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
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.
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