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].