Question

A certain reaction has value of $K_p=0.0260$ at ${25}^{o}C$ and ${\Delta }_{rxn}{H}^o=32.4kJmo{l}^{-1}$. Calculate the value of $K_p$ at ${57}^{o}C$.
antilog 0.55 = 3.55

• A. 0.1431
• B. 0.0924
• C. 0.2431
• D. 0.0431

Answer 1

The correct option is B 0.0924

Given, $T_1={25}^{o}C=25+273=298K,$ $T_2={57}^{o}C=57+273=330K,$ ${\Delta }_{rxn}{H}^o=32.4kJmo{l}^{-1}=32400kJmo{l}^{-1}$, $K_{p_1}=0.0260$
we know the van't Hoff equation is given by $log\frac{K_{p_2}}{K_{p_1}}=\frac{△H}{2.303R}\left[\frac{T_2-T_1}{T_1{T}_2}\right]$
putting the values we get,
$log\frac{K_{p_2}}{0.0260}=\frac{32400}{2.303\times 8.314}\left[\frac{330-298}{298\times 330}\right]$
$log\frac{K_{p_2}}{0.0260}=\frac{32400}{19.147}\left[\frac{32}{92380}\right]$
$log\frac{K_{p_2}}{0.0260}=0.2198$
$K_{p_2}=0.0924$