Question

Sucrose, C12H22O11, is common table sugar. The enthalpy change at 25°C and 1 atm for the complete burning of 7 mol of sucrose in oxygen to give CO2(g) and H2O(l) is -3.949 × 10^4 kJ. From this and from data given below: ?Hf°(H2O(l)) = -285.8 kJ/mol ?Hf°(CO2(g)) = -393.5 kJ/mol ?Hf°(O2(g)) = 0 kJ/mol Calculate the standard enthalpy of formation of sucrose. ?Hf° = kJ/mol

          Sucrose, C12H22O11, is common table sugar. The enthalpy change at 25°C and 1 atm for the complete burning of 7 mol of sucrose in oxygen to give CO2(g) and H2O(l) is -3.949 × 10^4 kJ. From this and from data given below:
?Hf°(H2O(l)) = -285.8 kJ/mol
?Hf°(CO2(g)) = -393.5 kJ/mol
?Hf°(O2(g)) = 0 kJ/mol
Calculate the standard enthalpy of formation of sucrose.
?Hf° = kJ/mol

Sucrose, C12H22O11, is common table sugar. The enthalpy change at 25°C and 1 atm for the complete burning of 7 mol of sucrose in oxygen to give CO2(g) and H2O(l) is -3.949 × 10^4 kJ. From this and from data given below:
?Hf°(H2O(l)) = -285.8 kJ/mol
?Hf°(CO2(g)) = -393.5 kJ/mol
?Hf°(O2(g)) = 0 kJ/mol
Calculate the standard enthalpy of formation of sucrose.
?Hf° = kJ/mol

Added by Joshua I.

Chemistry: Structure and Properties

Nivaldo Tro 2nd Edition

Instant Answer

Solved by Expert David Collins

10/28/2022

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949 x 10^4 kJ for 7 moles of sucrose. Show more…

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Sucrose, C12H22O11, is common table sugar. The enthalpy change at 25°C and 1 atm for the complete burning of 7 mol of sucrose in oxygen to give CO2 (g) and H2O (l) is -3.949 x 10^4 kJ. From this and from data given below: ΔHf° (H2O(l)) = -285.8 kJ/mol ΔHf° (CO2(g)) = -393.5 kJ/mol ΔHf° (O2(g)) = 0 kJ/mol Calculate the standard enthalpy of formation of sucrose, ΔHf°, in kJ/mol.