For the reaction, X(s) leftharpoons Y(s)+Z(g) , the plot of p_ Z p^ θ versus 10^ 4 T is given below (in solid line), where p_ Z is the pressure (in bar) of the gas Z at temperature T and p^ θ =1 bar. (Given, d( K)

Correct answer: 0 According to graph when T = 1000K and 10 ^ 4 T = 10 then ln p _ z p ^ ϕ = - 3 Now, Δ G ^ o = Δ H ^ o - T Δ S ^ o - RT lnK _ p = Δ H ^ o - T Δ S ^ o Rln K _ p = Δ S ^ o - Δ H ^ o T Δ S ^ o = RlnK _ p + Δ H ^ o T lnK _ p = ln p _ z p ^ ϕ = - 3 Δ S ^ o = - 3R + 166 .28 × 10 ^ 3 J 1000 166 .28 - 3 × 8 .314 Δ S ^ o = 141 .34

JEE Advanced 2021Paper 1

Exam: JEE Advanced
Year: 2021
Paper: Paper 1
Subject: Chemistry
Chapter: Thermodynamics (C)

JEE Advanced 2021 · Paper 1 · Chemistry · Thermodynamics (C)

For the reaction, X(s) leftharpoons Y(s)+Z(g) , the plot of p_ Z p^...

For the reaction,

\mathbf{X}(s) \rightleftharpoons \mathbf{Y}(s)+\mathbf{Z}(g)

, the plot of

\ln \frac{p_{\mathbf{Z}}}{p^{\theta}}

versus

\frac{10^{4}}{T}

is given below (in solid line), where

p_{\mathbf{Z}}

is the pressure (in bar) of the gas

\mathbf{Z}

at temperature

T

and

p^{\theta}=1

bar.

(Given,

\frac{\mathrm{d}(\ln K)}{\mathrm{d}\left(\frac{1}{T}\right)}=-\frac{\Delta H^{\theta}}{R}

, where the equilibrium constant,

K=\frac{p_{z}}{p^{\theta}}

and the gas constant,

\mathrm{R}=8.314 \mathrm{~J} \mathrm{~K}^{-1} \mathrm{~mol}^{-1}

)

The value of $∆ {S}^{o}$ (in $kJ{mol}^{- 1}$ ) for the given reaction, at $1000K$ is ___.

Correct answer

Step-by-step solution

According to graph when

T = 1000K

and

\dfrac{{10}^{4}}{T} = 10

then

ln\dfrac{{p}_{z}}{{p}^{ϕ}} = - 3

Now, ${\Delta G}^{o} = {\Delta H}^{o} - T{\Delta S}^{o}$

$- RT{lnK}_{p} = {\Delta H}^{o} - T{\Delta S}^{o}$

Rln{K}_{p} = {\Delta S}^{o} - \dfrac{{\Delta H}^{o}}{T} $$ {\Delta S}^{o} = {RlnK}_{p} + \dfrac{{\Delta H}^{o}}{T} $$ {lnK}_{p} = ln\dfrac{{p}_{z}}{{p}^{ϕ}} = - 3 $$ {\Delta S}^{o} = - 3R + \dfrac{166 .28 \times {10}^{3}J}{1000} $$ 166 .28 - 3 \times 8 .314 $$ {\Delta S}^{o} = 141 .34

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