Question

CHEMWORK The preparation of NO2(g) from N2(g) and O2(g) is an endothermic reaction: N2(g) + O2(g) + NO2(g) (unbalanced) The enthalpy change of reaction for the balanced equation (with lowest whole-number coefficients) is AH = 67.7 kJ. If 304 ml N2 (9) at 100ºC and 3.32 atm and 410 ml 0,(9) at 100°C and 3.32 atm are mixed, what amount of heat is necessary to synthesize the maximum yield of NO2(g) ? ​

Answer 1

Heat required : 1.523 kJ

Further explanation

Reaction

N₂+2O₂⇒2NO₂ AH = 67.7 kJ

mol of N₂:(use Pv=nRT)

`\tt n=(3.32* 0.304)/(0.082* 373)=0.033`

mol of O₂:

`\tt n=(3.32* 0.41)/(0.082* 373)=0.045`

Limiting reactant :

N₂ : O₂

`\tt (0.033)/(1)/ (0.045)/(2)=0.033:0.0225`

Limiting reactant : O₂(smaller ratio)

mol NO₂ = mol O₂ = 0.045

2 mol ⇒ 67.7 kJ, so for 0.045 mol, heat required :

`\tt (0.045)/(2)* 67.7=1.523~kJ`