D - G - S reaction - dT - G - H - T - S - G - H - T d - G - dT PCalculate - S for the given cell reaction in the previous question- A- -73-53 JK -1 mol -1- 83-53 JK -1 mol -1C- 100 JK -1 mol -1D- None of these

The correct option is A 73.53 JK^ 1mol^ 1d(Δ G) = Δ H TΔ S Δ G = nFE 2× 96500 × 1.015 = 21800 300 Δ S 195895 + 21800 = 300 Δ S 22105/300 = Δ S Δ S = ...

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Byju's Answer

Standard XII

Chemistry

Gibb's Free Energy

d Δ G =-Δ S r...

Question

$d (Δ G) = - (Δ S_{reaction}) . d T$ $Δ G = Δ H - T Δ S$ $Δ G = Δ H + T {(\frac{d (Δ G)}{d T})}_{P}$
Calculate $Δ S$ for the given cell reaction in the previous question.

- 73.53 {JK}^{- 1} {mol}^{- 1}

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83.53 {JK}^{- 1} {mol}^{- 1}

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100 {JK}^{- 1} {mol}^{- 1}

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None of these

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Solution

The correct option is A $- 73.53 {JK}^{- 1} {mol}^{- 1}$
$d (Δ G) = Δ H - T Δ S$

$Δ G = - n F E$

$- 2 \times 96500 \times 1.015 = - 21800 - 300 Δ S$

$- 195895 + 21800 = - 300 Δ S$

$\frac{- 22105}{300} = Δ S$

$Δ S = - 73.53 J K^{- 1} m o l^{- 1}$

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Similar questions

d (Δ G) = - (Δ S_{reaction}) . d T

Δ G = Δ H - T Δ S

Δ G = Δ H + T {(\frac{d (Δ G)}{d T})}_{P}

Calculate

Δ S

for the given cell reaction in the previous question.

Q. The cell potential

(E_{c e l l})

of a reaction is related as

Δ G = n F E_{c e l l}

, where

Δ G

represents maximum useful electrical work.
n = no. of moles of electrons exchanged during the reaction.
For reversible cell reaction

d (Δ G) = (Δ V) d p - (Δ S) d T

At const. P,

d (Δ G) = - (Δ S) d T

At const. P,

Δ G = Δ H - T Δ S

..(1)

∴ Δ G = Δ H + T {(\frac{d (Δ G)}{Δ T})}_{P}

. (2)

{(\frac{d E_{c e l l}}{d T})}_{p}

is known as temperature coefficient of the emf of the cell.

The

Δ S

for the cell reaction

Z n + 2 A g C l \to Z n C l_{2} + 2 A g

d (Δ G) = - (Δ S_{reaction}) . d T

Δ G = Δ H - T Δ S

Δ G = Δ H + T {(\frac{d (Δ G)}{d T})}_{P}

The temperature coefficient of the e.m.f of a cell,

{(\frac{d E}{d T})}_{P}

is given by:

d (Δ G) = - (Δ S_{reaction}) . d T

Δ G = Δ H - T Δ S

Δ G = Δ H + T {(\frac{d (Δ G)}{d T})}_{P}

300 K, Δ H

for the reaction,

Z n_{(S)} + 2 A g C l_{(S)} \to Z n C l_{2_{(a q)}} + 2 A g_{(S)}

- 218 kJ/mol

while the e.m.f of the cell is

1.015 V

{(\frac{d E}{d T})}_{P}

of the cell is:

d (Δ G) = - (Δ S_{reaction}) . d T

Δ G = Δ H - T Δ S

Δ G = Δ H + T {(\frac{d (Δ G)}{d T})}_{P}

The temperature coefficient of the e.m.f of a cell,

{(\frac{d E}{d T})}_{P}

is given by:

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d(ΔG)=−(ΔSreaction).dT ΔG=ΔH−TΔS ΔG=ΔH+T(d(ΔG)dT)P Calculate ΔS for the given cell reaction in the previous question.

The correct option is A −73.53 JK−1mol−1d(ΔG) = ΔH − TΔS ΔG = −nFE −2× 96500 × 1.015 = −21800 − 300 ΔS −195895 + 21800 = −300 ΔS −22105300 = ΔS ΔS = −73.53JK−1mol−1

$d (Δ G) = - (Δ S_{reaction}) . d T$ $Δ G = Δ H - T Δ S$ $Δ G = Δ H + T {(\frac{d (Δ G)}{d T})}_{P}$
Calculate $Δ S$ for the given cell reaction in the previous question.

The correct option is A $- 73.53 {JK}^{- 1} {mol}^{- 1}$
$d (Δ G) = Δ H - T Δ S$

$Δ G = - n F E$

$- 2 \times 96500 \times 1.015 = - 21800 - 300 Δ S$

$- 195895 + 21800 = - 300 Δ S$

$\frac{- 22105}{300} = Δ S$

$Δ S = - 73.53 J K^{- 1} m o l^{- 1}$