Question

The reaction 2a → a2​​​​​ was experimentally determined to be second order with a rate constant, k, equal to 0.0265 m–1min–1. if the initial concentration of a was 4.25 m, what was the concentration of a (in m) after 180.0 min?

Answer 1

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

Answer 2

Answer : The final concentration was 0.199 M

Explanation :

The expression used for second order kinetics is:

$kt=\frac{1}{[A_t]}-\frac{1}{[A_o]}$

where,

k = rate constant = $0.0265M^{-1}min^{-1}$

t = time = 180.0 min

$[A_t]$ = final concentration = ?

$[A_o]$ = initial concentration = 4.25 M

Now put all the given values in the above expression, we get:

$0.0265\times 180.0=\frac{1}{[A_t]}-\frac{1}{4.25}$

$[A_t]=0.199M$

Therefore, the final concentration was 0.199 M