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Question #44842

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The Question

A gaseous reactant \ce{A(g)} is continuously bubbled into an inert solvent at a constant pressure of $1 \text{ bar}$ , maintaining a saturated solution. In the dissolved state, \ce{A(aq)} undergoes a first-order decomposition. The half-life of this decomposition is independent of the initial concentration of \ce{A(aq)}.

The steady-state rate of product formation ( $r$ ) is monitored across a range of temperatures. A plot of $\ln(r)$ versus $1/T$ yields a straight line with a slope of $-7217 \text{ K}$ .

The Henry's law constant ( $K_H$ , defined as $P = K_H [\ce{A(aq)}]$ ) for the dissolution of \ce{A(g)} in this solvent varies with temperature according to the relation: $\ln K_H = 10.5 - \frac{3007}{T}$

Determine the true activation energy ( $E_a$ ) for the decomposition of \ce{A(aq)}. (Given: $R = 8.314 \text{ J K}^{-1} \text{ mol}^{-1}$ )

Options

A.
$85.0 \text{ kJ mol}^{-1}$
B.
$60.0 \text{ kJ mol}^{-1}$
C.
$35.0 \text{ kJ mol}^{-1}$
D.
$110.0 \text{ kJ mol}^{-1}$

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Correct Answer

Step-by-Step Solution

Explanation

How the Examiner Thinks

Numerical_Precision

Improve calculation stamina and avoid rounding errors in high-stakes competitive examinations.

Concepts Behind This Question

Graphical Slopes Analysis|First Order Half-Life Independence|Henry’s Law Constant (Kh) Trends

Why This Question Is Tricky

Common Mistakes

This problem tests the integration of Chemical Kinetics (Arrhenius equation) with Thermodynamics (Henry's Law temperature dependence).

The 'Hidden Variable' trap is that students often treat the concentration of the dissolved gas as a constant across different temperatures, ignoring that the solubility (and thus $K_H$ ) is highly temperature-dependent.

By failing to account for the $d(\ln K_H)/d(1/T)$ term, students will incorrectly equate the graphical slope directly to $-E_a/R$ , leading to the trap answer of $60.0 \text{ kJ/mol}$ .

Furthermore, misinterpreting the definition of $K_H$ (e.g., assuming $K_H = [\ce{A(aq)}]/P$ ) leads to the $35.0 \text{ kJ/mol}$ trap.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Chemical Kinetics - Class 12
Concept	Graphical Slopes Analysis
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free