Q58 · CSIR NET Chemistry, December 2017
Paper: CSIR NET December 2017 · Subject: Physical Chemistry · Chapter: Chemical Kinetics · Topic: Rate Laws Order · Marks: 2 · Difficulty: Medium
Reaction between A and B is carried out for different initial concentrations and the corresponding half-life times are measured. The data listed in the table:
| Entry | $\left[\mathrm{A}_{0}\right](\mu \mathrm{M})$ | $\left[\mathrm{B}_{0}\right](\mu \mathrm{M})$ | $\mathrm{t}_{1/2}(\mathrm{sec})$ |
|---|---|---|---|
| 1 | 500 | 10 | 60 |
| 2 | 500 | 20 | 60 |
| 3 | 10 | 500 | 60 |
| 4 | 20 | 500 | 30 |
(a)$\mathrm{k}[\mathrm{A}][\mathrm{B}]^{2}$
(b)$\mathrm{k}[\mathrm{A}]^{2}$
(c)$\mathrm{k}[\mathrm{A}]^{2}[\mathrm{B}]$
(d)$\mathrm{k}[\mathrm{A}][\mathrm{B}]^{2}$
Answer
Answer: C ✓ checked by 4AB · confidence high
The source book printed no answer; this one was worked out and checked — see the explanation.
Explanation
With A in excess, doubling [B]₀ leaves t₁/₂ unchanged — first order in B. With B in excess, doubling [A]₀ halves t₁/₂ — second order in A. Rate = k[A]²[B].
Study loop for Chemical Kinetics