Question

Consider the following reaction:
2H2S(g)+SO2(g)⇌3S(s)+2H2O(g)
A reaction mixture initially containing 0.520 M H2Sand 0.520 M SO2 was found to contain 1.3×10⁻³ MH2O at a certain temperature. A second reaction mixture at the same temperature initially contains [H2S]= 0.245 M and [SO2]= 0.335 M .

Calculate the equilibrium concentration of H2O in the second mixture at this temperature.

Express your answer using two significant figures.

Answer 1

First, calculate the K expression

K = [H2O]^2[S]^3/[H2S]^2[SO2]

since S is solid

then

K = [H2O]^2/[H2S]^2[SO2]

K =((1.3*10^-3)^3) / ((0.52^2)(0.52))

K = 1.5625*10^-8

then

second reaction

[H2S] = 0.245

[SO2] = 0.335

[H2O] = ?

apply

K = [H2O]^2/[H2S]^2[SO2]

K is constant so

1.5625*10^-8 =(H2O^2) /(0.245^2*0.335)

solve for H2O

( 1.5625*10^-8)*(0.245^2)(0.335) = H2O^2

3.14193*10^-10 = H2O^2

[H2O] = sqrt(3.14193*10^-10 = 0.00001772549 = 1.7*10^-5

[H2O] =1.7*10^-5 M