Answer:
Rate constant k = 1.57*10⁻⁵ s⁻¹
Explanation:
Given reaction:
[tex]A\rightarrow B[/tex]
Expt [A] M [B] M Rate [M/s]
1 3.40 4.16 1.82*10^-4
2 4.59 4.16 3.32*10^-4
3. 3.40 5.46 1.82*10^-4
[tex]Rate = k[A]^{x}[B]^{y}[/tex]
where k = rate constant
x and y are the orders wrt to A and B
To find x:
Divide rate of expt 2 by expt 1
[tex]\frac{3.32*10^{-4} }{1.82*10^{-4} } =\frac{[4.59]^{x} [4.16]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\x =2[/tex]
To find y:
Divide rate of expt 3 by expt 1
[tex]\frac{1.82*10^{-4} }{1.82*10^{-4} } =\frac{[3.40]^{x} [5.46]^{y} }{[3.40]^{x} [4.16]^{y} }\\\\y =0[/tex]
Therefore: x = 2, y = 0
[tex]Rate = k[A]^{2}[B]^{0}[/tex]
To find k
Use rate for expt 1:
[tex]k = \frac{Rate1}{[A]^{2} } =\frac{1.82*10^{-4}M/s }{[3.40]^{2} } =1.57*10^{-5} s-1[/tex]
