Given the reactions below, answer the following questions.
cl_2(g) + f_2(g) rlhar 2clf(g) delta g degree_rxn = 115.4 kj/mol
cl_2(g) + br_2(g) rlhar 2clbr(g) delta g degree_rxn = -2.0 kj/mol
calculate the delta g degree_rxn for 2clf(g) + br_2(g) rlhar 2clbr(g) + f_2(g) kj/mol

Explanation:

Gibbs free energy is an additive property.

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So, standard gibbs free enrgy change for the given reaction is -117.4 kJ/mol

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.

it is a solid

explanation: