Determine if reaction is spontaneous under standard conditions, if not at what temperature will it be spontaneous? 3 A (g) + 5 B (s) → 3 AB (s) + B2 (g) ∆H° = 112.7 kJ, ∆S° = 78.3 J/K.
A
1439.3 K
B
274.6 K
C
1.439 K
D
719.65 K
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1
Identify the given values: the change in enthalpy (∆H°) is 112.7 kJ and the change in entropy (∆S°) is 78.3 J/K. Note that ∆H° is given in kJ, so it may be necessary to convert it to J for consistency with ∆S°.
Recall the Gibbs free energy equation: ΔG° = ΔH° - TΔS°. A reaction is spontaneous under standard conditions if ΔG° < 0.
Convert ∆H° from kJ to J by multiplying by 1000, so ∆H° = 112,700 J.
Set up the inequality for spontaneity: 0 > 112,700 J - T(78.3 J/K). Solve for T to find the temperature at which the reaction becomes spontaneous.
Rearrange the inequality to solve for T: T > 112,700 J / 78.3 J/K. Calculate this value to find the minimum temperature for spontaneity.