Question

At a certain temperature, the equilibrium constant for the following chemical equation is 2.40. SO2(g)+NO2(g) --> SO3 (g) + NO(g) At this temperature, calculate the number of moles of NO2(g) that must be added to 2.42 mol of SO2(g) in order to form 1.10 mol of SO3(g) at equilibrium.

Answer 1

consider 1 L of solution

intially

[S02] = 2.42

let [N02] = y

at equilibrium

[S03] = 1.1

now

consider the given reaction

S02 + N02 --> S03 + N0

using ICE table

[S03]eq = x

[NO]eq = x

[S02] eq = 2.42 - x

[N02]eq = y - x

now

given

[S03]eq = 1.1

so

x = 1.1

so

[N0]eq = x = 1.1

[S02]eq = 2.42 - x = 2.42 - 1.1 = 1.32

[N02]eq = y -x = y - 1.1

now

S02 + N02 --> S03 + N0

Kc = [S03] [N0] / [S02] [N02]

2.4 = [1.1] [1.1] / [ 1.32 ] [ y - 1.1]

solving

we get

y = 1.48

so

1.48 moles of N02 must be added