Question

The standard reaction enthalpy of Zn(s) + H2O(g) →ZnO(s) + H2(g) is known to be HR 0 = 224 kJ and is approximately constant from 920 K up to 1280 K. The standard reaction free energy is +33 kJ at 1280 K. Calculate the equilibrium constant at 1280 K and then calculate the temperature at which the equilibrium constant becomes greater than 1.

Answer 1

Answer:

K = 0.0450, T > 1501 K

Explanation:

We may apply the Gibbs free energy equation relating the Gibbs free energy to an equilibrium constant of a reaction. The relationship is described by the following equation:

$\Delta G^o = -RT ln(K)$

Rearrange the equation for the equilibrium constant:

$ln(K) = -\frac{\Delta G^o}{RT}\therefore K = e^{-\frac{\Delta G^o}{RT}}$

Given the temperature $T = 1280 K$ and the ideal gas law constant  $R = 8.314 \frac{J}{K mol}$, we obtain:

$K = e^{-\frac{33\cdot 10^3 J}{8.314 \frac{J}{K mol}\cdot 1280 K}} = 0.0450$

Now notice if $K > 1$, then $ln(K) > 1$ and $\Delta G^o < 0$.

We may firstly solve for the entropy change of this reaction using the following equation:

$\Delta G^o = \Delta H^o - T\Delta S^o \therefore \Delta S^o = \frac{\Delta H^o - \Delta G^o}{T} = \frac{224\cdot 10^3 J - 33\cdot 10^3 J}{1280 K} = 149.2 \frac{J}{K}$

Using the same equation, solve when the change in the Gibbs free energy is negative:

$\Delta H^o - T\Delta S^o < 0\therefore T > \frac{\Delta H^o}{\Delta S^o} = \frac{224\cdot 10^3 J}{149.2 \frac{J}{K}} = 1501 K$