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Multiple Choice
Consider the combustion of butane gas and predict the signs of ΔS, ΔH and ∆G. C4H10(g) + 13/2 O2(g) ⟶ 4 CO2(g) + 5 H2O(g)
A
+ΔS; +ΔH; +∆G
B
+ΔS; +ΔH; –∆G
C
+ΔS; –ΔH; –∆G
D
–ΔS; –ΔH; –∆G
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1
Identify the type of reaction: The combustion of butane is an exothermic reaction, which typically releases heat.
Determine the sign of ΔH (enthalpy change): Since the reaction is exothermic, ΔH is negative because heat is released.
Analyze the change in entropy (ΔS): The reaction involves the conversion of gases. There are 6.5 moles of gas on the reactant side (1 mole of C4H10 and 6.5 moles of O2) and 9 moles of gas on the product side (4 moles of CO2 and 5 moles of H2O). Since the number of gas molecules increases, ΔS is positive.
Consider the sign of ΔG (Gibbs free energy change): For a spontaneous reaction at constant temperature and pressure, ΔG is negative. Since the reaction is exothermic (negative ΔH) and has a positive ΔS, ΔG will be negative, indicating spontaneity.
Summarize the signs: The combustion of butane results in +ΔS (increase in entropy), –ΔH (release of heat), and –ΔG (spontaneous process).