5.95 atm, which is not among the options. However, given the options, it is likely that Kp = 6.8 × 10^-2 (a common value). With Kp = 0.068, solving x/(0.2 - x) = 0.068 gives x ≈ 0.01273. Then P_H2S = (xRT)/V, with R = 0.0821 L·atm/mol·K, T = 363 K, V = 1 L, giving P_H2S ≈ 0.379 atm.

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Step-by-step solution

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