Question

Chemistry 452 2016-10-10 Glucose-Lactate Under anaarobic conditions glucose is broken down in the muscle tissue to form lactic acid according to the reaction Part D GGH1206 -2CHзсH (он соон Calculate the molar enthalny change the reaction at T-310K For answering the questions below the following data will be use valid at T-298K and P 101325 Pa usetul. These data a (JKmal Glucose -1273.1 219 2092 Lactic Acid 673.6 127.6 192.1 Value Units Note: Give your answer for Parts A-G to three significant fiqures. Request Answer Submit Part E Calculate the molar entropy change for the chemicl reaction at T-310K Part A T-298K Using the data in the table, caloulate the molar enthalpy change for the chemical reaction (i.e. defined as the "system" ( a Value Units Value Units Request Answer Submit Request Answer Submit Part F Part B Calculate the malar entropy change of the surroundings. Assume the surroundings are at a temperature of T-310K. Using the intormation in the table above, calculate the molar entropy change for e chemical reaction at T-298K. 23 Value Units Value Units Request Answer Submit Request Answer Submit Part G Part C Calculate the molar entropy change of the universe Usina the data in the table abpve, calculate the chanpe in mplar heat capacity for the chemical reaction ACe ( sa .

Physical Chemistry, Please show work so I can understand, Thank You!

Answer 1

Part-A:

H^o = (H_f^o of all products) - (H_f^o of all reactants)

=> H^o = [2 mol*H_f^o(CH3CH(OH)COOH)] - [1 mol*H_f^o(C6H12O6)]

=> H^o = [2 mol * (- 673.6 kJ/mol] - [1 mol * (- 1273.1 kJ/mol)]

=> H^o = - 1347.2 kJ + 1273.1 kJ

=> H^o = - 74.1 kJ (answer)

Part-B:

S^o = (S^o of all products) - (S^o of all reactants)

=> S^o = [2 mol*S^o(CH3CH(OH)COOH)] - [1 mol*S^o(C6H12O6)]

=> S^o = [2 mol * (192.1 J/mol.K] - [1 mol * (209.2J/mol.K)]

=> S^o = 384.2 J/K - 209.2J/K

=> S^o = 175 J/K (answer)

Part-C:

C_P = (C_P of all products) - (C_P of all reactants)

=> C_P = [2 mol*C_P(CH3CH(OH)COOH)] - [1 mol*C_P(C6H12O6)]

=> C_P = [2 mol * (127.6 J/mol.K] - [1 mol * (219.2J/mol.K)]

=> C_P = 255.2 J/K - 219.2J/K

=> C_P = 36.0 J/K (answer)

Part-D:

H(310 K) = H(298 K) + C_P * T

=> H(310 K) = - 74.1 kJ + 36.0 J/K * (310 - 298) K

=> H(310 K) = - 74.1 kJ + 432 J * (1kJ/1000 J)

=> H(310 K) = - 73.7 kJ (Answer)