Question

$A\left(s\right)\rightleftharpoons B\left(g\right)+C\left(g\right)$ $K_p=40{atm}^2$
$X\left(s\right)\rightleftharpoons B\left(g\right)+E\left(g\right)$
Above equilibrium is allowed to attain in a closed container and pressure of $B$ was found to be $10atm$. Calculate standard Gibb's free energy change for $X\left(s\right)\rightleftharpoons B\left(g\right)+E\left(g\right)$ at $300K$ $(takeR=2cal/K/mol)$

• A. $3.5Kcal/mol$
• B. $3Kcal/mol$
• C. $2.5Kcal/mol$
• D. $2Kcal/mol$

Answer 1

The correct option is C $2.5Kcal/mol$

$A\left(s\right)\rightleftharpoons \underset{x+y}{B\left(g\right)}+\underset{x}{\left(g\right)}{K}_p=40atm$

$X\left(s\right)\rightleftharpoons \underset{y+x}{B\left(g\right)}+\underset{y}{E\left(g\right)}K=K_2$

$P_B=10$

$\therefore P_B\times P_C=40$

$\Rightarrow P_c=4atm$

$x+y=10$ $x=4$

$\Rightarrow y=6$

$\therefore K_2=(y+x)\times y$

$=10\times 6$

$=60$

$\therefore △G=-RTlnK$

$=-2.303\times 2\times 300\times \log 60$

$=-4.606\times 300\times (1+\log 6)$

$=-1381.8\times (1+0.78)$

$=-1381.8\times 1.78$

$=-2459.6cal$

$=\approx -2.5KCal$