At a certain temperature, Kc for the following reaction between sulphur dioxide

Question

At a certain temperature, Kc for the following reaction between sulphur dioxide and nitrogen dioxide is 4.8

SO2(g) and NO2(g) have the same initial concentration:0.36 mol/L. What amount of SO3(g) is present in a 5.0L container at equilibrium?

At a certain temperature, Kc for the following reaction between sulphur dioxide and nitrogen dioxide is 4.8 SO2(g) + NO2(g) doubleheadarrow NO(g) + SO3(g) SO2(g) and NO2(g) have the same initial concentration:0.36 mol/L. What amount of SO3(g) is present in a 5.0L container at equilibrium?

Explanation / Answer

[SO2]= [NO2] = 0.36-x and [NO] = [SO3] = x

Kc = 4.8 = [NO][SO3]/ [SO2][NO2] = x^2 / (0.36-x)^2

if we take the square root of both sides of this equation we get

2.2 = x / 0.36-x
0.79 - 2.2 x = x
0.79 = 3.2 x
x = 0.25 M = [SO3]

moles SO3 = 0.25 M x 5.0 L=1.25
mass SO3 = 1.25 mol x 96 g/mol=120 g

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