## Chapter 7

## Q. 7.6

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.

## Step-by-Step

## Verified Solution

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.