Nitrogen and hydrogen combine to form ammonia in the haber process.

calculate (in kj) the standard enthalpy change image from custom entry toolh for the reaction written below, using the bond energies given.

n2(g) + 3h2(g) > 2nh3 (g)

bond energy (kj/mol): nimage from custom entry tooln 945; h - h 432; n h 391

-105 kJ

Explanation:

The enthalpy change of a reaction is the sum of the energy of the bonds of the reactants and the products. The bonds at the reactants are being broken, so it's an endothermic reaction, so the bond energy must be positive.

The bonds at the products are being formed, so the process is exothermic, and the bond energy must be negative. There are being broken 1 N≡N bond and 3 H-H bonds, and are being formed 6 N-H bonds:

Reactants: 945 + 3*432 = 2241 kJ

Products: 6*(-391) = -2346 kJ

ΔH = 2241 - 2346

ΔH = -105 kJ

answer: - 3.) the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.

solution: - as per the rules, oxidation number of oxygen in its compounds is -2. the sum of oxidation numbers of all the elements of a molecule or compound is zero.

also, the oxidation number in elemental form is zero.

in no, the oxidation number of o is -2 so the oxidation number of n is +2 so that the sum is zero.

similarly, the oxidation number of c in co is +2 and o is -2.

on product side the oxidation number of n is 0 as it is in the elemental form() .

oxidation number of c in carbon dioxide is +4 as there are two oxygen in it and we know that each oxygen is -2.

so, oxidation number of n is changing from +2 to 0 and the oxidation number of c is changing from +2 to +4.

since nitrogen is reduced and hence no is acting as an oxidizing agent. carbon is oxidized and so co is acting as a reducing agent.

so, the right choice is number 3.)the oxidation state of nitrogen in no changes from +2 to 0, and the oxidation state of carbon in co changes from +2 to +4 as the reaction proceeds.