Go To Your Study Dashboard

Get Answers to all your Questions

header-bg

Find the equilibrium constant for the reaction given at $\mathrm{350\; K}$ if $\mathrm{\Delta H^{0}=75.3\; kJ\; mol^{-1}}$ and $\mathrm{\Delta S^{0}=120\; J\; K^{-1}\; mol^{-1}}$ .

$\mathrm{PCl_{5}\left ( g \right )\rightarrow PCl_{3}\left ( g \right )+Cl_{2}(g)}$
Option: 1
$1.958\times 10^{-4}$

Option: 2
$-1.958\times 10^{-4}$

Option: 3
$1.074\times 10^{-5}$

Option: 4
$-1.074\times 10^{5}$

Answers (1)

best_answer

As we have learnt,

$\begin{array}{l}{\text { Relationship between } \Delta \mathbf{G}^{\circ} \text { and Equilibrium }} {\text { constant }\left(\mathbf{K}_{\mathrm{eq}}\right)}\end{array}$

$\mathrm{\Delta G^0 = -RTlnK_{eq}}$

We know at standard conditions

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

Given data:

$\mathrm{\Delta H^{0}=75.3\times 10^{3}J\; mol^{-1},\; \; T=350\; K}$

$\mathrm{\Delta S^{0}=120\; J\; K^{-1}mol^{-1},\; \; T=350\; K}$

So, $\mathrm{\Delta G^{0}=75300-350\times 120=33300}$

Now,

$\mathrm{\Delta G^{0}=-2.303\; RT\; log\; K_{p}}$

$\mathrm{-33300=2.303\times 8.314\times 350\times log\; K_{p}}$

Hence, $\mathrm{K_{p}=1.074\times 10^{-5}}$

Therefore, Option (3) is correct.

Posted by

Ritika Jonwal

View full answer

Similar Questions

0.1 M acetic acid solution having pH =3 is titrated against 0.05 M NaOH solution. Calculate pH at 1/4th stage of neutralization of acid.Option: 1

0.25×60+0.10x=0.15×(60+x)

Latest Question