Assume that the partial pressure of sulfur dioxide, P_SO_2, is equal to the part

Question

Assume that the partial pressure of sulfur dioxide, P_SO_2, is equal to the partial pressure of dihydrogen sulfide, P_H_2S}, and therefore P_SO_2 = P_H_2S.

****If the vapor pressure of water is 24 torr, calculate the equilibrium partial pressure of SO_2 (P_SO_2) in the system at 298 K.****

Express the pressure in atmospheres to two significant figures.
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The reaction

{ m SO_2 (g) + 2 H_2S (g) ightleftharpoons 3 S(s) + 2 H_2O(g)}
is the basis of a suggested method for removal of { m SO_2} from power-plant stack gases.

The values below may be helpful when answering questions about the process.

Substance Delta G_{ m f}^circ
{(kJ/mol)} Delta H_{ m f}^circ
{(kJ/mol)}
{H_2O(g)} -228.6 -241.8
{H_2O(l)} -237.1 -285.8
SO_2(g)} -300.4 -296.9
SO_3(g)} -370.4 -395.2
H_2S(g)} -33.01 -20.17
S(s) 0 0

Explanation / Answer

Ans: -convert the given pressure of water to atm, which is: .030 -Kp is 8.0*10^15 Kp=P(H2O)^2/P(SO2)P(H2S)^2 then: 8.0*10^15 = (0.030)^2/x^3 8.0*10^15x^3=9.0*10^-4 so i took the cube root of both sides and got 8.0*10^15x=.09654 x=1.21*10^-17

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