Question

Sodium carbonate can be prepared by heating sodium bicarbonate. 2 NaHCO3(s) ----> Na2CO3(s) + CO2(g) + H2O(g). Given that DHo = 128.9 kJ and DSo = 321 J/K·mol for this reaction, at what temperature will K = 1?

Answer 1

Answer:

T = 401.6 K

Explanation:

Given reaction:

$NaHCO3(s) ----> Na2CO3(s) + CO2(g) + H2O(g)$

The thermodynamic parameters; ΔG°, ΔH° and ΔS° are related via the Gibbs -Helmholtz equation given as:

$\Delta G^{0}=\Delta H^{0}-T\Delta S^{0}----(1)$

For a given reaction, the gibbs free energy change ΔG° is related to the equilibrium constant K as:

$\Delta G^{0}=-RTln(K) ---(2)$

It is given that K = 1,

Therefore,

$\Delta G^{0}=-RTln(1) = 0$

Substituting for ΔG° in equation (1)

$0=\Delta H^{0}-T\Delta S^{0}$

$\Delta H = T\Delta S$

$T = \Delta H^{0}/\Delta S^{0}=\frac{128.9 kJ}{0.321 kj/mol-K}=401.6 K$