At which temperature would a reaction with H = -92 kJ/mol, S = -0.199 kJ/(molK) be spontaneous?A.400 KB.500 KC.600 KD.700 KSUBMITarrow_backPREVIOUS

The spontaneity of a reaction can be determined using the Gibbs Free Energy equation:

ΔG = ΔH - TΔS

Where: ΔG is the change in Gibbs Free Energy ΔH is the change in enthalpy T is the temperature in Kelvin ΔS is the change in entropy

For a reaction to be spontaneous, ΔG must be less than 0.

Given: ΔH = -92 kJ/mol ΔS = -0.199 kJ/(molK)

We can substitute these values into the equation and solve for T:

0 = -92 - T(-0.199)

Rearranging the equation to solve for T gives:

T = -92 / -0.199

Calculating this gives T ≈ 462 K

Therefore, the reaction would be spontaneous at temperatures above 462 K. Looking at the options provided, the reaction would be spontaneous at 500 K, 600 K, and 700 K. So the correct answers are B, C, and D.