Question

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 1

c...........................................................

Answer 2

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.