6. Sodium carbonate can be made by heating sodium bicarbonate: 2 NaHCO3(s) ? NaC

Question

6. Sodium carbonate can be made by heating sodium bicarbonate: 2 NaHCO3(s) ? NaCols) co,(g) H2O(g) + + a. Calculate AH° using the standard enthalpies of formation. b. Predict the sign of AS. Calculate AS° using the standard entropies. Are your results consistent with your prediction? Support your answer with a 1 sentence explanation. c. Calculate the standard free energy using the answers from a. and b. d. Calculate AG using the standard free energies of formation d. Will the reaction be spontaneous at 25°C? Support your answer with a 1 sentence explanation. e. At what temperatures will the reaction be spontaneous? 7. Calculate the equilibrium constant Kp at 227°C for the following reaction: N2O,(g) 2 NO,(g)

Explanation / Answer

6.

2NaHCO3(s) --> Na2CO3(s) + CO2(g) + H2O(g)

a. dHo = dHo(products) - dHo(reactants)

            = (-1130.94 - 393.51 - 241.82) - (2 x -950.8)

            = 135.33 kJ/mol

b. dSo entropy change would be positive as we are forming gaseous products from solid reactant.

dSo = dSo(products) - dSo(reactants)

       = (135.98 + 213.68 + 188.72) - (2 x 101.7)

       = 334.98 J/K.mol

c. dGo = dHo - TdSo

            = 135.33 - 298 x 0.33498

            = 35.506 kJ/mol

d. dGo = dGo(products) - dGo(reactants)

            = (-1047.67 - 393.38 - 228.59) - (2 x -837.8)

            = 5.96 kJ/mol

The reaction is non-spontaneous as dGo is positive.

e. T = 135.33/0.33498 = 404 K

so above 404 K, the reaction would be spontaneous.

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7. dGo = dGo(products) - dGo(reactants)

            = (2 x 51.3) - (97.8)

            = 4.8 kJ/mol

dGo = -RTlnKp

T = 227 oC + 273 = 500 K

R = gas constant

So,

4800 = -8.314 x 500 lnKp

Kp = 0.315

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