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The rate constant for the decomposition of gaseous NO2 to NO and O2 is 4.7/(M?s) at...

The rate constant for the decomposition of gaseous NO2 to NO and O2 is 4.7/(M?s) at 383 ?C. Consider the decomposition of a sample of pure NO2 having an initial pressure of 752mmHg in a 5.50L reaction vessel at 383 ?C.

It is a second order reaction.

B. What is the initial rate of formation of O2 in g/(L?s)? USE TWO SIGNIFICANT FIGURES

C. What is the mass of O2 in the vessel after a reaction time of 2.00min ?

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Answer #1

B. What is the initial rate of formation of O2 in g/(L?s)?

NO2 has initial pressure of 752mmHg 752/760 = 0.99 atm in a

5.50L reaction vessel at

383 ?C = 383+273 = 656K

R = 0.082 atm K-1 M-1

(C_NO2)0 = P/(R*T) = 0.99/(0.082*656) = 0.0184 M

rNO2/(-1) = rO2/(1/2)

It is a second order reaction.

rNO2 = -k*(C_NO2)^2

k = 4.7/(M?s)

MW of O2 = 32

rO2 =- 2*rNO2 = 2*4.7*(.0184)^2 = 0.0032 moles/(L . s) = 32*0.0032 gm/(L . s) = 0.102 gm/(L . s)

C. What is the mass of O2 in the vessel after a reaction time of 2.00min ?

(C_NO2)0 = 0.0184M

k = 4.7/(M?s)

t = 2min = 120s

1/C_NO2 = k*t + 1/(C_NO2)0 = 4.7*120 + 1/0.0184 = 618.35

C_NO2 = 0.00162M

NO2 reacted = .0184-.00162 = 0.0168M

NO2 = NO + 1/2O2

1M NO2 reacted to give 1/2 M O2

.0168M NO2 reacted to give 1/2*.0168 = 0.0084 M O2

Volume = 5.5L

Moles of O2 produced =5.5* .0084 = 0.046 mol

Mass of O2 produced = 32*.046 = 2.477gm

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