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.
___________________________________

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

Thank you!!

Explanation / Answer

the chemical reaction which takes place is

SO2(g) + 2H2S(g) -----------> 3S(s) + 2H2O(g)

so

Kp = [PH2O]2/[[PH2S]2[PSO2]

GIVEN

PH2S = PSO2 = x ..........(let) and PH2O = 24 torr

so,

Kp = (24*24)/x3 = 576/x3

now,

G(rxn) = 2G(f,H2O(g)) - 2G(f,H2S(g)) - G(f,SO2(g))

= 2(-228.6) - 2(-33.01) - (-300.4) = -90.78KJ = -90780J

so

G(rxn) = -RTln(Kp)

so

ln(Kp) = -G(rxn)/RT

put

G(rxn) = -90780

R = 8.3145

T = 298

such that ln(Kp) = 36.64

so, Kp = 8.165e15 = 576/x3

solve for x

you will get,

x = 4.132e-5 torr

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