Question

For a hypothetical chemical reaction \( A_2 + 3B_2 \rightarrow 2AB_3 + \text{Heat} \), which one of the following combinations of state variables would support the spontaneity of the reaction at a particular temperature?

• A. \( \Delta H 0, \Delta G > 0 \)
• B. \( \Delta H > 0, \Delta S > 0, \Delta G < 0 \)
• C. \( \Delta H > 0, \Delta S < 0, \Delta G < 0 \)
• D. \( \Delta H > 0, \Delta S 0 \)

Hint:

For spontaneity, remember the Gibbs free energy equation: \( \Delta G = \Delta H - T \Delta S \). Spontaneity occurs when \( \Delta G \) is negative, and this depends on both enthalpy and entropy.

Answer 1

The Correct Option is B

Step 1: Analyze the Gibbs free energy equation.
The spontaneity of a reaction is determined by the Gibbs free energy change, \( \Delta G \), which is given by:
\[ \Delta G = \Delta H - T \Delta S \]
where: - \( \Delta H \) is the change in enthalpy,
- \( \Delta S \) is the change in entropy,
- \( T \) is the temperature in Kelvin.
For the reaction to be spontaneous:
- \( \Delta G \) must be negative (\( \Delta G < 0 \)).

Step 2: Analyze the reaction.
The given reaction is exothermic (\( \Delta H < 0 \)) since heat is released. Therefore, a negative \( \Delta H \) supports spontaneity.
The change in entropy (\( \Delta S \)) for this reaction is positive (\( \Delta S > 0 \)) because there is an increase in the number of molecules from \( A_2 + 3 B_2 \) to \( 2 AB_3 \), increasing disorder.
Thus, the reaction can be spontaneous at high temperatures when \( \Delta H \) and \( \Delta S \) work together to make \( \Delta G \) negative.

Step 3: Conclusion.
Hence, the correct answer is (B), where \( \Delta H > 0 \), \( \Delta S > 0 \), and \( \Delta G < 0 \).