Question

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.

Answer 1

A on edge

Step-by-step explanation:

Answer 2

Answer: The enthalpy of reaction is -900.8 kJ

Step-by-step 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* \Delta H^o_f_((H_2O(g))))+(4* \Delta H^o_f_((NO(g))))]-[(4* \Delta H^o_f_((NH_3(g))))+(5* \Delta H^o_f_((O_2(g))))]`

We are given:

`\Delta H^o_f_((H_2O(g)))=-241.8kJ/mol\\\Delta H^o_f_((O_2(g)))=0kJ/mol\\\Delta H^o_f_((NH_3(g)))=-46.2kJ/mol\\\Delta H^o_((NO(g)))=91.3kJ`

Putting values in above equation, we get:

`\Delta H^o_(rxn)=[(6* (-241.8))+(4* (91.3))]-[(4* (-46.2))})+(5* (0))]`

`\Delta H^o_(rxn)=-1085.6kJ-(-184.8)kJ=-900.8kJ`

The enthalpy of reaction is -900.8 kJ