For a given reaction, ΔH = -26.6 kJ/mol and ΔS = -77.0 J/K⋅mol. The reaction is spontaneous ________. Assume that ΔH and ΔS do not vary with temperature.
For a given reaction, = -26.6 kJ/mol and = -77.0 J/Kmol. The reaction is spontaneous ________. Assume that and do not vary with temperature.
at T > 298 K | |
at all temperatures | |
at T < 345 K | |
at T < 298 K | |
at T > 345 K |
reaction is spontaneous if delta G is less than zero
The equation for calculating ΔG is the following:
ΔG = ΔH – T(ΔS)
Gibbs Free Energy and must be negative for a spontaneous reaction. Set delta G = 0 (for equilibrium) and solve for T
ΔH = -26.6 kJ/mol = -26600 j/mol
ΔS = -77.0 J/K⋅mol
ΔG = -26600 -T(-77.0)
0 = -26600 +(77.0)T
26600= 77.0T
T = 26600/77
T= 345
delta S is negative, so (-T delta S) is positive and G is more
positive for higher temperatures
This reaction is spontaneous for temperatures less than 345 K
so .at T < 345 K reaction is spontaneous
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