Question

A reaction mixture was prepared by mixing 0.100 mol SO2, 0.200 mol NO2, 0.100 mol NO,...

A reaction mixture was prepared by mixing 0.100 mol SO2, 0.200 mol NO2, 0.100 mol NO, and 0.150 mol SO3 in a 5.00 L reaction vessel. The reaction is allowed to reach equilibrium at 460 K, when Kc = 85.0. What is the equilibrium concentration of each substance? SO2(g) + NO2(g) NO(g) + SO3(g)

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Answer #1

SO2(g) + NO2(g) ---------------> NO(g) + SO3(g) Kc = 85

Initial 0.1 mol/5L 0.2 mol/5L 0.1 mol/5L 0.15 mol/ 5L

= 0.02 M = 0.04 M = 0.02 M = 0.05 M

at equilibrium 0.02 -x 0.04-x 0.02 +x 0.05+x

Kc = ( 0.02 +x ) ( 0.05+x)/ (  0.02 -x ) ( 0.04-x )

85 =  ( 0.02 +x ) ( 0.05+x)/ ( 0.02 -x ) ( 0.04-x )

85 ( 0.02 -x ) ( 0.04-x ) =   ( 0.02 +x ) ( 0.05+x)

0.068 - 5.1 x + 85 x2 = x2 + 0.07x + 0.001

84 x2 - 5.17x + 0.067 = 0

On solving,

x = 0.0185 M

Therefore, equilibrium concentrations are

[SO2] =   0.02 -x = 0.02-0.0185 = 0.0015 M

[NO2] =   0.04 -x = 0.04-0.0185 = 0.0215 M

[NO] = 0.02 + x = 0.02 + 0.0185 = 0.0385 M

[SO3] = 0.05+x = 0.05 + 0.0185 = 0.0685 M

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