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.

Respuesta :

Answer:

a) The rate law of the reaction is ;

[tex]=k[NH_4^{+}]^1[NO_2^{-}]^1[/tex]

b) The value of the rate constant : [tex]0.0003 M^{-1} s^{-1}[/tex]

c) [tex]6.084\times 10^{-6} M/s[/tex] is the reaction rate when the concentrations are [tex][NH_4^{+}][/tex] = 0.39 M and [tex][NO_2^{-}][/tex] = 0.052 M.

Explanation:

Rate of the reaction will be given by :

[tex]=k[NH_4^{+}]^x[NO_2^{-}]^y[/tex]

a) Rate of the reaction when[tex][NH_4^{+}][/tex]  and [tex][NO_2^{-}][/tex] is 0.24 M and 0.1 M respectively = R

R = [tex]7.2\times 10^{-6} M/s[/tex]

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

Rate of the reaction when[tex][NH_4^{+}][/tex]  and [tex][NO_2^{-}][/tex] is 0.12 M and 0.1 M respectively = R'

R' = [tex]3.6\times 10^{-6} M/s[/tex]

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

Rate of the reaction when[tex][NH_4^{+}][/tex]  and [tex][NO_2^{-}][/tex] is 0.12 M and 0.15 M respectively = R'

R'' = [tex]5.4\times 10^{-6} M/s[/tex]

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

Dividing [1] and [2]

[tex]\frac{R}{R'}=\frac{k[0.24 M]^x[0.1]^y}{k[0.12 M]^x[0.1]^y}[/tex]

[tex]\frac{7.2\times 10^{-6} M/s}{3.6\times 10^{-6} M/s}=\frac{k[0.24 M]^x[0.1 M]^y}{k[0.12 M]^x[0.1 M]^y}[/tex]

[tex]2=2^x[/tex]

x = 1

Dividing [2] and [3]

[tex]\frac{R'}{R''}=\frac{k[0.12 M]^x[0.1 M]^y}{k[0.12 M]^x[0.15]^y}[/tex]

[tex]\frac{3.6\times 10^{-6} M/s}{5.4\times 10^{-6} M/s}=\frac{k[0.12 M]^x[0.1 M]^y}{k[0.12 M]^x[0.15 M]^y}[/tex]

[tex]\frac{2}{3}=(\frac{2}{3})^y[/tex]

y = 1

The rate law of the reaction is ;

[tex]=k[NH_4^{+}]^1[NO_2^{-}]^1[/tex]

b)  The value of the rate constant :

From [1]:

[tex]7.2\times 10^{-6} M/s=k[0.24 M]^1[0.1 M]^1[/tex]...[1]

[tex]k=\frac{7.2\times 10^{-6} M/s}{[0.24 M]^1[0.1 M]^1}=0.0003 M^{-1} s^{-1}[/tex]

The value of the rate constant : [tex]0.0003 M^{-1} s^{-1}[/tex]

c) Rate of the reaction when[tex][NH_4^{+}][/tex]  and [tex][NO_2^{-}][/tex] is 0.39 M and 0.052 M respectively = R

[tex]R=0.0003 M^{-1} s^{-1}\times [0.39 M][0.052 M][/tex]

[tex]R=6.084\times 10^{-6} M/s[/tex]

[tex]6.084\times 10^{-6} M/s[/tex] is the reaction rate when the concentrations are [tex][NH_4^{+}][/tex] = 0.39 M and [tex][NO_2^{-}][/tex] = 0.052 M.