Answer:
The value of the constant is k = 0.042 M⁻¹ min⁻¹
Explanation:
The equation of velocity for this reaction can be written as follows:
v = k[A]²
where:
v = velocity
[A] = concentration of A ([A] = 0.065 mol / 0.1 l = 0.65 M)
The velocity of this reaction can also be written as follows:
v = -d[A]/dt
where d[A] and dt are infinitesimal variation of [A] and t respectively.
Then:
-d[A]/dt = k[A]² (separating variables)
-d[A]/[A]² = k dt
( integrating both sides from [A] (concentration fo A at time t) to [A]₀ (initial concentration) and from t = t to t = 0):
1/[A] - 1/[A]₀ = kt
Solving for k:
(1/[A] - 1/[A]₀) / t = k
Now we can replace with the data provided, for example:
at time t = 10 min, [A] = 0.051 mol / 0.1 l = 0.51 M
Then:
k = (1/0.51 M - 1/0.65 M) / 10 min = 0.042 M⁻¹ min⁻¹
The same should be done with every pair of data and the average of k can be calculated:
k = (1/0.42 M - 1/0.65) / 20 min = 0.042 M⁻¹ min⁻¹
k = (1/0.36 M - 1/0.65) / 30 min = 0.041 M⁻¹ min⁻¹
k = (1/0.31 M - 1/0.65) / 40 min = 0.042 M⁻¹ min⁻¹
The average value of k is:
k = 0.042 M⁻¹ min⁻¹
The same could be done plotting 1/[A] vs t and the slope of the straight line will be k (see attachment):
1/[A] = kt + 1/[A]₀
y = mx + b
