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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

Question

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

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Solution

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

A reaction is spontaneous if ΔG is negative.

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

We can rearrange the equation to solve for T:

T = ΔH / ΔS

Substituting the given values:

T = -(-92 kJ/mol) / -0.199 kJ/(molK) T = 462 K

Therefore, the reaction would be spontaneous at temperatures below 462 K.

So, the correct answer is A. 400 K.

This problem has been solved

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