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Question #44710

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The Question

A gaseous reactant \ce{A} is continuously bubbled through an aqueous solution containing $0.20 \text{ M}$ of reactant \ce{B} at $298 \text{ K}$ . The partial pressure of \ce{A(g)} above the solution is maintained constant at $0.60 \text{ bar}$ . The Henry's law constant for \ce{A} in water at $298 \text{ K}$ is given as $K_H = 12.0 \text{ bar L mol}^{-1}$ .

The overall reaction is: $\ce{2A(aq) + 3B(aq) -> A2B3(aq)}$

The proposed mechanism for this reaction is: Step 1: \ce{2A(aq) <=> A2(aq)} \quad (Fast equilibrium, $K_c = 4.0 \text{ L mol}^{-1}$ ) Step 2: \ce{A2(aq) + B(aq) -> A2B(aq)} \quad (Slow, $k_1 = 2.5 \times 10^{-2} \text{ L mol}^{-1} \text{ s}^{-1}$ ) Step 3: \ce{A2B(aq) + 2B(aq) -> A2B3(aq)} \quad (Fast)

Assuming ideal behavior, what is the initial rate of disappearance of \ce{B}?

Options

A.
$1.5 \times 10^{-4} \text{ mol L}^{-1} \text{ s}^{-1}$
B.
$5.0 \times 10^{-5} \text{ mol L}^{-1} \text{ s}^{-1}$
C.
$4.6 \times 10^{-1} \text{ mol L}^{-1} \text{ s}^{-1}$
D.
$3.1 \times 10^{0} \text{ mol L}^{-1} \text{ s}^{-1}$

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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

Henry’s Law Constant (Kh) Trends|Stoichiometric Coefficient Division

Why This Question Is Tricky

Common Mistakes

The question tests two major traps.

First, the 'Unit Conversion Trap' in Henry's Law: students often blindly use $P = K_H x$ and multiply the resulting mole fraction by the molarity of water ( $55.5 \text{ M}$ ) to find concentration, ignoring that $K_H$ is explicitly given in $\text{bar L mol}^{-1}$ , which directly relates pressure to molarity ( $P = K_H C$ ).

Second, the 'Stoichiometric Coefficient Division' trap: students frequently equate the rate of the rate-determining step directly to the rate of disappearance of a reactant, failing to account for the additional moles of the reactant consumed in subsequent fast steps.

Here, the stoichiometry of the mechanism dictates that the rate of disappearance of B is three times the rate of the slow step.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Solutions - Class 12
Concept	Henry’s Law Constant (Kh) Trends
Difficulty	8
Cognitive Load	7
Target Solve Time	160 sec
Access	Free