Top 50 Most Repeated THERMODYNAMICS PYQs | JEE MAINS
A curated collection of the most important questions from THERMODYNAMICS, fully solved with step-by-step concepts to prepare for JEE MAINS.
A curated collection of the most important questions from THERMODYNAMICS, fully solved with step-by-step concepts to prepare for JEE MAINS.
Use Hess's law: $\Delta H = [\Delta H_f(CO_2)] - [2\times\Delta H_f(CO)] = -393.5 - 2(-110.5) = -172.5$ kJ....
Read Full Step-by-Step Solution →ΔG° = ΔH° − TΔS° = 30000 − 298×40 = 18080 J/mol ≈ 18 kJ/mol....
Read Full Step-by-Step Solution →State functions depend only on initial and final states. Internal energy is a state function....
Read Full Step-by-Step Solution →$\Delta G = \Delta H - T\Delta S = 50 - 298 \times 0.15 = 50 - 44.7 = 5.3 > 0$ → non-spontaneous....
Read Full Step-by-Step Solution →$\Delta G^\circ < 0$ implies spontaneous under standard conditions....
Read Full Step-by-Step Solution →Using the first law of thermodynamics, ΔU = q + w. Heat absorbed by the system is +500 kJ. Work done by the system is taken as –200 kJ. Hence, ΔU = 50...
Read Full Step-by-Step Solution →Using $\Delta G^{\circ} = -RT\ln K$, we have $\ln K = -\Delta G^{\circ}/(RT) = 33000/(8.314\times298) \approx 13.33$, so $K = e^{13.33} \approx 6.0\ti...
Read Full Step-by-Step Solution →ΔG° = ΔH° – TΔS°. Convert ΔS° to kJ K⁻¹ (–0.198 kJ K⁻¹). Then ΔG° = –92.4 kJ – (298 K)(–0.198 kJ K⁻¹) = –92.4 kJ + 59.0 kJ = –33.4 kJ. The negative si...
Read Full Step-by-Step Solution →Reverse reaction (1) and (2) to obtain reactants on the right. ΔH for reversed steps changes sign. Adding the reversed ΔH values gives ΔH₃ = (+393.5) ...
Read Full Step-by-Step Solution →First compute ΔCₚ = ΣνCₚ(products) – ΣνCₚ(reactants) = (2·(29.9+0.012T)) – [(29.1+0.01T)+(29.4+0.015T)] = 1.3 – 0.001T. Integrate ΔCₚ from 298 K to 50...
Read Full Step-by-Step Solution →Reaction becomes spontaneous when ΔG < 0 → T > ΔH/ΔS = 60000/200 = 300 K....
Read Full Step-by-Step Solution →Bond dissociation energy for 1 mol H₂ = 436 kJ; for 0.5 mol = 218 kJ....
Read Full Step-by-Step Solution →Use Hess's law: $\Delta H = [\Delta H_f(CO_2)] - [2\times\Delta H_f(CO)] = -393.5 - 2(-110.5) = -172.5$ kJ....
Read Full Step-by-Step Solution →ΔG° = ΔH° − TΔS° = 30000 − 298×40 = 18080 J/mol ≈ 18 kJ/mol....
Read Full Step-by-Step Solution →State functions depend only on initial and final states. Internal energy is a state function....
Read Full Step-by-Step Solution →$\Delta G = \Delta H - T\Delta S = 50 - 298 \times 0.15 = 50 - 44.7 = 5.3 > 0$ → non-spontaneous....
Read Full Step-by-Step Solution →$\Delta G^\circ < 0$ implies spontaneous under standard conditions....
Read Full Step-by-Step Solution →Using the first law of thermodynamics, ΔU = q + w. Heat absorbed by the system is +500 kJ. Work done by the system is taken as –200 kJ. Hence, ΔU = 50...
Read Full Step-by-Step Solution →Using $\Delta G^{\circ} = -RT\ln K$, we have $\ln K = -\Delta G^{\circ}/(RT) = 33000/(8.314\times298) \approx 13.33$, so $K = e^{13.33} \approx 6.0\ti...
Read Full Step-by-Step Solution →ΔG° = ΔH° – TΔS°. Convert ΔS° to kJ K⁻¹ (–0.198 kJ K⁻¹). Then ΔG° = –92.4 kJ – (298 K)(–0.198 kJ K⁻¹) = –92.4 kJ + 59.0 kJ = –33.4 kJ. The negative si...
Read Full Step-by-Step Solution →Reverse reaction (1) and (2) to obtain reactants on the right. ΔH for reversed steps changes sign. Adding the reversed ΔH values gives ΔH₃ = (+393.5) ...
Read Full Step-by-Step Solution →First compute ΔCₚ = ΣνCₚ(products) – ΣνCₚ(reactants) = (2·(29.9+0.012T)) – [(29.1+0.01T)+(29.4+0.015T)] = 1.3 – 0.001T. Integrate ΔCₚ from 298 K to 50...
Read Full Step-by-Step Solution →Reaction becomes spontaneous when ΔG < 0 → T > ΔH/ΔS = 60000/200 = 300 K....
Read Full Step-by-Step Solution →Bond dissociation energy for 1 mol H₂ = 436 kJ; for 0.5 mol = 218 kJ....
