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Question #44230

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The Question

The standard enthalpy of formation of gaseous Molybdenum hexachloride, $\ce{MoCl6}(g)$ , is $-840 \text{ kJ mol}^{-1}$ and that of Molybdenum tetrachloride, $\ce{MoCl4}(g)$ , is $-380 \text{ kJ mol}^{-1}$ at $298 \text{ K}$ . The standard enthalpy of atomization of solid Molybdenum, $\Delta_{atom}H^\circ(\ce{Mo}, s)$ , is $660 \text{ kJ mol}^{-1}$ and the bond dissociation enthalpy of $\ce{Cl2}(g)$ is $240 \text{ kJ mol}^{-1}$ .

A student calculates the enthalpy of the decomposition reaction: $\ce{MoCl6}(g) \rightarrow \ce{MoCl4}(g) + \ce{Cl2}(g)$ by erroneously assuming that the average $\ce{Mo-Cl}$ bond enthalpy remains constant across different oxidation states and is equal to its value in $\ce{MoCl6}(g)$ .

What is the true standard enthalpy of the decomposition reaction, and what is the value calculated by the student, respectively?

Options

A.
$+460 \text{ kJ mol}^{-1}$ and $+500 \text{ kJ mol}^{-1}$
B.
$+460 \text{ kJ mol}^{-1}$ and $+460 \text{ kJ mol}^{-1}$
C.
$+500 \text{ kJ mol}^{-1}$ and $+460 \text{ kJ mol}^{-1}$
D.
$-460 \text{ kJ mol}^{-1}$ and $-500 \text{ kJ mol}^{-1}$

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

Hess's Law Application|Bond Enthalpy Pitfall|Variable Oxidation State Stability

Why This Question Is Tricky

Common Mistakes

This problem tests the limitation of using average bond enthalpies.

Students often treat average bond enthalpies as universal constants for a given pair of atoms.

However, for transition metals with variable oxidation states (like Mo in +6 vs +4), the bond dissociation energy depends heavily on the oxidation state.

As the oxidation state changes, the effective nuclear charge and d-orbital contraction alter the covalent overlap, changing the bond enthalpy (here, 370 kJ/mol in MoCl6 vs 380 kJ/mol in MoCl4).

The trap exposes the failure of the formula $\Delta H = \sum BE_{broken} - \sum BE_{formed}$ when applied across different oxidation states, reinforcing reliance on rigorous Hess's Law state functions.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Thermodynamics - Class 11 Part 1
Concept	Hess's Law Application
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free