Question

The reaction 2HCl(g) H2(g) + Cl2(g) has Kc = 3.2 × 10–34 at 25 °C. If a reaction vessel contains initially 0.0500 mol L–1 of HCl and then reacts to reach equilibrium, what will be the concentrations of H2 and Cl2?

Answer 1

2HCl(g) ---------> H2(g) + Cl2(g)

Initially 0.05 0 0

At equ. 0.05-2x x x

From the equation we can see we have equal concentration of Hydrogen and chlorine at equilbrium.The amount of hydrogen and chlorine will be half of the disassociated HCl

So now,

K_{c =} [H2][Cl2] / [HCl]²

So, 3.2 *10^{-34 =} x²/ (0.05-2x)²

Now since the value on Left side is very small , So , we can neglect x as it should be very small

Hence

3.2 *10^{-34 =} x²/0.0025

0.008 * 10^{-34 =} x²

x = 8.9 * 10^-19

So concentration of Cl2 = 8.9 * 10^-19 mole/Litre

So concentration of H2 = 8.9 * 10^-19 mole/Litre