Read Full Step-by-Step Solution →Use Hess's law: $\Delta H = [\Delta H_f(CO_2)] - [2\times\Delta H_f(CO)] = -393.5 - 2(-110.5) = -172.5$ kJ....
Read Full Step-by-Step Solution →ΔG° = ΔH° − TΔS° = 30000 − 298×40 = 18080 J/mol ≈ 18 kJ/mol....
Read Full Step-by-Step Solution →State functions depend only on initial and final states. Internal energy is a state function....
Read Full Step-by-Step Solution →$\Delta G = \Delta H - T\Delta S = 50 - 298 \times 0.15 = 50 - 44.7 = 5.3 > 0$ → non-spontaneous....
Read Full Step-by-Step Solution →$\Delta G^\circ < 0$ implies spontaneous under standard conditions....
Read Full Step-by-Step Solution →Using the first law of thermodynamics, ΔU = q + w. Heat absorbed by the system is +500 kJ. Work done by the system is taken as –200 kJ. Hence, ΔU = 50...
Read Full Step-by-Step Solution →Using $\Delta G^{\circ} = -RT\ln K$, we have $\ln K = -\Delta G^{\circ}/(RT) = 33000/(8.314\times298) \approx 13.33$, so $K = e^{13.33} \approx 6.0\ti...
Read Full Step-by-Step Solution →ΔG° = ΔH° – TΔS°. Convert ΔS° to kJ K⁻¹ (–0.198 kJ K⁻¹). Then ΔG° = –92.4 kJ – (298 K)(–0.198 kJ K⁻¹) = –92.4 kJ + 59.0 kJ = –33.4 kJ. The negative si...
Read Full Step-by-Step Solution →Reverse reaction (1) and (2) to obtain reactants on the right. ΔH for reversed steps changes sign. Adding the reversed ΔH values gives ΔH₃ = (+393.5) ...
Read Full Step-by-Step Solution →First compute ΔCₚ = ΣνCₚ(products) – ΣνCₚ(reactants) = (2·(29.9+0.012T)) – [(29.1+0.01T)+(29.4+0.015T)] = 1.3 – 0.001T. Integrate ΔCₚ from 298 K to 50...
Read Full Step-by-Step Solution →Reaction becomes spontaneous when ΔG < 0 → T > ΔH/ΔS = 60000/200 = 300 K....
Read Full Step-by-Step Solution →Bond dissociation energy for 1 mol H₂ = 436 kJ; for 0.5 mol = 218 kJ....
Read Full Step-by-Step Solution →Use Hess's law: $\Delta H = [\Delta H_f(CO_2)] - [2\times\Delta H_f(CO)] = -393.5 - 2(-110.5) = -172.5$ kJ....
Read Full Step-by-Step Solution →ΔG° = ΔH° − TΔS° = 30000 − 298×40 = 18080 J/mol ≈ 18 kJ/mol....
Read Full Step-by-Step Solution →State functions depend only on initial and final states. Internal energy is a state function....
Read Full Step-by-Step Solution →$\Delta G = \Delta H - T\Delta S = 50 - 298 \times 0.15 = 50 - 44.7 = 5.3 > 0$ → non-spontaneous....
Read Full Step-by-Step Solution →$\Delta G^\circ < 0$ implies spontaneous under standard conditions....
Read Full Step-by-Step Solution →Using the first law of thermodynamics, ΔU = q + w. Heat absorbed by the system is +500 kJ. Work done by the system is taken as –200 kJ. Hence, ΔU = 50...
Read Full Step-by-Step Solution →Using $\Delta G^{\circ} = -RT\ln K$, we have $\ln K = -\Delta G^{\circ}/(RT) = 33000/(8.314\times298) \approx 13.33$, so $K = e^{13.33} \approx 6.0\ti...
Read Full Step-by-Step Solution →ΔG° = ΔH° – TΔS°. Convert ΔS° to kJ K⁻¹ (–0.198 kJ K⁻¹). Then ΔG° = –92.4 kJ – (298 K)(–0.198 kJ K⁻¹) = –92.4 kJ + 59.0 kJ = –33.4 kJ. The negative si...
Read Full Step-by-Step Solution →Reverse reaction (1) and (2) to obtain reactants on the right. ΔH for reversed steps changes sign. Adding the reversed ΔH values gives ΔH₃ = (+393.5) ...
Read Full Step-by-Step Solution →First compute ΔCₚ = ΣνCₚ(products) – ΣνCₚ(reactants) = (2·(29.9+0.012T)) – [(29.1+0.01T)+(29.4+0.015T)] = 1.3 – 0.001T. Integrate ΔCₚ from 298 K to 50...
Read Full Step-by-Step Solution →Reaction becomes spontaneous when ΔG < 0 → T > ΔH/ΔS = 60000/200 = 300 K....
Read Full Step-by-Step Solution →Bond dissociation energy for 1 mol H₂ = 436 kJ; for 0.5 mol = 218 kJ....
Read Full Step-by-Step Solution →Use Hess's law: $\Delta H = [\Delta H_f(CO_2)] - [2\times\Delta H_f(CO)] = -393.5 - 2(-110.5) = -172.5$ kJ....
Read Full Step-by-Step Solution →ΔG° = ΔH° − TΔS° = 30000 − 298×40 = 18080 J/mol ≈ 18 kJ/mol....
Read Full Step-by-Step Solution →