Consider the following reaction:
C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l) ΔH = – 2808 kJ / mol
Under what conditions will this reaction proceed spontaneously?
A. Low temperature, low pressure
B. Low temperature, any pressure
C. High temperature, any pressure
D. Any temperature, any pressure
Click for Explanation
Spontaneity is determined by a negative ΔG value in Gibbs free energy equation:
ΔG = ΔH – TΔS
A reaction with a negative ΔH and a positive ΔS will thus be spontaneous under any conditions. The reaction given has a negative ΔH. It also involves converting 1 mole of solid and 6 moles of gas into 6 moles of liquid and 6 moles of gas. That phase change means entropy is increasing (solid to liquid). Thus, with a negative ΔH and a positive ΔS this reaction is spontaneous under any conditions and (D) is the right answer.
A, B: Reactions with a negative ΔH and a negative ΔS are spontaneous at low temperatures (e.g. the freezing of water from liquid to solid).
C: Reactions with a positive ΔH and a positive ΔS are spontaneous at high temperatures (e.g. the boiling of liquid water into steam)
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