Question

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

Ratio of their partial pressures at $298\, K$ 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, $K_{P}=\frac{p_{(1-\text { butyne })}}{p_{(1,3-\text { butadiene })}}$
$\Delta G^{\circ}=-2.303\, RT\, \log\, K_{ P }$
$\Delta G_{( net )}^{\circ}=\Delta G ^{\circ}$($1$- butyne )$-\Delta G ^{\circ}$($1,3$- butadiene)
$=201.7-152.4=49.3\, kJ$
$\therefore \log\, K _{ p }=-\frac{\Delta G^{o}}{2.303\, RT}$
$=\frac{-49.3\, kJ }{2.303 \times 8.314 \times 10^{-3}\, kJ\, mol ^{-1}\, K^{-1}\, \times 298\, K}=-8.6402$
$\therefore K _{ P }=2.3 \times 10^{-9}$