Question

37.Enter your answer in the provided box. A quantity of 9.50 g of SO2Cl2 was placed in a 2.00 L flask. At 648 K, there is 0.0345 mole of SO2 present. Calculate Kc for the reaction. SO₂Cl₂(g) ⇆ SO₂(g) + Cl₂(g)

Answer 1

initial,
[SO2Cl2] = mass/(molar mass * volume)
= 9.50 g / (135 g/mol * 2 L)
=0.0352 M

at equilibrium,
[SO2] = number of mol / volume
= 0.0345 mol / 2 L
= 0.01725 M

SO2Cl2(g) <-----> SO2(g) + Cl2(g)
0.0352              0        0         (initial)
0.0352-x            x        x         (at equilibrium)

[SO2] = x = 0.01725 M

Kc = [SO2] [Cl2] / [SO2Cl2]
Kc = x*x/(0.0352-x)
= (0.01725 * 0.01725) / (0.0352 - 0.01725)
= 0.0166

Answer: 0.0166