Nitrogen dioxide is one of the many oxides of nitrogen (often collectively calle

Question

Nitrogen dioxide is one of the many oxides of nitrogen (often collectively called "NOx") that are of interest to atmospheric chemistry. It can react with itself to form another form of NOx, dinitrogen tetroxide A chemical engineer studying this reaction fills a 1.5 L flask at 11· with 0.51 atm of nitrogen dioxide gas. He then raises the temperature considerably, and when the mixture has come to equilibrium determines that it contains 0.29 atm of nitrogen dioxide gas. The engineer then adds another 0.26 atm of nitrogen dioxide, and allows the mixture to come to equilibrium again. Calculate the pressure of dinitrogen tetroxide after equilibrium is reached the second time. Round your answer to 2 significant digits. atm x10

Explanation / Answer

The reaction is 2NO2<--->N2O4

concentration of NO2 can be calculated from gas law, PV= nRT or n/V= P/RT= 0.51/(0.0821*(11+273)= 0.022M

moles of NO2= 0.022*V=0.022*1.5=0.033 moles

let x= partial pressure of N2O4 at equilibrium, let P =pressure of corresponding gas at equilibrium.

at equilibrum, PN2O4=x and PNO2= 0.51-2x=0.29

2x=0.51-0.29=0.22 , x=0.11

hence at equilibrium, PN2O4= 0.11 atm and PNO2= 0.29 atm

KP= equilibrium constant at equilibrium, = PN2O4/(PNO2)2= 0.11/(0.29*0.29)= 1.31

when 0.26 atm of NO2 is added, new equilibrium is established.

New PNO2=0.11+0.26=0.37

hence Q= reaction coefficient = PN2O4/ (PNO2)2= 0.11/(0.37*0.37)= 0.8<K. So the reaction proceeds in the forward reaction so as to increase K.

hence let x=change in pressure of N2O4 to reach equilibrium

At new equilibrium, PN2O4= 0.11+x and PNO2= 0.37-2x

Kp = 1.31= (0.11+x)/ (0.37-2x)2 when solved for x using excel, x= 0.0248 atm

new equilibrium , PN2O4= 0.11+0.0248=0.1348 atm

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