Question

please explain your solutions. thank you so much

3. The chemistry of nitrogen oxides is very versatile, Given the following reactions and their standard enthalpy changes (1) NO(g) + NO:(g) → N;O;(g) AH° = -39,8 kJ (2) NO(g) + NO:(g) 0:(g) → N;O:(g) AH° = -1125 kJ (3) 2NO:(g) → N;O(g) AH° --57.2 k (4) 2NO(g) + 02 (g) → 2NO:(g) AH° = -114.2 kJ AH° - 54.1 kJ (5) N,Os(s) (8FO°N Calculate the standard enthalpy change of reaction for (3 points) N;O,(g) + N;O(s) → 2N;O.g)

Answer 1

solution: e enthalpy is extensive property therefore we add directaly when adding the reaction eguation. Given reaction is. N₂O₂ (g) + N2 Os(s) 2 N204 g Several To get the above equation, we follow the steps which is given below. - First of all we revese the equation and and Add (sign will be changed N₂0319 No (q + NO2(g) 64°= 39.8kJ N2Os (9) NO(g) + NO2(g) + O2(g) on 112.sky them, we get, N2O3 (gj + N2 Os(9) ► 2009 + 2NO2(g) + O2(g), 04 = 152.3 kg Now, add the equation 6 6 and 4, we get 200 g) + O2lg 2 NO2G on=-114.2kg N20563) N₂O5 (9) onz54.1kJ N 03 (8) + N2 918 2NO(g) + 2NO, 19+02 g on' = 152.3 kg N203 (9) + N₂O5 (5) 4 NO2 ig , on=92.2kg #

Now, multiply the equation ② by 2 , we get, 4 Nolg) 2 N₂O4 (g) , who =2x-57.2kJ - 114.4 KJ (** finally add the equation # and **, we get the desine equation as. N₂O₂(g) + N₂O5 (8) 4 NO₂ ), on=92.2kg 4NO2(g) 2 N2O4 (9 042-116.4 kg 2N₂O4 g), TH=-22.2kg NO₂ (s) + N₂O5 (s) Ans