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JEE Advanced · Chemistry · 6. Thermodynamics (C)

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, Ho (in kJ mol-1) for the given reaction is ___.

  1. A 166.28
  2. B 166.25
  3. C 166.01
  4. D 152.25
Verified Solution

Answer & Solution

Correct Answer

(A) 166.28

Step-by-step Solution

Detailed explanation

Slop g the Given graph d lnpzpϕd 104T=7+31210=2 d lnpzpϕd1T2×104=ΔHoR
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