Calculate the enthalpy change of the reaction H_2 (g) + Cl_2(g) 2HCl(g) if bond energies (kJ mol^ -1 ) H-H = 436 Cl-Cl = 242, H-Cl = 431

MHT CET 2026 Chemistry question

Calculate the enthalpy change of the reaction

H (g) + Cl(g) 2HCl(g)

if bond energies (kJ mol)

H-H = 436 Cl-Cl = 242, H-Cl = 431

(A)-184 kJ/molCorrect
(B)-246 kJ/mol
(C)-242 kJ/mol
(D)-431 kJ/mol

Answer: (A)

Solution

= (energy of bonds broken) - (energy of bonds formed). Bonds broken: 1 H-H (436) + 1 Cl-Cl (242) = 678 kJ. Bonds formed: 2 H-Cl = 2 431 = 862 kJ. So $Δ

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Question ID VK-097725