Write the expression for the equilibrium constant for each of the following chem

Question

Write the expression for the equilibrium constant for each of the following chemical reactions. FE exam manual) 1) C_(s) + CO_2(g) doubleheadarrow CO_(g) 2) HNO_3 + H_2O doubleheadarrow H_3O^+ + NO^+_3 3) CaCO_3(s) doubleheadarrow CaO_(s) + CO_2 4) BaO_(s) + CO_2(g) doubleheadarrow BaCO_3(s) H_2PO^-_4 is in equilibrium with HPO^2-_4 in water: H_2PO^-_4 doubleheadarrow H^+ + HPO^2-_4 T = 25 degree C; [H_2PO^-_4] = 9.698 times 10^-5 M; [H^+] = [HPO^2 _4] = 3.018 times 10^6 M Using the extended Debye-Huckel approximation equation, calculate the following 1) when mu = 0 a) the activity coefficients for H_2PO^-_4, HPO^2-_4, and H^+ b) the equilibrium constant, Ka_2 c) the equilibrium constant corrected for ionic strength, Ka^_2 d) pKa_2 and pKa^c_2 2) when mu = 0.01 a) the activity coefficients for H_2PO^-_4, HPO^2-_4, and H^+ b) the equilibrium constant, Ka_2 c) the equilibrium constant corrected for ionic strength, Ka^c_2 d_ pKa_2 and pKa^c_2 key: Ka_2 = {H^+} {HPO^2+_}/{H_2PO^-_4} = gamma [H^+] gamma_H [HPO^2-_4]/gamma [H_2PO^-_4] [H^+] [HPO^2-_4]/[H_2PI^-] = (gamma_H_2PO^-_4)/(gamma_N) (gamma_HPO) Ka_2 = Ka^_2 - log Ka^_2 = pKa^_2

Explanation / Answer

1.

1)

C (s) + CO2 (g)    <——>       2CO(g)

Solid is not considered while writing Kc

So,

Kc = [CO]^2 / [CO2]

2)

HNO3 (aq) + H2O (l) <—> H3O+ (aq) + NO3- (aq)

Liquid is not considered while writing Kc

So,

Kc = [H3O+][NO3-] / [HNO3]

3)

CaCO3 (s) <—> CaO (s) + CO2 (g)

Kc = [CO2]

4)

BaO(s) + CO2 (g) <——> BaCO3 (s)

Kc = 1/[CO2]

I am allowed to answer only 1 question at a time

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