Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
Manipal MET2021ChemistryThermodynamics (C)

Calculate the temperature at which G=-5.2 ~kJ ~mol ⁻¹, H=145.6 ~kJ ~mol ⁻¹ and S=216 Jk ⁻¹ ~mol ⁻¹ for a chemical reaction :

Options

  1. A698.1^ C
  2. B698.1 K
  3. C130 K
  4. D130^ C

Correct answer

B. 698.1 K

Step-by-step solution

aligned G₁ & = H-T S or T= H- G S & = [145.6-(-5.2)] 10^3 216 =698.1 ~K aligned

Practice Thermodynamics (C) on Quantrex Academy →

More from Thermodynamics (C)

Identify the INCORRECT statement 2026The standard enthalpies of formation of CH₄ (g), CO₂ (g) and H₂ O(l) are -74.8 kJ mol ⁻¹ , -393.5 kJ mol ⁻¹ and -285.8 kJ mol ⁻¹ respectively. Then the enthalpy change for the give 2026For an ideal gas undergoing an isothermal change, there is 2026Ozone is formed by the reaction O _ 2(g) + O _ (g) O _ 3(g) , H = -107.2 kJ . Given O=O bond energy is 498.0 kJ mol ⁻¹ , the average bond energy of ozone is: 2026H and S for a reaction are 35.5 kJ mol ⁻¹ and 83.6 J K ⁻¹ respectively. Assuming that H and S do not vary with temperature, the reaction is spontaneous when: 2026The heat of combustion of carbon to CO ₂ is -393.5 kJ mol ⁻¹ . The heat released on the formation of 35.2 g of CO ₂ by combustion of C is: 20265 moles of a gas is allowed to pass through a series of changes as shown in the graph, in a cyclic process. The processes C A , B C and A B respectively are 20251 mole of an ideal gas is allowed to expand isothermally and reversibly from 1 L to 5 L at 300 K. The change in enthalpy (in kJ ) is ( R =8.3 ~J ~K ⁻¹ ~mol ⁻¹ ) 2025 Full Thermodynamics (C) list All Manipal MET PYQs