Question

SO3(g) + NO(g) NO2(g) + SO2(g) Kc was found to be 0.500 at a certain temperature. A reaction mixture is prepared in which 0.189 mol NO2 and 0.189 mol of SO2 are placed in a 4.00 L vessel. After the system reaches equilibrium what will be the equilibrium concentrations of all four gases?

Answer 1

SO3(g) + NO(g) ⇌ NO2(g) + SO2(g)

Kc = 0.5

initial concnetrations

[NO2] = 0.189 /4 = 0.0473

[SO2] = 0.189 /4 = 0.0473

in equilbirium?

Assume x is the extent of reaction

[SO2] = 0.0473-x

[NO] = 0.0473- x

[SO3] = 0+x

[NO2] = 0+x

Substitute in equilibrium expression

K = [SO2][NO2]/[SO3][NO]

K = (0.0473- x)^2 / (x)(x)

0.5 = (0.0473- x)^2 / x^2

sqr(0.5 ) = (0.0473- x)/x

0.707*x = (0.0473-x)

x = 0.0277

Substitute:

[SO2] = 0.0478 -0.0277= 0.0201

[NO] = 0.0478 - 0.0277= 0.0201

[SO3] = 0+0.0277= 0.0277

[NO2] = 0+0.028 = 0.0277