Estimate Delta H for CH4(g) + Cl2(g) -> CH3Cl(g) + HCl(g) using bond enthalpies in kJ mol^-1: C-H=414, Cl-Cl=243, C-Cl=339, H-Cl=431.

JEE Main levelNumerical answerDifficulty 2 of 3td130

Estimate $Δ H$ for $CH X_{4} (g) + Cl X_{2} (g) CH X_{3} Cl (g) + HCl (g)$ using bond enthalpies in $kJ mo l^{- 1}$ : $C - H = 414$ , $Cl - Cl = 243$ , $C - Cl = 339$ , $H - Cl = 431$ .

Numerical answer type, answer in kJ/mol, tolerance 0.5.

Δ H = \sum (bonds broken) - \sum (bonds formed)

. Broken: one

C - H

and one

Cl - Cl

414 + 243 = 657

. Formed: one

C - Cl

and one

H - Cl

339 + 431 = 770

Δ H = 657 - 770 = - 113 kJ mo l^{- 1}