At 298 K, H0 = -314 kJ/mol and S0 = -0.372 kJ/(K•mol). What is the Gibbs free energy of the reaction?A.-425 kJB.0.393 kJC.34,900 kJD.-203 kJ
Question
At 298 K, H0 = -314 kJ/mol and S0 = -0.372 kJ/(K•mol). What is the Gibbs free energy of the reaction?A.-425 kJB.0.393 kJC.34,900 kJD.-203 kJ
Solution
The Gibbs free energy (ΔG) of a reaction at any temperature can be calculated using the equation:
ΔG = ΔH - TΔS
where: ΔH is the change in enthalpy (heat content), T is the absolute temperature in Kelvin (K), and ΔS is the change in entropy (disorder).
Given in the problem, ΔH = -314 kJ/mol, T = 298 K, and ΔS = -0.372 kJ/(K•mol).
Substituting these values into the equation gives:
ΔG = -314 kJ/mol - (298 K * -0.372 kJ/(K•mol))
Solving this gives:
ΔG = -314 kJ/mol - (-110.736 kJ/mol)
ΔG = -314 kJ/mol + 110.736 kJ/mol
ΔG = -203.264 kJ/mol
So, the closest answer is D. -203 kJ.
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