Find the ΔG° for the combustion of half a mole of glucose C6H12O6 in kJ

-1435 kJ

Explanation:

The Gibbs free energy, ΔG is an indication the amount of (useful) work that can be obtained from a chemical reaction involving reacting species and it is negative for spontaneous reaction

ΔG° = ΔH° - TΔS

ΔG = ΔG° + RT㏑Q

ΔG = G (reactants) - G(products)

Given that the combustion reaction of glucose is as follows;

C₆H₁₂O₆ + 6O₂ → 6CO₂ + 6H₂O

The above reaction has a ΔG° value of -2870 kJ

Therefore, the free energy for half mole of glucose = -2870 kJ/2 = -1435 kJ

n+h→nh3

we must remember that nitrogen and hydrogen are both diatomic molecules in their standard gas form. this adjusts the equation to

n2+h2→nh3

now we need to adjust coefficients in order to balance the atoms on each side of the equation. currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.

we can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.

n2+3h2→2nh3

this gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.