Question

The initial rate data at 25 oC are listed for the reaction

NH4+ (aq) + NO2-1 -> N2(g) + H2O(l)

Experiment Initial [NH4+1] Initial [NO2-1] Initial rate (M/s)
1 0.24 0.1 7.2 X10^-6
2 0.12 0.1 3.6 X 10^-6
3 0.12 0.15 5.4 X 10^-6

a. Determine the rate law
b. Determine the value of the rate constant.

c. What is the reaction rate when the concentrations are [NH4+1] = 0.39 M and [NO2-1] = 0.052 M.

Answer 1

a) The rate law of the reaction is ;

`=k[NH_4^(+)]^1[NO_2^(-)]^1`

b) The value of the rate constant :
`0.0003 M^(-1) s^(-1)`

c)
`6.084* 10^(-6) M/s` is the reaction rate when the concentrations are
`[NH_4^(+)]` = 0.39 M and
`[NO_2^(-)]` = 0.052 M.

Step-by-step explanation:

Rate of the reaction will be given by :

`=k[NH_4^(+)]^x[NO_2^(-)]^y`

a) Rate of the reaction when
`[NH_4^(+)]` and
`[NO_2^(-)]` is 0.24 M and 0.1 M respectively = R

R =
`7.2* 10^(-6) M/s`

`R=k[0.24 M]^x[0.1 M]^y`...[1]

Rate of the reaction when
`[NH_4^(+)]` and
`[NO_2^(-)]` is 0.12 M and 0.1 M respectively = R'

R' =
`3.6* 10^(-6) M/s`

`R'=k[0.12 M]^x[0.1 M]^y`...[2]

Rate of the reaction when
`[NH_4^(+)]` and
`[NO_2^(-)]` is 0.12 M and 0.15 M respectively = R'

R'' =
`5.4* 10^(-6) M/s`

`R''=k[0.12 M]^x[0.15 M]^y`...[3]

Dividing [1] and [2]

`(R)/(R')=(k[0.24 M]^x[0.1]^y)/(k[0.12 M]^x[0.1]^y)`

`(7.2* 10^(-6) M/s)/(3.6* 10^(-6) M/s)=(k[0.24 M]^x[0.1 M]^y)/(k[0.12 M]^x[0.1 M]^y)`

`2=2^x`

x = 1

Dividing [2] and [3]

`(R')/(R'')=(k[0.12 M]^x[0.1 M]^y)/(k[0.12 M]^x[0.15]^y)`

`(3.6* 10^(-6) M/s)/(5.4* 10^(-6) M/s)=(k[0.12 M]^x[0.1 M]^y)/(k[0.12 M]^x[0.15 M]^y)`

`(2)/(3)=((2)/(3))^y`

y = 1

The rate law of the reaction is ;

`=k[NH_4^(+)]^1[NO_2^(-)]^1`

b) The value of the rate constant :

From [1]:

`7.2* 10^(-6) M/s=k[0.24 M]^1[0.1 M]^1`...[1]

`k=(7.2* 10^(-6) M/s)/([0.24 M]^1[0.1 M]^1)=0.0003 M^(-1) s^(-1)`

The value of the rate constant :
`0.0003 M^(-1) s^(-1)`

c) Rate of the reaction when
`[NH_4^(+)]` and
`[NO_2^(-)]` is 0.39 M and 0.052 M respectively = R

`R=0.0003 M^(-1) s^(-1)* [0.39 M][0.052 M]`

`R=6.084* 10^(-6) M/s`

`6.084* 10^(-6) M/s` is the reaction rate when the concentrations are
`[NH_4^(+)]` = 0.39 M and
`[NO_2^(-)]` = 0.052 M.