What is the change in enthalpy for the following reaction?
4nh3(g) + 302(g) 2n2(g) + 6h2o(l)
given:
nh3: ∆h= -46.2 kj
given: h2o: ∆h= -286 kj
a. -1530.4 kj
b. -1020.3 kj
c. 1530.4 kj
d. 1020.3 kj
e. -510.1 kj

Answer;

A. -1530.4 kJ

Explanation;Enthalpy change is the amount of heat evolved or absorbed in a chemical reaction at a constant pressure. It is abbreviated as delta H or ΔH.The enthalpy change is a result of the difference between the sum of internal energy of the products and the sum of internal energy of the reactants.

Since; NH3: ∆H= -46.2 kJ

          H2O: ∆H= -286 kJ

Then; ΔH = 4(-46.2) + 3(-286)

      = -1530.4 kJ

This means the reaction is exothermic as ΔH is negative.