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Question
In a reaction between A and B, the initial rate of reaction (r0) was measured for different initial concentrations of A and B as given below:
| A/mol L−1 | 0.20 | 0.20 | 0.40 |
| B/mol L−1 | 0.30 | 0.10 | 0.05 |
| r0/mol L−1 s−1 | 5.07 × 10−5 | 5.07 × 10−5 | 1.43 × 10−4 |
What is the order of the reaction with respect to A and B?
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Solution
Suppose the order of the reaction with respect to A is m and with respect to B is n.
∴ Rate (r0) = k[A]m [B]n
For the given data, we have,
Rate (r1) = 5.07 × 10−5
= k (0.20)m × (0.30)n ...(i)
Rate (r2) = 5.07 × 10−5
= k (0.20)m × (0.10)n ...(ii)
Rate (r3) = 1.43 × 10−4
= k (0.40)m × (0.05)n ...(iii)
Dividing eq. (i) by eq. (ii), we have,
`(5.07 xx 10^-5)/(5.07 xx 10^-5) = (k (0.20)^m xx (0.30)^n)/(k (0.20)^m xx (0.10)^n)`
⇒ 1 = (3)n
⇒ n = 0
Dividing eq. (iii) by eq. (ii) and putting the value of n, we get,
`(1.43 xx 10^-4)/(5.07 xx 10^-5) = (k (0.40)^m xx (0.05)^0)/(k (0.20)^m xx (0.10)^0)`
⇒ `(1.43 xx 10^-4)/(5.07 xx 10^-5)` = 2m
⇒ (2)m = 2.82
⇒ m log10 2 = log10 2.82 = 0.45
⇒ m = `0.45/(log_10 2)`
= `0.45/0.3010`
= 1.49
= 1.50
Hence, the order with respect to A is 1.5, and the order with respect to B is 0.
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