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Chemistry - Thermodynamics - II Question with Solution | TestHub

ChemistryThermodynamics - IIFree Energy and EquilibriumHard2 minPYQ_2021

ChemistryHardnumerical

Passage / Comprehension

For the reaction, $X (s) ⇌ Y (s) + Z (g)$ , the plot of $ln \frac{p _{Z}}{p ^{θ}}$ versus $\frac{1 0 ^{4}}{T}$ 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, $\frac{d ( l n K )}{d ( \frac{1}{T} )} = - \frac{Δ H ^{θ}}{R}$ , where the equilibrium constant, $K = \frac{p _{z}}{p ^{θ}}$ and the gas constant, $R = 8.314 J K^{- 1} mol^{- 1}$ )

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

Question diagram: The value of ∆ S o (in kJ mol - 1 ) for the given reaction,

Answer:

141.34

Solution:

According to graph when $T = 1000 K$ and $\frac{10^{4}}{T} = 10$ then $\ln \frac{p_{z}}{p^{ϕ}} = - 3$

Now, ${ΔG}^{o} = {ΔH}^{o} - T {ΔS}^{o}$

$- RT {lnK}_{p} = {ΔH}^{o} - T {ΔS}^{o}$

$R \ln K_{p} = {ΔS}^{o} - \frac{{ΔH}^{o}}{T}$

${ΔS}^{o} = {RlnK}_{p} + \frac{{ΔH}^{o}}{T}$

${lnK}_{p} = \ln \frac{p_{z}}{p^{ϕ}} = - 3$

${ΔS}^{o} = - 3 R + \frac{166 .28 \times 10^{3} J}{1000}$

$166 .28 - 3 \times 8 .314$

${ΔS}^{o} = 141 .34$

Stream:JEE_ADVSubject:ChemistryTopic:Thermodynamics - IISubtopic:Free Energy and Equilibrium

⏱ 2mℹ️ Source: PYQ_2021

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