For the reactionA + B → Ccalculate ΔG in kJ.mol–1 at temperature 298 K using the following thermochemical data for the reaction:ΔH = –54.4 kJ.mol–1 ΔS = 144 J.mol–1.K–1. Enter your answer to 3 significant figures. Do not include units. Include the sign.Is this reaction spontaneous?
Question
For the reactionA + B → Ccalculate ΔG in kJ.mol–1 at temperature 298 K using the following thermochemical data for the reaction:ΔH = –54.4 kJ.mol–1 ΔS = 144 J.mol–1.K–1. Enter your answer to 3 significant figures. Do not include units. Include the sign.Is this reaction spontaneous?
Solution
To calculate ΔG (Gibbs free energy change), we can use the formula:
ΔG = ΔH - TΔS
where: ΔH is the change in enthalpy, T is the temperature in Kelvin, and ΔS is the change in entropy.
Given: ΔH = -54.4 kJ/mol = -54.4 x 10^3 J/mol (since 1 kJ = 10^3 J), T = 298 K, and ΔS = 144 J/mol.K.
Substituting these values into the formula, we get:
ΔG = -54.4 x 10^3 J/mol - (298 K x 144 J/mol.K) = -54.4 x 10^3 J/mol - 42.9 x 10^3 J/mol = -97.3 x 10^3 J/mol = -97.3 kJ/mol
So, ΔG = -97.3 kJ/mol to 3 significant figures.
The sign of ΔG determines whether the reaction is spontaneous. If ΔG is negative, the reaction is spontaneous. If ΔG is positive, the reaction is non-spontaneous. Since ΔG is negative in this case, the reaction is spontaneous.
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