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Question #44235

Difficulty: 8/10 | 🟢 Free

The Question

Calculate the standard enthalpy of formation ( $\Delta_f H^\circ$ ) of liquid tetrachloromethane, $\ce{CCl4(l)}$ , at $298\text{ K}$ and $1\text{ bar}$ , given the following thermodynamic data:

Enthalpy of sublimation of $\ce{C(graphite)} = 715\text{ kJ mol}^{-1}$ Enthalpy of vaporization of $\ce{CCl4(l)} = 30\text{ kJ mol}^{-1}$ Bond dissociation enthalpy of $\ce{Cl-Cl(g)} = 242\text{ kJ mol}^{-1}$ Average bond enthalpy of $\ce{C-Cl(g)} = 328\text{ kJ mol}^{-1}$

Options

A.
$-143\text{ kJ mol}^{-1}$
B.
$-113\text{ kJ mol}^{-1}$
C.
$-83\text{ kJ mol}^{-1}$
D.
$-828\text{ kJ mol}^{-1}$

Check Your Answer

Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Mechanism_Trap

Don't get tricked by the process! Learn the mechanical traps used in competitive exam questions.

Concepts Behind This Question

Bond Enthalpy Pitfall|Standard State Conditions

Why This Question Is Tricky

Common Mistakes

This problem tests the 'State Function' pitfall in thermodynamics.

Students frequently misapply the formula $\Delta H = \sum BE(\text{reactants}) - \sum BE(\text{products})$ directly to condensed phases (yielding the trap answer -828 kJ/mol).

Bond dissociation enthalpies are exclusively defined for homolytic cleavage in the gas phase.

The hidden trap is the necessity to incorporate the enthalpy of sublimation for solid graphite and the enthalpy of condensation (negative vaporization) for liquid $\ce{CCl4}$ to complete the thermodynamic cycle.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Thermodynamics - Class 11 Part 1
Concept	Bond Enthalpy Pitfall
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free