Question

Calculate the carbon carbon double bond energy in ethane from the following reaction, H2C=CH2(g) + H2(g) –> H3C−CH3(g) ΔH = −138 kJ/mol. If Bond enthalpies are: C−C = 348; H−H = 436; C−H = 412 in KJ/mol.

(a) 498 KJ/mol

(b) 593 KJ/mol

(c) 508 KJ/mol

(d) 598 KJ/mol

This question was posed to me by my college director while I was bunking the class.

This interesting question is from Thermodynamics in division Thermodynamics of Chemistry – Class 11

Answer 1

Right option is (d) 598 KJ/mol

Easiest explanation: By following the rule ΔHr = Σ ΔHf[products] – ∑ ΔHf[reactants], we get -(x + 4(412) + 432) + (348 + 6(412)) = -138 KJ/mol; x = 598 KJ/mol. So the carbon carbon double bond energy in Ethane is given as 598 KJ per Mol.