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 Slop g the Given graph dln p _ z p ^ ϕ d( 10 ^ 4 T ) = (- 7 + 3)/(12 - 10) = - 2 dln p _ z p ^ ϕ d(1)/(T) - 2 × 10 ^ 4 = - Δ H ^ o R Δ H ^ o = 2 × 10 ^ 4 × R = 2 × 8 .314 × 10kJ / mol = 166 . 28

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 standard enthalpy, $∆ {H}^{o}$ (in $kJ{mol}^{- 1}$ ) for the given reaction is ___.

Correct answer

Step-by-step solution

Slop

g

the Given graph

$\dfrac{dln\dfrac{{p}_{z}}{{p}^{ϕ}}}{d\left(\dfrac{{10}^{4}}{T}\right)} = \dfrac{- 7 + 3}{12 - 10} = - 2$

$\dfrac{dln\dfrac{{p}_{z}}{{p}^{ϕ}}}{d\dfrac{1}{T}} - 2 \times {10}^{4} = - \dfrac{{\Delta H}^{o}}{R}$

${\Delta H}^{o} = 2 \times {10}^{4} \times R$

$= 2 \times 8 .314 \times 10kJ / mol$

$= 166 . 28$

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