Answer:
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.
Given the statements in the question, the order of reaction is 2.
Given that the initial pressure of the gas is Po
Since; r = kPo^n
n = order of reaction
Po = initial pressure
k = rate constant
9.71 Pa s-1 = k(0.9 × 12.6 kPa)^n -------- (1)
7.67 Pa s-1 = k(0.8 × 12.6 kPa)^n -------- (2)
If we divide equation (1) by (2)
9.71 Pa s-1/7.67 Pa s-1 = k(0.9 × 12.6 kPa)^n/k(0.8 × 12.6 kPa)^n
1.27 = 1.1125^n
Taking natural logarithm of both sides;
ln(1.27) = n ln1.125
n = ln(1.27)/ln1.125
n = 2
The order of reaction is 2.
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