Standard Gibb’s energy of reaction ( _ r G ^ ) at a certain temperature can be computed…
Chemistry · JEE Advanced · NTA Exams — Chemical Thermodynamics
Standard Gibb’s energy of reaction (\(\Delta_{\mathrm{r}} \mathrm{G}^{\circ}\)) at a certain temperature can be computed as \(\Delta_{r} G^{o}=\Delta_{r} H^{o}-T \cdot \Delta_{r} S^{o}\)and the change in the value of \(\Delta_{\mathrm{r}} \mathrm{H}^{\circ}\)and \(\Delta_{\mathrm{r}} \mathrm{~S}^{0}\)for a reaction with temperature can be computed as follow:
\[\begin{array}{l} \Delta_{\mathrm{r}} \mathrm{H}_{\mathrm{T}_{2}}^{\circ}-\Delta_{\mathrm{r}} \mathrm{H}_{\mathrm{T}_{1}}^{\circ}=\Delta_{\mathrm{r}} \mathrm{C}_{\mathrm{p}}^{\circ}\left(\mathrm{T}_{2}-\mathrm{T}_{1}\right) \\ \Delta_{\mathrm{r}} \mathrm{~S}_{\mathrm{T}_{2}}^{\circ}-\Delta_{\mathrm{r}} \mathrm{~S}_{\mathrm{T}_{1}}^{\circ}=\Delta_{\mathrm{r}} \mathrm{C}_{\mathrm{p}}^{\circ} \ln \left(\frac{\mathrm{T}_{2}}{\mathrm{~T}_{1}}\right) \end{array}\]
\[\begin{array}{l} \Delta_{\mathrm{r}} \mathrm{G}^{\circ}=\Delta_{\mathrm{r}} \mathrm{H}^{\circ}-\mathrm{T} \cdot \Delta_{\mathrm{r}} \mathrm{~S}^{\circ}\\ \text { and by } \Delta_{\mathrm{r}} \mathrm{G}^{\circ}=-\mathrm{RT} \ln \mathrm{~K}_{\mathrm{eq}} \text {. } \end{array}\]
Consider the following reaction:
\(\mathrm{CO}(\mathrm{~g})+2 \mathrm{H}_{2}(\mathrm{~g}) \rightleftharpoons \mathrm{CH}_{3} \mathrm{OH}(\mathrm{~g})\)
Given: \(\Delta_{f} \mathrm{H}^{\circ}\left(\mathrm{CH}_{3} \mathrm{OH}, \mathrm{~g}\right)\)= –201 kJ/mol; \(\Delta_{f} \mathrm{H}^{\circ}(\mathrm{CO}, \mathrm{~g})\)= –144 kJ/mol
So(CH3OH, g) = 240 J/K-mol; So(H2, g) = 29 JK–1mol–1
So(CO, g) = 198 J/mol-K; \(\mathrm{C}_{\mathrm{p}, \mathrm{~m}}^{o}\left(\mathrm{H}_{2}\right)\)=28.8 J/mol-K
\(\mathrm{C}_{\mathrm{p}, \mathrm{~m}}^{o}(\mathrm{CO})\)=29.4 J/mol-K; \(\mathrm{C}_{\mathrm{p}, \mathrm{~m}}^{\mathrm{o}}\left(\mathrm{CH}_{3} \mathrm{OH}\right)\) = 44 J/mol-K
and \(\ln \left(\frac{320}{300}\right)=0.06\), all data at 300 K
\(\Delta_{\mathrm{r}} \mathrm{H}^{\circ}\)at 300 K for the reaction is :
- –87 kJ/mol
- 87 kJ/mol
- –315 kJ/mol
- –288 kJ/mol
Answer
(A) –87 kJ/mol
Sign up free on Edukali to view the step-by-step worked solution and practice thousands of similar questions.
Related practice questions
- At 5 × 10 5 bar pressure, density of diamond and graphite are 3 g/cc and 2 g/cc respectively, at certain…
- Choose the reaction(s) from the following options, for which the standard enthalpy of reaction is equal to…
- Given are the following entropy values at 298 K and 1 atm: H 2 (g) = 130.6 J K –1 mol –1 ; Cl 2 (g) = 223.0 J…
- An ideal gas at constant temperature and pressure expands, then its
- For the process: H 2 O ( l ) → H 2 O ( g ) at T = 100 °C and 1 atmosphere pressure, the correct choice is
- What is the change in molar entropy of helium in the following process ? 1 He (298 K, 1 bar) → 1 He (100 K…
- An ideal gas initially at P 1 , V 1 is expanded to P 2 , V 2 and then compressed adiabatically to the same…
- Choose the reaction(s) from the following options, for which the standard enthalpy of reaction is equal to…
More Chemical Thermodynamics questions · Browse all practice questions