Question

a. Calculate (i) the standard enthalpy change, AH0 (ii) the standard entropy change, ASO (iii) the standard free energy change at 30°C and 500°C for the reaction: Fe203(s) + 3H2(g) ------> 2Fe(s) + 3H2O(g) Given that the standard entropies of Fe203(s), H2, Fe, and H20 are 90, 131, 27 and 189J/K/mol respectively and the standard enthalpies are -822, 0.0, 0.0, and -242KJ/mol respectively​

Answer 1

(i) ΔH° = [2(0.0) + 3(−242)] − [−822 + 3(0.0)] = −1644 + 822 = −822 kJ/mol

(ii) ΔS° = [2(27) + 3(189)] − [90 + 3(131)] = 621 − 483 = 138 J/K/mol

(iii) ΔG° = −822 − (773)(0.138) = −928 kJ/mol

Step-by-step explanation:

• To calculate the standard enthalpy change (ΔH°), we need to use the formula: ΔH° = ΣnΔH° f (products) − ΣmΔH° f (reactants), where n and m are the stoichiometric coefficients and ΔH° f are the standard enthalpies of formation of the compounds. Using the given values, we get: ΔH° = [2(0.0) + 3(−242)] − [−822 + 3(0.0)] = −1644 + 822 = −822 kJ/mol
• To calculate the standard entropy change (ΔS°), we need to use the formula: ΔS° = ΣnS°(products) − ΣmS°(reactants), where n and m are the stoichiometric coefficients and S° are the standard entropies of the compounds. Using the given values, we get: ΔS° = [2(27) + 3(189)] − [90 + 3(131)] = 621 − 483 = 138 J/K/mol
• To calculate the standard free energy change (ΔG°), we need to use the formula: ΔG° = ΔH° − TΔS°, where T is the absolute temperature in kelvins. For 30°C, which is equivalent to 303 K, we get: ΔG° = −822 − (303)(0.138) = −863 kJ/mol12. For 500°C, which is equivalent to 773 K, we get: ΔG° = −822 − (773)(0.138) = −928 kJ/mol