Question

Calculate the value of the equilibrium constant, Kc, for the reaction below, if 0.208 moles of sulfur dioxide gas, 0.208 moles of oxygen gas and 0.425 moles of sulfur trioxide gas are present in a 1.50-liter reaction vessel at equilibrium.
2SO2(g) +O2 (g) <--> 2SO3 (g)

A. Kc=3.32 x 10^-2
B. Kc=2.01 x 10^1 (I know for a fact this one is wrong.)
C. Kc=2.98 x 10^1
D. Kc=1.47 x 10^1

Please help!

Answer 1

First, we convert the moles of each substance into the concentration using the volume of the reactor.
[SO₃] = 0.425/1.5 = 0.283 M
[SO₂] = 0.208 / 1.5 = 0.139 M
[O₂] = 0.208/1.5 = 0.139 M
The equilibrium constant is calculated by:
Kc = [SO₃]² / [O₂][SO₂]²
Kc = (0.283)²/(0.139)(0.139)²
Kc = 29.8 = 2.98 x 10¹

The answer is C