The chemistry of nitrogen oxides is very versatile. Consider the following react

Question

The chemistry of nitrogen oxides is very versatile. Consider the following reactions: NO_(g) + NO_2(g) rightarrow N_2O_3(g) Delta H degree_rxn = -39.8 kJ NO_(g) + NO_2(g) + O_2(g) rightarrow N_2O_5(g) Delta H degree_rxn = -112.5 kJ 2NO_2(g) rightarrow N_2O_4(g) Delta H degree_rxn = -57.2 kJ 2NO_(g) + O_2(g) rightarrow 2NO_2(g) Delta H degree_rxn = -114.2 kJ N_2O_5(g) rightarrow N_2O_5(g) Delta H degree sub = +54.1 kJ Calculate the heat of reaction for the following reaction and clearly state if the system is exothermic or endothermic: N_2O_3(g) + N_2O_5(g) rightarrow 2N_2O_4(g)

Explanation / Answer

Ans. Using Hess’s Law, the reaction N2O3(g) + N2O5 (s) ------> 2N2O4 can be written as the sum of following reactions-

N2O3(g) ---------> NO(g) + NO2(g)         ; dHrxn = +39.8 kJ       - Reverse of Rxn 1

N2O5(s) ---------> N2O5(g)                       ; dHrxn = +54.1 kJ       - Rxn 5

N2O5(g) --------> NO(g) + NO2(g) + O2(g); dHrxn = +112.5 kJ - Reverse of Rxn 2

4 NO2(g) --------> 2 N2O4                        ; dHrxn = - 114.4 kJ       -2 x (Rxn 3)

2NO(g) + O2(g) --> 2 NO2(g)                     ; dHrxn = -114.2 kJ         - Rxn 4

N2O3(g) + N2O5(s) ---> 2 N2O4(g)        ; dHrxn = ?                      - Rxn 6

dHrxn (Rxn 6) = + 39.8 kJ + 54.1 kJ + 112.5 kJ – 114.4 kJ – 114.2 kJ

                        = - 22.2 kJ

Since, dHrxn (Rxn 6) is “negative ( - 22.2 kJ)”, the reaction is exothermic, i.e. heat is released during reaction.

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.