Question

The enthalpy of combustion of carbon, hydrogen and sucrose are –393.5, –286.2 and –5644.2 kJ/mol respectively. Calculate the enthalpy of formation of sucrose :-
1. –6323.9 kJ
2. –2226 kJ
3. +2226 kJ
4. can't predict
Fastest correct answer will be marked as brainliest.

Answer 1

Answer : 2

Explanation:

This question can be solved using Hess law. See the image for better understanding.

Answer 2

Given: Enthalpy of combustion of carbon = -393.5 kJ/mole

           Enthalpy of combustion of hydrogen = -286.2 kJ/mole

           Enthalpy of combustion 0f sucrose = -5644.2 kJ/mole

To find: Enthalpy of formation of sucrose

Solution :

Enthalpy of combustion of  the following equation :

C + O₂ → CO₂                     ΔH₁ = -393.5 kJ/mole

H₂ + 1/2O₂→ H₂O                ΔH₂ = -286.2kJ/mole    

C₁₂H₂₂O₁₁ + 12O₂→ 12CO₂ + 11H₂O      ΔH₃ = -5644.2 kJ/mole

Now , enthalpy of formation of sucrose can be written as -

12C + 11H₂ + 11/2O₂ → C₁₂H₂₂O₁₁

Using Hess law, we can get this equation as

C + O₂ → CO₂    ×12

H₂ + 1/2O₂→ H₂O     ×11

C₁₂H₂₂O₁₁ + 12O₂ → 12CO₂ + 11H₂0       ×-1

Now, ΔH₁×12 + ΔH₂×11  - 1×ΔH₃

     = 12×(-393.5) + 11×(-286.2) - 1×(-5644.2)

     = -2226kJ

Therefore, the correct option is 3 which is -2226 kJ