88. Calculate the entropy of the reaction 2NaHCO3 (cr) → Na2CO3 (cr) H2O (g) CO2 (g) at temperatures of 298.15 and 500K based on the absolute entropies of substances at T = 298.15K and p = 1 atm .: S0Na2CO3 = 138, 80 J / mol ∙ K, S0H2O = 188.72 J / mol ∙ K, S0CO2 = 213.66 J / mol ∙ K, S0NaHCO3 = 102.10 J / mol ∙ K. The heat of the reaction participants is: Ср (Na2CO3) = 〖70.63 135.6 ∙ 10〗 ^ (- 3) TJ / mol ∙ K
Ср (H2O) = 〖30.00 10.71 ∙ 10〗 ^ (- 3) T 0.33 ∙ 〖10〗 ^ 5 T ^ (- 2) J / mol ∙ K
Ср (CO2) = 〖44.14 9.04 ∙ 10〗 ^ (- 3) T-8.54 ∙ 〖10〗 ^ 5 T ^ (- 2) J / mol ∙ K
Ср (NaHCO3) = 〖44.89 143.89 ∙ 10〗 ^ (- 3) T J / mol ∙ K
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