For a particular reaction, Delta H Degree is -16.1 kJ/mol and Delta S Degree is

Question

For a particular reaction, Delta H Degree is -16.1 kJ/mol and Delta S Degree is -21.8 J/(mol- K). Assuming these values change very little with temperature, over what temperature range is the reaction spontaneous in the forward direction? The reaction is spontaneous for temperatures the change in Gibbs free energy is given by Delta G= Delta H- T Delta S where Delta G is the change in Gibbs free energy, Delta H is the change in enthalpy, Delta S is the change in entropy, and T is the temperature. A reaction is spontaneous when Delta G 0, so Delta G = 0 represents the "cutoff" point for a spontaneous reaction. What is the value of T when Delta G = 0?

Explanation / Answer

Given that

H= -16.1 kJ = -16.1 x 103 J

S= -21.8 J/K

For any reaction to be spontaneous, G < 0

The relation between G, H, S is

G = H - TS

G < 0 ( condition for spontaneity)

H - TS < 0

H < TS

TS > H

T > H/S

T > (-16.1 x 103J) / ( -21.8 J/K )

T > 738.5 K

To become spontaneous, temperature should be greater than 738.5 K.

Therefore,

For temperatures above 738.5 K , the given reaction will become spontaneous.

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Ans = Greater than , T = 738.5 K

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