For the reaction $\mathbf{X}(\mathrm{s}) \rightleftharpoons \mathbf{Y}(\mathrm{s})+\mathbf{Z}(\mathrm{g})$, the plot of $\ln \frac{\mathrm{p}_{\mathrm{z}}}{\mathrm{p}^{\ominus}}$ versus $\frac{10^{4}}{T}$ is given below (in solid line), where $\mathrm{p}_{\mathrm{z}}$ is the pressure (in bar) of the gas $\mathbf{Z}$ at temperature $T$ and $\mathrm{P}^{\theta}=1$ bar. (Given, $\frac{\mathrm{d}(\ln \mathrm{K})}{\mathrm{d}\left(\frac{1}{T}\right)}=-\frac{\Delta H^{\ominus}}{\mathrm{R}}$, where the equilibrium constant, $\mathrm{K}=\frac{\mathrm{p}_{\mathrm{z}}}{\mathrm{p}^{\ominus}}$ and the gas constant, $\mathrm{R}=8.314$ $\mathrm{J~K}^{-1}~\mathrm{mol}^{-1}$) The value of standard enthalpy, $\Delta \mathrm{H}^{\ominus}$ (in $\mathrm{kJ~mol}^{-1}$) for the reaction is ____.

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Chemical Thermodynamics

For the reaction

X (s) ⇌ Y (s) + Z (g)

, the plot of

ln \frac{p _{z}}{p ^{⊖}}

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 ^{⊖}}{R}

, where the equilibrium constant,

K = \frac{p _{z}}{p ^{⊖}}

and the gas constant,

R = 8.314

J K^{- 1} mol^{- 1}

)

The value of standard enthalpy,

Δ H^{⊖}

(in

kJ mol^{- 1}

) for the reaction is ____.

Answer: 166.28

From the van't Hoff equation:

ln K = - \frac{Δ H ^{⊖}}{R T} + \frac{Δ S ^{⊖}}{R}

The slope of the plot of

ln (p_{z} / p^{⊖})

1 0^{4} / T

is:
Slope =

- \frac{Δ H ^{⊖}}{R} \times 1 0^{- 4}

From the graph, the slope =

\frac{- 4 - 0}{2 - 0} \times 1 0^{4} = - 2 \times 1 0^{4}

Therefore:

- \frac{Δ H ^{⊖}}{R} = - 2 \times 1 0^{4}

Δ H^{⊖} = 2 \times 1 0^{4} \times R = 2 \times 1 0^{4} \times 8.314 = 166280

J/mol = 166.28 kJ/mol

JEE Advanced · 2021 · 03 Oct 2021 · Paper 1 · Numerical answer