A chemistry graduate student is studying the rate of this reaction: NH4OH(aq)→NH3(aq)+H2O(aq)

She fills a reaction vessel with and measures its concentration as the reaction proceeds:

Time (minutes) NH4OH
0 0.200M
1.0 0.0895M
2.0 0.577M
3.0 0.0426M
4.0 0.0337M

Use this data to answer the following questions.

a. Write the rate law for this reaction.
b. Calculate the value of the rate constant.

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:

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)

The value is constant, so this is the true value of k.

Hope this helps

i think so it’s will be b yes i believe its

the answer is d. 74.02