Question

At which temperature would a reaction with h = -220 kj/mol and s = -0.05 kj/(mol * k) be spontaneous?

Answer 1

The reaction with given enthalpy and entropy changes is spontaneous at temperatures below 4400 K, as determined by calculating the Gibbs free energy change.

Step-by-step explanation:

The student's question pertains to the spontaneity of a reaction at a given temperature, which can be determined by using the Gibbs free energy formula ΔG = ΔH - TΔS. Here, ΔH represents the change in enthalpy and ΔS represents the change in entropy. For a reaction to be spontaneous, ΔG should be negative.

The question specifies that ΔH = -220 kJ/mol and ΔS = -0.05 kJ/(mol•K). To find the temperature at which the reaction becomes nonspontaneous (where ΔG = 0), we set up the equation 0 = ΔH - TΔS and solve for T:

0 = (-220 kJ/mol) - T(-0.05 kJ/(mol•K))
T = (-220 kJ/mol) / (-0.05 kJ/(mol•K))
T = 4400 K

Therefore, the reaction would be spontaneous at temperatures below 4400 K.