Δ H and Δ S for the reaction, Ag_2 O(s) 2 Ag(s)+(1)/(2) O_2(~g) , are 30.56 ~kJ ~mol^ -1 and 66.00 JK^ -1 ~mol^ -1 respectively. The temperature at which the free energy change for the reaction will be zero, is

Correct answer: B. 463 K From the Gibb's - Helmholtz equation, Δ G=Δ H-T Δ S At equilibrium, Δ G=0 Hence, Δ H-T Δ S=0 or, Δ H=T Δ S Or, T=(Δ H)/(Δ S) Given that, Δ H=30.56 ~kJ ~mol^ -1 aligned & =30560 ~J ~mol^ -1 Δ S & =66.0 ~J ~K^ -1 ~mol^ -1 T & =(30560)/(66.0)=463 ~K aligned

JIPMER 2015

Exam: JIPMER
Year: 2015
Subject: Chemistry
Chapter: Hydrocarbons

JIPMER 2015 · Chemistry · Hydrocarbons

Δ H and Δ S for the reaction, Ag_2 O(s) 2 Ag(s)+(1)/(2) O_2(~g) ,...

\Delta H

and

\Delta S

for the reaction,

\mathrm{Ag}_2 \mathrm{O}(\mathrm{s}) \longrightarrow 2 \mathrm{Ag}(s)+\frac{1}{2} \mathrm{O}_2(\mathrm{~g})

, are

30.56 \mathrm{~kJ} \mathrm{~mol}^{-1}

and

66.00 \mathrm{JK}^{-1} \mathrm{~mol}^{-1}

respectively. The temperature at which the free energy change for the reaction will be zero, is

Options

A3528 K
B463 K
C73 K
D144 K

Correct answer

B. 463 K

Step-by-step solution

From the Gibb's - Helmholtz equation,

\Delta G=\Delta H-T \Delta S

At equilibrium,

\Delta G=0

Hence,

\quad \Delta H-T \Delta S=0

or,

\Delta H=T \Delta S

Or,

T=\frac{\Delta H}{\Delta S}

Given that,

\Delta H=30.56 \mathrm{~kJ} \mathrm{~mol}^{-1}

\begin{aligned} & =30560 \mathrm{~J} \mathrm{~mol}^{-1} \\ \therefore \quad \Delta S & =66.0 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1} \\ \therefore \quad T & =\frac{30560}{66.0}=463 \mathrm{~K}\end{aligned}

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