Question 7.6: Spontaneity of Reactions: Enthalpy, Entropy, and Free Energy......
Spontaneity of Reactions: Enthalpy, Entropy, and Free Energy The industrial method for synthesizing hydrogen by reaction of carbon with water has ΔH = +31.3 kcal/mol (+131 kJ/mol) and ΔS = +32 cal/ [mol • K](+134 J/ [mol • K]). What is the value of ΔG (in kcal and kJ) for the reaction at 27 °C (300 K)?
Is the reaction spontaneous or nonspontaneous at this temperature?
C(s) + H_2O(l) → CO(g) + H_2(g)
ANALYSIS The reaction is endothermic (ΔH positive) and does not favor spontaneity, whereas the ΔS indicates an increase in disorder (ΔS positive), which does favor spontaneity. Calculate ΔG to determine spontaneity.
BALLPARK ESTIMATE The unfavorable ΔH (+31.3 kcal/mol) is 1000 times greater than the favorable ΔS (+32 cal/mol • K), so the reaction will be spontaneous (ΔG negative) only when the temperature is high enough to make the T ΔS term in the equation for ΔG larger than the ΔH term. This happens at T ≥ 1000 K. Since T = 300 K, expect ΔG to be positive and the reaction to be nonspontaneous.
Use the free-energy equation to determine the value of ΔG at this temperature. (Remember that ΔS has units of calories per mole-kelvin or joules per mole-kelvin, not kilocalories per mole-kelvin or kilojoules per mole-kelvin.)
ΔG = ΔH – T ΔS
ΔG = +31.3 \frac{kcal}{mol} – (300 \cancel{K}) \left( +32 \frac{\cancel{cal}}{mol • \cancel{K}} \right ) \left(\frac{1\ kcal}{1000\ \cancel{cal}} \right) = +21.7 \frac{kcal}{mol}
ΔG = +131 \frac{kJ}{mol} – (300 \cancel{K}) \left( +134 \frac{\cancel{J}}{mol • \cancel{K}} \right ) \left(\frac{1\ kJ}{1000\ \cancel{J}} \right) = +90.8 \frac{kJ}{mol}
BALLPARK CHECK Because ΔG is positive, the reaction is nonspontaneous at 300 K, consistent with our estimate.