At 400 K, the rate of decomposition of a gaseous compound initially at a pressure of 12.6 kPa, was 9.71 Pa s-1 when 10.0 per cent had reacted and 7.67 Pa s-1 when 20.0 percent had reacted. Determine the order of the reaction.

The order of the reaction with respect to the gas = 2

Explanation:

Let the original gas pressure be [G₀]

Initial rate of reaction is given as

r = k [G₀]ⁿ

When 10% had reacted, amount of gas left = [0.9G₀], r = 9.71 Pa/s

r = k [0.9G₀]ⁿ = 9.71 (eqn 1)

when 20% had reacted, amount of gas left = [0.8G₀], r = 7.67 Pa/s

r = k [0.8G₀]ⁿ = 7.67 (eqn 2)

Dividing (eqn 1) by (eqn 2)

(9.71/7.67) = [0.9/0.8]ⁿ

1.266 = 1.125ⁿ

1.125ⁿ = 1.266

Take natural logarithms of both sides

n (In 1.125) = In 1.266

n = 0.236/0.118

n = 2.

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