Question

The reaction 2NO(g) + O2(g) ⇌ 2NO2(g) has Kp = 1.65 × 1011 at 25°C. To a 1.00 L flask, 1.03 g of NO and 532 mL of O2 measured at 29°C and 772 torr are mixed. When the mixture comes to equilibrium at 25°C, what is the concentration of NO(g)? If needed, use “E” for scientific notation. Do not enter units as part of your answer. Report your answer to the correct number of significant figures.

Answer 1

molar mass of NO= mass of NO/Molar mass= 1.03/28= 0.037 moles

moles of oxygen can be calculated from gas law equation

n=PV/RT, R= 0.0821 L.atm/mole.K, P= pressrue in atm= 772 Torr= 772/760 atm =1.015 atm

V= 532ml, V in L= 532/1000L=0.532 L, T= 29 deg.c= 29+273= 302K

moles of oxygen= 1.015*0.532/(0.0821*302)= 0.02178

Concentrations of gases initially = moles/L

Concentrations : NO= 0.037/1=0.037M, O2=0.02178/1=0.02178M

since KP=KC*(RT)^delta. deltan= change in no of moles during the reaction= 2-3=-1

Kp=KC*(0.0821*(25+273)^-1

KC=KP*24.46=1.65*10¹¹* 24.46=40.4*10¹¹

let x= drop in concentration of Oxygen

At equilibrium [NO]=0.037-2x, [O2]=0.02178-x, [NO2]=2x

hence KC= [NO2]²/ [NO]² [O2] = 4x²/(0.037-2x)²*(0.02178-x)=40.4*10¹¹

when solved using excel, x= 0.01849984 M, [NO]=0.037-2*0.01849984= 3.2*10^-7 M