Question

If a reaction has an enthalpy change (ΔH) of -890 kJ/mol and an entropy change (ΔS) of -0.24 kJ/(molK), what conclusion can be drawn?

A. It is never spontaneous.
B. It is at equilibrium at 371 K.
C. It is always spontaneous.
D. It is spontaneous at 2000 K

Answer 1

The reaction with ΔH of -890 kJ/mol and ΔS of -0.24 kJ/(molK) becomes spontaneous at temperatures above 3711 K. Therefore, it is not spontaneous at 2000 K, contradicting option D, which states it is spontaneous at 2000 K. correct option is D.

Step-by-step explanation:

If a reaction has an enthalpy change (ΔH) of -890 kJ/mol and an entropy change (ΔS) of -0.24 kJ/(molK), we must consider both factors to determine the reaction's spontaneity at different temperatures. The spontaneity can be assessed using the Gibbs free energy change (ΔG), which is calculated using the equation ΔG = ΔH - TΔS, where T is the temperature in Kelvin (K). If ΔG is negative, the reaction is spontaneous.

To find out at which temperature the reaction is spontaneous, we can set ΔG to zero and solve for T as follows:

0 = (-890 kJ/mol) - T(-0.24 kJ/(molK))

T = 890 kJ/mol / 0.24 kJ/(molK)

T ≈ 3711 K

This means the reaction becomes spontaneous at temperatures greater than 3711 K. Thus, at 2000 K, the reaction is not spontaneous because it is below the threshold temperature. Therefore, the reaction is not spontaneous at 2000 K and this temperature is too low for the reaction to become spontaneous. In summary, the correct answer is option D: It is spontaneous at 2000 K.