Answer to Question #87469 in General Chemistry for jay

Question

Given the standard enthalpy changes for the following two reactions:

(1) N2(g) + 2O2(g)N2O4(g)...... ΔH° = 9.2 kJ

(2) 2N2O(g)2N2(g) + O2(g)......ΔH° = -164.2 kJ

what is the standard enthalpy change for the reaction:

(3) 2N2O(g) + 3O2(g) -------> 2N2O4(g) ΔH° = ?

__________ kJ

Answer 1

Solution.

delta H reaction = (2)+(1)*2

2N2O - 2N2 - O2 + 2N2 + 4O2 - 2N2O4

2N2O + 3O2 = 2N2O4

delta H (O2) = 0

delta H reaction = -164.2 + 9.2*2 = -145.8 kJ

Answer:

delta H reaction = -164.2 + 9.2*2 = -145.8 kJ