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Question #105465

Difficulty: 8/10 | 🟢 Free

The Question

The dissolution of a sparingly soluble gas $\ce{A}$ in water is represented by the equilibrium: $\ce{A(g) <=> A(aq)}$ . At $300\text{ K}$ , the Henry's law constant ( $K_H$ ) for gas $\ce{A}$ is $1.0 \times 10^5\text{ bar}$ . The standard enthalpy of dissolution is $\Delta H^\circ = -31.365\text{ kJ mol}^{-1}$ . Assuming $\Delta H^\circ$ and the standard entropy of dissolution ( $\Delta S^\circ$ ) are independent of temperature, calculate the Gibbs free energy change ( $\Delta G$ ) for the dissolution of gas $\ce{A}$ at $350\text{ K}$ when its partial pressure is $2.0\text{ bar}$ and its mole fraction in the aqueous solution is $1.0 \times 10^{-5}$ .

(Given: $R = 8.3\text{ J K}^{-1}\text{ mol}^{-1}$ , $\ln 10 = 2.3$ , $\ln 2 = 0.69$ . The density of the aqueous solution is $1.02\text{ g mL}^{-1}$ )

Options

A.
$+3.22\text{ kJ mol}^{-1}$
B.
$+74.05\text{ kJ mol}^{-1}$
C.
$-6.78\text{ kJ mol}^{-1}$
D.
$-63.59\text{ kJ mol}^{-1}$

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

Henry’s Law Constant (Kh) Trends|Gibbs Energy and Spontaneity

Why This Question Is Tricky

Examiner Levers

Extraneous Data

Provides unnecessary variables (Red Herrings) to test conceptual focus.

Difficulty Boost : +3

Approximation Bait

Options are extremely close; tests precision vs. early rounding.

Difficulty Boost : +3

Dimensional Analysis

Tests if the student verifies units/dimensions before selecting.

Difficulty Boost : +2

Common Mistakes

This problem tests the thermodynamic inversion of gas solubility and the correct formulation of the reaction quotient.

Students often erroneously treat $K_H$ as an equilibrium constant $K_{eq}$ , leading to a sign error in $\Delta G^\circ$ (using $\Delta G^\circ = -RT \ln K_H$ instead of $+RT \ln K_H$ ).

Furthermore, the trap exploits the entropy-driven nature of gas dissolution: because $\Delta S^\circ$ is highly negative, increasing the temperature makes $\Delta G^\circ$ significantly more positive (less spontaneous).

Distractor B traps students who invert the reaction quotient ( $Q = p/x$ instead of $x/p$ ).

Distractor C traps those who fail to recalculate $\Delta G^\circ$ at the new temperature ( $350\text{ K}$ ).

Distractor D traps those who use the wrong sign for the $\Delta G^\circ$ and $K_H$ relationship.

Quick Facts About This Question

Question Information

Subject	Chemistry
Chapter	Solutions - Class 12
Concept	Henry’s Law Constant (Kh) Trends
Difficulty	8
Cognitive Load	8
Target Solve Time	180 sec
Access	Free