Respuesta :
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:
[tex]2HNO3(aq) + NO(g) \rightarrow 3 NO2(g) + H2O(l)[/tex]
The equation that relates the standard free energy change ΔG° to the equilibrium constant K is:
[tex]\Delta G^{0}= -RTlnK[/tex]
(or) [tex]K = e^{-\Delta G^{0}/RT}----(1)[/tex]
where R = gas constant = 8.314 J/mol-K
T = temperature in Kelvin
[tex]\Delta G^{0}=\sum n_{p}\Delta G^{0}f(products)-\sum n_{r}\Delta G^{0}f(reactants)[/tex]
where n(p) and n(r) are the number of moles of the products and reactants respectively
Therefore for the given reaction:
[tex]\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)][/tex]
Substituting the given values for ΔG°f:
[tex]\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)][/tex]
ΔG° = + 51 kJ
Substituting the calculated ΔG° in equation (1) at T = 298 K gives:
[tex]K = e^{-\51000/8.314*298}=1.15*10^{-9}[/tex]
