Question

The enthalpy change of reaction for the balanced N2(g)+O2(g)--> No2(g) (Unbalanced) equation (with lowest whole-number coefficients) is ∆? = 67.7 kJ. If 2.50 x 103 mL N2(g) at 100.0 C and 2.50 atm and 4.50 x 103 mL O2(g) at 100.0 C and 2.50 atm are mixed, what amount of heat is necessary to synthesize the maximum yield of NO2(g)?

Answer 1

We know that ideal gas equation is PV = nRT

Where

T = Temperature = 100Oc= 100+273 = 373 K

P = pressure = 2.50atm

n = No . of moles = ?

R = gas constant = 0.0821 L atm / mol - K

V= Volume of the gas =

For N2 , n =(PV)/(RT) = 0.204 mol

For O2 , n' =(PV)/(RT) = 0.367 mol

The balanced reaction is N2(g) + 2O2(g) ---> 2NO2(g) : ∆? = 67.7 kJ

2 moles of O2 reacts with 1mole of N2

0.367 moles of O2 reacts with 0.367/2=0.184 moles of N2

So 0.204-0.184 = 0.02 moles of N2 left unreacted which is excess reactant

Since all of O2 completly reacted O2 is limiting reactant

2 moles of O2 requires 67.7 kJ

0.367 moles of O2 requires (0.367x67.7)/2 =12.4 kJ