Question

At a certain temperature, the equilibrium constant for the following chemical equation is 3.20.

SO₂ (g) + NO_{2 (g)} ⇌ SO₃ (g) + NO (g)

At this temperature, calculate the number of moles of NO2(g) that must be added to 2.40 mol of SO2(g) in order to form 1.00 mol of SO3(g) at equilibrium.

Answer 1

Answer – We are given, Kc = 3.20 , moles of SO₂(g) = 2.40 moles ,

At equilibrium , moles SO₃(g) = 1.00 moles

So ICE chart as follow –

    SO₂(g) + NO₂(g) <-----> SO₃(g) + NO(g)

I     2.40           Y                  0             0

C     -x           -x                  +x          +x

E 2.40-x   Y-x                  1.00           +x

So, x = 1.00

So, at equilibrium, moles of SO₂ = 2.40-1.00 = 1.40 moles

Moles of NO₂(g) = Y-1.00 , moles of SO₃ = 1.00 , moles of NO = 1.00

So, assume volume is 1 L

K =[ SO₃(g)] [NO(g)] / [ SO₂(g)] [NO₂(g)]

3.20 = 1.00*1.00 / 1.40 * (Y-1.00)

3.20 [1.40Y -1.40] = 1

4.48Y – 4.48 = 1

4.48Y = 1 + 4.48

Y = 1.22

So, the 1.22 moles of NO₂(g) that must be added to 2.40 mol of SO2(g) in order to form 1.00 mol of SO3(g) at equilibrium.