Question

3.25 moles of NOCl were placed in a 2.50 L reaction chamber at 427°C. After equilibrium was reached, 1.10 moles of NOCl remained. Calculate the equilibrium constant, Kc, for the reaction 2NOCl(g)⇋2NO(g)+Cl2(g).

Answer 1

Step 1

2NOCl(g) ⇋ 2NO(g) + Cl2(g) (balanced)

Initial 1.3 M 0 0

Reacts -2x 2x x

Equilibrium 1.3-2x 2x x

Equilibrium 0.44 M 2x0.43= 0.86M 0.43M

Concentrations:

Volume of the chamber = 2.50 L

[NOCl] initial = 3.25 moles/2.50 L = 1.3 M

After equilibrium was reached, 1.10 moles of NOCl remained. Therefore,

1.3 - 2x = 1.10 moles/2.50 L = 0.44 M

(1.3 - 0.44)/2 = x = 0.43 M

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Step 2

`Kc\text{ = }(\lbrack Cl\rbrack\lbrack NO\rbrack^2)/(\lbrack NOCl\rbrack^2)=((0.43M)(0.86M)^2)/((0.44M)^2)=1.642`

Answer: Kc = 1.642