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Multiple Choice
The following reaction is spontaneous at room temperature: C3H8(g) + 5O2(g) → 3CO2(g) + 4H2O(g). Which of the following is true about the enthalpy (ΔH) and entropy (ΔS) changes for this reaction?
A
ΔH is negative and ΔS is positive
B
ΔH is negative and ΔS is negative
C
ΔH is positive and ΔS is negative
D
ΔH is positive and ΔS is positive
Verified step by step guidance
1
Step 1: Understand the concept of spontaneity in chemical reactions. A spontaneous reaction is one that occurs naturally under given conditions without external influence. For a reaction to be spontaneous at room temperature, the Gibbs free energy change (ΔG) must be negative.
Step 2: Recall the Gibbs free energy equation: ΔG = ΔH - TΔS, where ΔG is the change in Gibbs free energy, ΔH is the change in enthalpy, T is the temperature in Kelvin, and ΔS is the change in entropy. For a reaction to be spontaneous, ΔG should be negative.
Step 3: Analyze the enthalpy change (ΔH). In combustion reactions like the one given, energy is released as heat, indicating that the reaction is exothermic. Therefore, ΔH is negative.
Step 4: Consider the entropy change (ΔS). Entropy is a measure of disorder or randomness. In the given reaction, the number of gas molecules increases from 6 (1 C3H8 + 5 O2) to 7 (3 CO2 + 4 H2O), suggesting an increase in disorder. Thus, ΔS is positive.
Step 5: Conclude based on the analysis: For the reaction to be spontaneous, ΔH is negative (exothermic) and ΔS is positive (increase in disorder), which aligns with the condition for a negative ΔG.