From the following thermochemical equations, find out bond dissociation enthalpy of CH3-H bond: CH3I(g) - CH3(g) + I(g); δ H = 54.0 kcal; (g) + (s) - CH3I(g) + (g); δ H = 29.0 kcal; (g) - H(g) + I(

Question

From the following thermochemical equations, find out bond dissociation enthalpy of CH3-H bond: CH3I(g) -> CH3(g) + I(g); δ H = 54.0 kcal; $CH_4$(g) + $I_2$(s) -> CH3I(g) + $HI$(g); δ H = 29.0 kcal; $HI$(g) -> H(g) + I(g); δ H = 79.8 kcal; $I_2$(s) -> 2I(g); δ H = 51.0 kcal

Prashant Kotian · Accepted Answer

Correct answer: D. Target: (g) -> CH3(g) + H(g). Sum: + -> CH3I + (29) + CH3I -> CH3 + I (54) + -> H + I (79.8) + 2I -> (-51). Cancelling: δ H = 29 + 54 + 79.8 - 51 = 111.8 kcal/mol.

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