Quantrex Quantrex AcademyJEE · NEET · NDA PYQs with solutions Open app
MHT CET202613 April 2026Morning ShiftChemistryThermodynamics (C)Actual

Calculate the enthalpy change of the reaction, H ₂ (g) + Cl ₂ (g) 2 HCl(g) if bond energies (kJ mol ⁻¹ ): H–H = 436, Cl–Cl = 242, H–Cl = 431

Options

  1. A-184 kJ/mol
  2. B-246 kJ/mol
  3. C-242 kJ/mol
  4. D-431 kJ/mol

Correct answer

A. -184 kJ/mol

Step-by-step solution

The enthalpy change of the reaction is given by the difference between the sum of bond energies of reactants and the sum of bond energies of products. H = BE(Reactants) - BE(Products) For the given reaction: H ₂ (g) + Cl ₂ (g) 2 HCl(g) H = [ BE(H-H) + BE(Cl-Cl) ] - [2 BE(H-Cl) ] Substituting the given values: H = (436 + 242) - (2 431) H = 678 - 862 H = -184 kJ mol ⁻¹ Answer: -184 kJ/mol

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 MHT CET PYQs