Data are given for the decomposition ofN 2 O 5 in CCl 4 solution at45 o C. 2 N 2
ID: 686184 • Letter: D
Question
Data are given for the decomposition ofN2O5 in CCl4 solution at45oC.2 N2O5(sol) --> 4NO2 (sol) + O2(g)
The rate is determined by measuring the volume ofO2 produced.
Concentration Initial Rate
[N2O5]0(M) (M O2 s-2)
0.900 5.58 x10-4
0.450 2.79 x10-4
0.150 9.30 x10-5
Write the complete rate-law equation for this reaction andcalculate the rate constant.
I dont understand why we havethree different rates. I know the equation for the rateconstant...concentration of the products over the concentration ofthe reactants....which gives the K constant.... how do i even begin with thisdata?
I dont understand why we havethree different rates. I know the equation for the rateconstant...concentration of the products over the concentration ofthe reactants....which gives the K constant.... how do i even begin with thisdata?
Explanation / Answer
2 N2O5(sol) --> 4NO2 (sol) + O2 (g)The rate is determined by measuring the volume ofO2 produced.
Concentration Initial Rate
[N2O5]0(M) (MO2 s-2)
0.900 5.58 x10-4
0.450 2.79 x10-4
0.150 9.30 x10-5
By carefull observation of the above data when the conc.of N2O5 is reduced to half the rate of reaction is also reduced tohalf. Concentration : 0.9 / 0.45 =2 Rate : 5.58x10-4 / 2.79x10-4 = 2 In the second case also when the conc. of N2O5 is reducedto1/6 th or the initial conc. ,then the rate of reactionis also reduced to 1/6 th Concentration : 0.9 /0.15=6 Rate : 5.58x10-4 / 9.30x10-5 = 6 So, rate law is rate = K [N2O5] By carefull observation of the above data when the conc.of N2O5 is reduced to half the rate of reaction is also reduced tohalf. Concentration : 0.9 / 0.45 =2 Rate : 5.58x10-4 / 2.79x10-4 = 2 In the second case also when the conc. of N2O5 is reducedto1/6 th or the initial conc. ,then the rate of reactionis also reduced to 1/6 th Concentration : 0.9 /0.15=6 Rate : 5.58x10-4 / 9.30x10-5 = 6 So, rate law is rate = K [N2O5]
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