Question

Enthalpy of formation (kJ/mol) C6H12O6(s)-1260 O2 (g)0 CO2 (g)-393.5 H2O (l)-285.8 Calculate the enthalpy of combustion per mole of C6H12O6. C6H12O6 (s) + 6 O2 (g) → 6 CO2 (g) + 6 H2O (l)

Answer 1

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Step-by-step explanation:

Answer 2

Answer : The enthalpy of combustion per mole of
`C_6H_(12)O_6` is -2815.8 kJ/mol

Explanation :

Enthalpy change : It is defined as the difference in enthalpies of all the product and the reactants each multiplied with their respective number of moles. It is represented as
`\Delta H^o`

The equation used to calculate enthalpy change is of a reaction is:

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

The equilibrium reaction follows:

`C_6H_(12)O_6(s)+6O_2(g)\rightleftharpoons 6CO_2(g)+6H_2O(l)`

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

`\Delta H^o_(rxn)=[(n_((CO_2))* \Delta H^o_f_((CO_2)))+(n_((H_2O))* \Delta H^o_f_((H_2O)))]-[(n_{(C_6H_(12)O_6)}* \Delta H^o_f_{(C_6H_(12)O_6)})+(n_((O_2))* \Delta H^o_f_((O_2)))]`

We are given:

`\Delta H^o_f_{(C_6H_(12)O_6(s))}=-1260kJ/mol\\\Delta H^o_f_((O_2(g)))=0kJ/mol\\\Delta H^o_f_((CO_2(g)))=-393.5kJ/mol\\\Delta H^o_f_((H_2O(l)))=-285.8kJ/mol`

Putting values in above equation, we get:

`\Delta H^o_(rxn)=[(6* -393.5)+(6* -285.8)]-[(1* -1260)+(6* 0)]=-2815.8kJ/mol`

Therefore, the enthalpy of combustion per mole of
`C_6H_(12)O_6` is -2815.8 kJ/mol