Question

$1,3$ -butadiene and $1$-butyne exist in equilibrium in gaseous phase at $298K$.

Ratio of their partial pressures at $298K$ is

• A. $2\times 10^{-9}$
• B. $1\times 10^{-9}$
• C. $8\times 10^{-7}$
• D. $8\times 10^{-6}$

Answer 1

Answer: (A)

We know, $K_P=\frac{p_{(1-\text{ butyne })}}{p_{(1,3-\text{ butadiene })}}$
$\Delta G^{\circ }=-2.303RT\log K_P$
$\Delta G_{(net)}^{\circ }=\Delta G^{\circ }$($1$- butyne )$-\Delta G^{\circ }$($1,3$- butadiene)
$=201.7-152.4=49.3kJ$
$\therefore \log K_p=-\frac{\Delta G^o}{2.303RT}$
$=\frac{-49.3kJ}{2.303\times 8.314\times 10^{-3}kJmo{l}^{-1}{K}^{-1}\times 298K}=-8.6402$
$\therefore K_P=2.3\times 10^{-9}$