Iron metal can be produced by reducing iron(III) oxide with hydrogen:
Fe_{2}O_{3}(s) + 3 H_{2}(g)\longrightarrow 2 Fe(s) + 3 H_{2}O(g) ΔH° = +98.8 kJ; ΔS° = +141.5 J/K
(a) Is this reaction spontaneous under standard-state conditions at 25 °C?
(b) At what temperature will the reaction become spontaneous?
STRATEGY
To determine whether the reaction is spontaneous at 25 °C, we need to determine the sign of ΔG = ΔH – TΔS. To find the crossover temperature at which the reaction becomes spontaneous, we use the equation T = ΔH/ΔS.
IDENTIFY | |
Known | Unknown |
ΔH° = +98.8 kJ | ΔG at 25 °C |
ΔS° = +141.5 J/K | Temperature (T) for spontaneity |
(a) At 25 °C (298 K), ΔG for the reaction is
ΔG = ΔH – TΔS = (98.8 kJ) – (298 K)(0.1415 kJ/K)
= (98.8 kJ) – (42.2 kJ)
= 56.6 kJ
Because the positive ΔH term is larger than the positive TΔS term, ΔG is positive and the reaction is nonspontaneous at 298 K.
(b) We can estimate the crossover temperature at which ΔG changes from positive to negative by substituting the values of ΔH and ΔS into the equation T = ΔH/ΔS:
T=\frac{\Delta H}{\Delta S}=\frac{98.8 kJ}{0.1415 kJ/K}=698 KBecause this calculation assumes the values of ΔH and ΔS are unchanged on going from the standard-state temperature of 298 K to 698 K, the calculated value of T is only an estimate.