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Question
For the reaction A + B → P.
If [B] is doubled at constant [A], the rate of reaction doubled. If [A] is triple and [B] is doubled, the rate of reaction increases by a factor of 6. Calculate the rate law equation.
Solution
Given:
A + B → P
∴ Rate of reaction,
rate = k[A]x [B]y ...(i)
If [B] is doubled at constant [A], the rate of reaction doubled.
∴ 2 × rate = k[A]x [2B]y
∴ 2 × rate = k[A]x 2y [2B]y ...(ii)
If [A] is tripled and [B] is doubled, the rate of reaction increased by a factor of 6.
∴ 6 × rate = k[3A]x [2B]y
6 × rate = 3x k[3A]x 2y [B]y ...(iii)
Divide equation (ii) by (i), we get
`(2 xx "rate")/"rate" = (k[A]^x 2^y[B]^y)/(k[A]^x [B]^y)`
21 = 2y
∴ y = 1
Now divide equation (iii) by (i), we get
`(6 xx "rate")/"rate" = (3^x k[A]^x 2^y [B]^y)/(k[A]^x [B]^y)`
6 = 3x2y
= 3x × 2 ...(∵ y = 1)
∴ x = 1
Now, x = 1 and y = 1 values should be written in equation (i)
∴ Rate law can be written as:
rate = k[A]1 [B]1
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