Question

calculate the standard enthalpy of combustion. The standard enthalpy of formation of sucrose is - 2226.1kj/mol

Answer 1

that would be 2226.109837560 sorry if i'm wrong have a nice day.

Answer 2

The question is incomplete, here is the complete question:

Calculate the standard enthalpy of combustion of sucrose. The standard enthalpy of formation of sucrose is -2226.1 kJ/mol

Answer: The standard enthalpy of combustion of sucrose is -5636.52 kJ

Step-by-step explanation:

Combustion reaction is defined as the chemical reaction in which a hydrocarbon reacts with oxygen gas to produce carbon dioxide gas and water molecule.

`\text{Hydrocarbon}+O_2\rightarrow CO_2+H_2O`

Enthalpy change is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. The equation used to calculate enthalpy change is of a reaction is:

`\Delta H^o_(rxn)=\sum [n* \Delta H_f^o_((product))]-\sum [n* \Delta H_f^o_((reactant)])`

The chemical equation for the combustion of sucrose follows:

`C_(12)H_(22)O_(11)(s)+12O_2(g)\rightarrow 12CO_2(g)+11H_2O(l)`

The equation for the enthalpy change of the above reaction is:

`\Delta H^o_(rxn)=[(12* \Delta H_f^o_((CO_2)))+(11* \Delta H_f^o_((H_2O)))]-[(1* \Delta H_f^o_{(C_(12)H_(22)O_(11))})+(12* \Delta H_f^o_((O_2)))]`

We are given:

`\Delta H_f^o_((H_2O))=-285.8kJ/mol\\\Delta H_f^o_((CO_2))=-393.51kJ/mol\\\Delta H_f^o_{(C_(12)H_(22)O_(11))}=-2226.1kJ/mol\\\Delta H_f^o_((O_2))=0kJ/mol`

Putting values in above equation, we get:

`\Delta H^o_(rxn)=[(12* (-393.51))+(11* (-285.8))]-[(1* (-2226.1))+(12* (0))]\\\\\Delta H^o_(rxn)=-5636.52kJ`

Hence, the standard enthalpy of combustion of sucrose is -5636.52 kJ