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) ?

The heat required to produce the maximum yield of NO2 is 1.52 kJ.

The balanced reaction equation is;

N2(g) + 2 O2(g) -> 2 NO2(g)

The number of moles of each gas is obtained from the ideal gas equation as follows;

For N2;

P = 3.32 atm

V = 304 ml or 0.304 L

T = 100ºC + 273 = 373 K

n = ?

R = 0.082 atmLK-1mol-1

From;

PV = nRT

n = PV/RT

n = 3.32 atm × 0.304 L/0.082 atmLK-1mol-1 × 373 K

n = 0.033 moles

For O2;

PV = nRT

n =  PV/RT

V = 410 ml or 0.410 L

P = 3.32 atm

R = 0.082 atmLK-1mol-1

T = 100°C + 273 = 373 K

n =  3.32 atm × 0.410 L/0.082 atmLK-1mol-1 × 373 K

n = 0.045 moles

1 mole of N2 reacts with 2 moles of O2

0.033 moles of N2 reacts with 0.033 moles × 2 moles/1 mole

= 0.066 moles

2 moles of O2 yields 2 moles of NO2

0.045 moles O2 yields 0.045 moles × 2 moles/2 moles = 0.045 moles

If 2 mole of NO2 requires 67.7 kJ

0.045 moles of NO2 requires 0.045 moles  × 67.7 kJ/2 mole

= 1.52 kJ

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Heat required : 1.523 kJ

Reaction

N₂+2O₂⇒2NO₂ AH = 67.7 kJ

mol of N₂:(use Pv=nRT)

mol of O₂:

Limiting reactant :

N₂ : O₂

Limiting reactant : O₂(smaller ratio)

mol NO₂ = mol O₂ = 0.045

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