Question

# For a given reaction, ΔH = -26.6 kJ/mol and ΔS = -77.0 J/K⋅mol. The reaction is...

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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