Question

A certain reaction has value of $K_p=0.03$ at 100 K and ${\Delta }_r{H}^o=921.2calmo{l}^{-1}.$
Calculate the value of $K_p$ at 200K.

• A. 0.3
• B. 0.003
• C. 0.9
• D. 0.05

Answer 1

The correct option is A 0.3

Given, $T_1=100K{T}_2=200K$

${\Delta }_r{H}^o=921.2cal{K}_{p_1}=0.03$

The Van't Hoff equation is given as:
$\log \frac{K_{p_2}}{K_{p_1}}=\frac{\Delta H^o}{2.303R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$
putting the values we get,
$\log \frac{K_{p_2}}{0.03}=\frac{921.2}{2.303\times 2}\left[\frac{200-100}{200\times 100}\right]$
$\log \frac{K_{p_2}}{0.03}=\frac{921.2}{4.606}\times \frac{1}{200}=\frac{921.2}{921.2}$

$log\frac{K_{p_2}}{0.03}=1$
$\frac{K_{p_2}}{0.03}={10}^1$
$K_{p_2}=0.03\times 10=0.3$