Answer :
according to the second order formula:
1/[At] = K t + 1/[Ao]
and when we have the K constant =0.0265 & we have t = 180 min & we have the initial concentration of A = 4.25 so by substitution:
1/[At] = 0.0265 X 180min + 1/4.25
1/[At] = 5
∴[At] = 1/5 = 0.2 m
1/[At] = K t + 1/[Ao]
and when we have the K constant =0.0265 & we have t = 180 min & we have the initial concentration of A = 4.25 so by substitution:
1/[At] = 0.0265 X 180min + 1/4.25
1/[At] = 5
∴[At] = 1/5 = 0.2 m
Answer : The final concentration was 0.199 M
Explanation :
The expression used for second order kinetics is:
[tex]kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}[/tex]
where,
k = rate constant = [tex]0.0265M^{-1}min^{-1}[/tex]
t = time = 180.0 min
[tex][A_t][/tex] = final concentration = ?
[tex][A_o][/tex] = initial concentration = 4.25 M
Now put all the given values in the above expression, we get:
[tex]0.0265\times 180.0=\frac{1}{[A_t]}-\frac{1}{4.25}[/tex]
[tex][A_t]=0.199M[/tex]
Therefore, the final concentration was 0.199 M