Answer:
Rate = k[NH₄OH]²
k = 6.17
Explanation:
We have concentrations of NH₄OH along with the given times. To determine the rate law of the reaction we need to determine first the order of reaction. This reaction can be order zero, first or second order. The expressions for each are the following:
Zero order:
k = [A₀] - [A] / t
First order:
k = 1/t * ln([A₀]/[A])
Second order:
k = (1/t) * (1/[A₀] - 1/[A])
And from here, the next part is easier. We just need to determine hat order is, calculating the value of k at two different times. If the value of k is constant, then we can say that the reaction is of that order.
Let's suppose its order zero (t = 1 and t = 2, [A₀] = 0.200 M):
k1 = 0.2 - 0.0895 / 1 = 0.1105
k2 = 0.2 - 0.577 / 2 = -0.1885
From this results we can conclude it's not zero order.
Let's suppose its order 1:
k1 = ln(0.2/0.0895) / 1 = 0.8041
k2 = ln(0.2/0.577) / 2 = 0.1733
It's not first order either, so we can conclude that this reaction is of 2nd order and the rate law would be:
<h2>
Rate = k[NH₄OH]² </h2>
Now that we know it's a second order reaction, we can determine the value of k using its expression:
k = (1/t) (1/[A] - 1/[A₀])
k = ln(1/0.0895 - 1/0.2) (1/1)
k = 6.17
And to confirm this value, let's calculate k for t = 2 s
k = (1/2) (1/0.0577 - 1/0.2)
<h2>
k = 6.17</h2>
The value is constant, so this is the true value of k.
Hope this helps
