Question

The equilibrium constant for the reaction given below is $2.0\times {10}^{-7}$ at $300$K. Calculate the standard entropy change for the reaction.
$(\Delta H^0=28.4kJmo{l}^{-1})$
$PC{l}_{5\left(g\right)}\rightleftharpoons PC{l}_{3\left(g\right)}+C{l}_{2\left(g\right)}$

• A. $-3.36Jmo{l}^{-1}$
• B. $-0.336Jmo{l}^{-1}$
• C. $-33.6Jmo{l}^{-1}$
• D. $-336Jmo{l}^{-1}$

Answer 1

Answer: (B)

$-0.336Jmo{l}^{-1}$

Given,

$T=300K$ $\Delta H=28.4$ KJ/mol

$K=2\times {10}^{-7}$

We have,

${\Delta G}^0=-RTlnK$

$=-8.314\times 300\times ln(2\times {10}^{-7})$

$=-38472.90J$

${\Delta G}^0=-38.47KJ/mol$

and now, we have

${\Delta G}^0=\Delta H-T\Delta S$

$-38.47=28.4-300\times \Delta S$

$\Delta S=\frac{-38.47-28.4}{-300}$

$\Delta S=0.33\frac{KJ}{molK}$

Ans :- Option B.