Question

Use the free energies of formation given below to calculate the equilibrium constant (K) for the following reaction at 298 K. 2 HNO3(aq) + NO(g) → 3 NO2(g) + H2O(l) K = ? Δ G0f(kJ/mol) -110.9 87.6 51.3 -237.1

Answer 1

Answer:

The equilibrium constant K = 1.15*10^-9

Explanation:

Given:

ΔG°f(HNO3) = -110.9 kj/mol

ΔG°f(NO) = 87.6 kj/mol

ΔG°f(NO2) = 51.3 kj/mol

ΔG°f(HNO3) = -237.1 kj/mol

To determine:

The equilibrium constant (K) for the given reaction

Calculation:

The chemical reaction is:

$2HNO3(aq) + NO(g) \rightarrow 3 NO2(g) + H2O(l)$

The equation that relates the standard free energy change ΔG° to the equilibrium constant K is:

$\Delta G^{0}= -RTlnK$

(or) $K = e^{-\Delta G^{0}/RT}----(1)$

where R = gas constant = 8.314 J/mol-K

T = temperature in Kelvin

$\Delta G^{0}=\sum n_{p}\Delta G^{0}f(products)-\sum n_{r}\Delta G^{0}f(reactants)$

where n(p) and n(r) are the number of moles of the products and reactants respectively

Therefore for the given reaction:

$\Delta G^{0}=[3\Delta G^{0}f(NO2)+3\Delta G^{0}f(H2O)]-[2\Delta G^{0}f(HNO3)+1\Delta G^{0}f(NO)]$

Substituting the given values for ΔG°f:

$\Delta G^{0}=[3\Delta G^{0}f(51.3)+3\Delta G^{0}f(-237.1)]-[2\Delta G^{0}f(-110.9)+1\Delta G^{0}f(87.6)]$

ΔG° = + 51 kJ

Substituting the calculated ΔG° in equation (1) at T = 298 K gives:

$K = e^{-\51000/8.314*298}=1.15*10^{-9}$