Question

If AH = -60.0 kJ and AS = -0.400 kJ/K, the reaction is spontaneous below a certain temperature. Calculate that temperature.

Options:
a. -150 K
b. -300 K
c. 150 K
d. 300 K

Answer 1

The temperature below which the reaction is spontaneous is 150 K. The calculation is based on setting the change in free energy to zero and solving for the transition temperature using the given enthalpy and entropy values.

Step-by-step explanation:

To determine the temperature below which the reaction is spontaneous, we apply the relationship that the change in free energy (ΔG) is zero at equilibrium or the transition temperature. For spontaneity, ΔG should be negative. The formula to use is ΔG = ΔH - TΔS where ΔH is the change in enthalpy, T is the temperature, and ΔS is the change in entropy. Given that ΔH = -60.0 kJ and ΔS = -0.400 kJ/K, we set ΔG = 0 and solve for T, the temperature at which the reaction becomes nonspontaneous:

ΔH = TΔS

Therefore:

-60.0 kJ = T (-0.400 kJ/K)

To solve for T, we can rearrange the equation:

T = ΔH / ΔS

T = (-60.0 kJ) / (-0.400 kJ/K)

T = 150 K

So, the reaction is spontaneous below 150 K.