To calculate the standard enthalpy of combustion for the given reaction, we need to know the standard enthalpies of formation (∆Hf°) for all the reactants and products involved in the reaction. The standard enthalpy of combustion (∆Hc°) can be calculated using the following formula:

∆Hc° = [∑ ∆Hf°(products)] - [∑ ∆Hf°(reactants)]

The standard enthalpies of formation for glucose (C6H12O6), oxygen (O2), carbon dioxide (CO2), and water (H2O) are -1273.3 kJ/mol, 0 kJ/mol, -393.5 kJ/mol, and -285.8 kJ/mol respectively.

Substituting these values into the formula, we get:

∆Hc° = [6(-393.5 kJ/mol) + 6(-285.8 kJ/mol)] - [-1273.3 kJ/mol + 6(0 kJ/mol)]
= [-2361 kJ + -1714.8 kJ] - [-1273.3 kJ]
= -4075.8 kJ + 1273.3 kJ
= -2802.5 kJ/mol

So, the standard enthalpy of combustion for the given reaction is -2802.5 kJ/mol.

Question

To calculate the standard enthalpy of combustion for the given reaction, we need to know the standard enthalpies of formation (∆Hf°) for all the reactants and products involved in the reaction. The standard enthalpy of combustion (∆Hc°) can be calculated using the following formula:

∆Hc° = [∑ ∆Hf°(products)] - [∑ ∆Hf°(reactants)]

The standard enthalpies of formation for glucose (C6H12O6), oxygen (O2), carbon dioxide (CO2), and water (H2O) are -1273.3 kJ/mol, 0 kJ/mol, -393.5 kJ/mol, and -285.8 kJ/mol respectively.

Substituting these values into the formula, we get:

∆Hc° = [6(-393.5 kJ/mol) + 6(-285.8 kJ/mol)] - [-1273.3 kJ/mol + 6(0 kJ/mol)]
     = [-2361 kJ + -1714.8 kJ] - [-1273.3 kJ]
     = -4075.8 kJ + 1273.3 kJ
     = -2802.5 kJ/mol

So, the standard enthalpy of combustion for the given reaction is -2802.5 kJ/mol.

Knowee AI · Accepted Answer

To calculate the standard enthalpy of combustion for the given reaction, we need to know the standard enthalpies of formation (∆Hf°) for all the reactants and products involved in the reaction. The standard enthalpy of combustion (∆Hc°) can be calculated using the following formula:

∆Hc° = [∑ ∆Hf°(products)] - [∑ ∆Hf°(reactants)]

The standard enthalpies of formation for glucose (C6H12O6), oxygen (O2), carbon dioxide (CO2), and water (H2O) are -1273.3 kJ/mol, 0 kJ/mol, -393.5 kJ/mol, and -285.8 kJ/mol respectively.

Substituting these values into the formula, we get:

∆Hc° = [6(-393.5 kJ/mol) + 6(-285.8 kJ/mol)] - [-1273.3 kJ/mol + 6(0 kJ/mol)]
     = [-2361 kJ + -1714.8 kJ] - [-1273.3 kJ]
     = -4075.8 kJ + 1273.3 kJ
     = -2802.5 kJ/mol

So, the standard enthalpy of combustion for the given reaction is -2802.5 kJ/mol.

Question 16Calculate the standard enthalpy of combustion for the following reaction:C6H12O6(s) + 6O2(g) --> 6CO2(g) + 6H2O(l)

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