Question

${\Delta }_f{G}^{}$ at 500 K for substance $S$ in liquid state and gaseous state are +100.7 kcal ${}^{-1}$ and +103 kcal ${}^{-1}$, respectively. Vapour pressure of liquid $S$ at 500 K is approximately equal to:
($R=2cal$ $K^{-1}{}^{-1}$)

• A. 0.1 atm
• B. 10 atm
• C. 100 atm
• D. 1 atm

Answer 1

The correct option is B 10 atm

Solution:- (B) $10atm$

As we know that, at equilibrium,

${\Delta G}^{\circ }=-2.303RT\log K_p.....\left(1\right)$

$S_{\left(l\right)}\rightleftharpoons S_{\left(g\right)}$

From the above reaction,

${\Delta G}^{\circ }={{\Delta G}_f}_{\left(\text{product}\right)}-{{\Delta G}^{\circ }}_{\left(\text{reactant}\right)}$

$\Rightarrow {\Delta G}^{\circ }=103-100.7=2.3kcal/mol=2.3\times {10}^{3}kcal/mol$

Given:-

$T=500K$

$R=2cal/mol-K$

From ${eq}^n\left(1\right)$, we have

$2.3\times {10}^3=-2.303\times 2\times 500\times \log K_p$

$\Rightarrow \log K_p=-1$

$\Rightarrow K_p={10}^{-1}atm$

Now, from the above reaction,

$K_p=\frac{1}{P_{S_{\left(l\right)}}}$

$\Rightarrow P_{S_{\left(l\right)}}=\frac{1}{{10}^{-1}}=10atm$

Hence the vapou pressure of liquid $S$ is approximately equal to $10atm$.