Part C
If ΔH = -80.0 kJ and ΔS = -0.200 kJ/K , the reaction is spontaneous below a certain temperature. Calculate that temperature.
Express your answer numerically in kelvins.
A reaction is spontaneous when G < 0
But,
G = H - T S
So,
H - T S < 0
-80.0 kJ - T K ( -0.2 kJ/K ) < 0
-80.0 kJ + T x 0.2 kJ < 0
T x 0.2 kJ < 80.0 kJ
T < (80.0 kJ) / (0.2 kJ)
T < 400
So, the temperature is 400 K, below 400 K the reaction is spontaneous.
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