Question

a chemistry graduate student is studying the rate of this reaction: 2nh₃(g)→n₂(g)+3h₂(g). she fills a reaction vessel with nh₃ and measures its concentration as the reaction proceeds: time (seconds) nh₃ 0 0.200m 10. 0.121m 20. 0.0732m 30. 0.0443m 40. 0.0268m. use this data to answer the following questions. write the rate law for this reaction. rate = k . calculate the value of the rate constant k. round your answer to 2 significant digits. also be sure your answer has the correct unit symbol. k =

Explanation:

Step1: Determine the reaction order

To find the reaction order with respect to $NH_3$, we can check if the reaction is first - order by plotting $\ln([NH_3])$ vs time or second - order by plotting $\frac{1}{[NH_3]}$ vs time. For a first - order reaction, the integrated rate law is $\ln([A]_t)=\ln([A]_0)-kt$, and for a second - order reaction, it is $\frac{1}{[A]_t}=\frac{1}{[A]_0}+kt$. Let's check the first - order case.
Calculate $\ln([NH_3])$ for each time point:
At $t = 0$, $\ln(0.200)\approx- 1.609$
At $t = 10$, $\ln(0.121)\approx-2.10$
At $t = 20$, $\ln(0.0732)\approx-2.61$
At $t = 30$, $\ln(0.0443)\approx-3.11$
At $t = 40$, $\ln(0.0268)\approx-3.62$
The plot of $\ln([NH_3])$ vs $t$ gives a straight - line, so the reaction is first - order with respect to $NH_3$. The rate law is rate = $k[NH_3]$.

Step2: Calculate the rate constant $k$

For a first - order reaction, the slope of the $\ln([A]_t)$ vs $t$ plot is equal to $-k$. We can use the two - point formula for the slope of a line $m=\frac{y_2 - y_1}{x_2 - x_1}$.
Let $(x_1,y_1)=(0,\ln(0.200))$ and $(x_2,y_2)=(10,\ln(0.121))$
$k=-\frac{\ln([NH_3]_2)-\ln([NH_3]_1)}{t_2 - t_1}=-\frac{\ln(0.121)-\ln(0.200)}{10 - 0}$
$k=-\frac{-2.10+1.609}{10}=0.0491\ s^{-1}\approx0.049\ s^{-1}$

Answer:

rate = $k[NH_3]$
$k = 0.049\ s^{-1}$