Question

For two I order reaction:
$A\to B;K_1={10}^{15}\times e^{-\frac{2\times {10}^3}{T}}$
$C\to D;K_2={10}^{14}\times e^{-\frac{{10}^3}{T}}$
The temperature at which both have same rate if $\left[A\right]=\left[C\right]$ at $t=0$:

• A. ${707.2}^{o}C$
• B. $434.2K$
• C. ${727}^{o}C$
• D. $707.2K$

Answer 1

Answer: (B)

$434.2K$

As we know,
$r_1=K_1\left[A\right]$
$r_2=K_2\left[C\right]$
$\because r_1=r_2$ and $\left[A\right]=\left[C\right]$ (at $t=0$)
$\therefore K_1=K_2$
$lo{g}_{10}{K}_1=15-\frac{2000}{2.303T}$
$lo{g}_{10}{K}_2=14-\frac{2000}{2.303T}$
$15-\frac{2000}{2.303T}=14-\frac{1000}{2.303T}$
$T=434.22K$