Heat required : 1.523 kJ
Reaction
N₂+2O₂⇒2NO₂ AH = 67.7 kJ
mol of N₂:(use Pv=nRT)
[tex]\tt n=\dfrac{3.32\times 0.304}{0.082\times 373}=0.033[/tex]
mol of O₂:
[tex]\tt n=\dfrac{3.32\times 0.41}{0.082\times 373}=0.045[/tex]
Limiting reactant :
N₂ : O₂
[tex]\tt \dfrac{0.033}{1}\div \dfrac{0.045}{2}=0.033:0.0225[/tex]
Limiting reactant : O₂(smaller ratio)
mol NO₂ = mol O₂ = 0.045
2 mol ⇒ 67.7 kJ, so for 0.045 mol, heat required :
[tex]\tt \dfrac{0.045}{2}\times 67.7=1.523~kJ[/tex]
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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