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Question #44211

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The Question

Calculate the standard enthalpy of formation ( $\Delta_f H^\circ$ ) of the methyl radical, $\ce{CH3^\bullet_{(g)}}$ , at $298\text{ K}$ .

Given the following thermodynamic data at $298\text{ K}$ : Standard enthalpy of formation of $\ce{CH4_{(g)}} = -75\text{ kJ mol}^{-1}$ Enthalpy of sublimation of $\ce{C_{(s, graphite)}} = 717\text{ kJ mol}^{-1}$ Bond dissociation enthalpy of $\ce{H2_{(g)}} = 436\text{ kJ mol}^{-1}$ First bond dissociation enthalpy of $\ce{CH4_{(g)}}$ ( $\ce{CH4_{(g)} -> CH3^\bullet_{(g)} + H^\bullet_{(g)}}$ ) $= 439\text{ kJ mol}^{-1}$

Options

A.
$+146\text{ kJ mol}^{-1}$
B.
$+123\text{ kJ mol}^{-1}$
C.
$+364\text{ kJ mol}^{-1}$
D.
$+54\text{ kJ mol}^{-1}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Exception_Exposure

Learn to identify edge cases and boundary exceptions that JEE examiners hide in math problems.

Concepts Behind This Question

IUPAC Sign Convention|Bond Enthalpy Pitfall|Bond Cleavage Energetics

Why This Question Is Tricky

Common Mistakes

This problem tests the critical distinction between 'Average Bond Enthalpy' and 'Successive Bond Dissociation Energy' (BDE) when dealing with radicals.

Students commonly fall into the trap of calculating the average C-H bond energy in methane ( $416\text{ kJ mol}^{-1}$ ) and using it via the standard formula $\Delta H = \sum BE(\text{reactants}) - \sum BE(\text{products})$ to find the formation enthalpy of the methyl radical, yielding the incorrect $+123\text{ kJ mol}^{-1}$ .

Furthermore, it tests the thermodynamic definition of formation, where the byproduct $\ce{H^\bullet}$ must be converted back to its standard state $\ce{1/2 H2_{(g)}}$ , trapping those who stop at $+364\text{ kJ mol}^{-1}$ .

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Thermodynamics - Class 11 Part 1
Concept	IUPAC Sign Convention
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free