Ammonia, NH3 (Delta.Hf = –46.2 kJ), reacts with oxygen to produce water (Delta.Hf = –241.8 kJ) and nitric oxide, NO (Delta.Hf =

Question

2 years ago

3

Ammonia, NH3 (Delta.Hf = –46.2 kJ), reacts with oxygen to produce water (Delta.Hf = –241.8 kJ) and nitric oxide, NO (Delta.Hf =

91.3 kJ), in the following reaction: 4 upper N upper H subscript 3 (g) plus 5 upper O subscript 2 (g) right arrow 6 upper H subscript 2 upper O (g) plus 4 upper N upper O (g). What is the enthalpy change for this reaction? Use Delta H r x n equals the sum of delta H f of all the products minus the sum of delta H f of all the reactants.

Chemistry

2 answers:

skelet666 [1.2K] · Answer 1 · 2023-05-15T12:19:01+03:00

skelet666 [1.2K]2 years ago

6 0

Answer: The enthalpy of reaction is -900.8 kJ

Explanation:

The chemical equation is as follows:

$4NH_3(g)+5O_2(g)\rightarrow 6H_2O(g)+4NO(g)$

The equation for the enthalpy change of the above reaction is:

$\Delta H^o_{rxn}=[(6\times \Delta H^o_f_{(H_2O(g))})+(4\times \Delta H^o_f_{(NO(g))})]-[(4\times \Delta H^o_f_{(NH_3(g))})+(5\times \Delta H^o_f_{(O_2(g))})]$