Question

Consider two $I^{st}$ order reations at 25${}^o$C with same initial concentrations of 1M.

I . $A\stackrel{t_{1/2}=40min}{\to }$ product (temperature coefficient = 2)
II . $B\stackrel{t_{1/2}=30min}{\to }$ product (temperature coefficient = 3)

If both reactions are carried out at 35${}^o$C. Find the ratio of concentrations of A & B after one hour

Answer 1

We know

$\frac{K_{t+10}}{K_t}=$ Temperature cofficient

I. $\frac{K_{35}}{K_{25}}=2\Rightarrow K_{35}=2\times \frac{ln2}{40}$

II. $\frac{K_{35}}{K_{25}}=3\Rightarrow K_{35}=3\times \frac{ln2}{30}$

${}^{t_{1/2}}{|}_{{35}^o}=20min$ ----- 1

${}^{t_{1/2}}{|}_{{35}^o}=10min$ ------ 2

No. of half-lives of A $=3$

No. of half-lives of B $=6$

$\therefore$ fraction of A left $=$ (1/2)${}^3$

$\therefore$ fraction of B left $=$ (1/2)${}^6$

Hence, $\frac{A_t}{B_t}=\frac{{\left(\frac{1}{2}\right)}^3}{{\left(\frac{1}{2}\right)}^6}=2^3=8$

The ratio of concentrations of A & B after one hour is 8.