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प्रश्न
Consider the data given below for the reaction:
\[\ce{A + B-> Product}\]
| S. No. | conc. of [A] mol L-1 | conc. of [B] mol L-1 | Rate; mol L-1 sec-1 |
| 1. | 0.1 | 0.1 | 4.0 × 10-4 |
| 2. | 0.2 | 0.2 | 1.6 × 10-3 |
| 3. | 0.5 | 0.1 | 1.0 × 10-2 |
| 4. | 0.5 | 0.5 | 1.0 × 10-2 |
Answer the following questions.
- What is the order of reaction with respect to A and B?
- Calculate the rate constant.
- Determine the reaction rate when the concentration of A and B are 0.2 mol L-1 sec-1 and 0.35 mol L-1 sec-1 respectively.
उत्तर
(i) Order with respect to [A] is 2 and with respect to [B] is 0.
(ii) Using any of the experimental data, we can now calculate the rate constant.
`"Rate" = "k"["A"]^2["B"]^0`
`4.0 xx 10^(-4) = "k"[0.1]^2`
`4.0 xx 10^(-4) = "k"[0.01]`
`"k" = (4.0 xx 10^(-4))/0.01`
k = 4.0 × 10-2 mol-1 Ls-1
∴ Rate constant k is 4.0 × 10-2 mol-1 Ls-1
(iii) To determine the reaction rate when the concentration of A is 0.2 mol L-1 and the concentration of B is 0.35 mol L-1 we can use,
The rate law for the reaction:
`"Rate" = "k"[A]^2[B]^0`
Substituting the given concentration of A into the rate law:
Rate = k[A]2
`"Rate" = (4.0 xx 10^(-2))(0.2)^2`4.0
`"Rate" = (4.0 xx 0^(-2))(0.04)`
Rate = 1.6 × 10-3 mol L-1 s-1
∴ reaction rate when the concentration of A is 0.2 mol L-1 and the concentration of B is 0.35 mol L-1 is 1.6 × 10-3 mol L-1 s-1
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\[\ce{A + B -> Product}\]
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