Question

a chemistry graduate student is studying the rate of this reaction: 2h₃po₄(aq)→p₂o₅(aq)+3h₂o(aq). she fills a reaction vessel with h₃po₄ and measures its concentration as the reaction proceeds:

time (seconds)	h₃po₄
10.	0.705m
20.	0.497m
30.	0.350m
40.	0.247m

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 $H_3PO_4$, we can check if the reaction is first - order by plotting $\ln[H_3PO_4]$ vs time or second - order by plotting $\frac{1}{[H_3PO_4]}$ 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.

time (s)	$[H_3PO_4]$ (M)	$\ln[H_3PO_4]$
10	0.705	- 0.349
20	0.497	- 0.700
30	0.350	- 1.05
40	0.247	- 1.39

The plot of $\ln[H_3PO_4]$ vs time gives a straight - line, so the reaction is first - order with respect to $H_3PO_4$. The rate law is rate = $k[H_3PO_4]$.

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 $m=\frac{y_2 - y_1}{x_2 - x_1}$. Let's take the first two points $(t_1 = 0,\ln[H_3PO_4]_1=0)$ and $(t_2 = 10,\ln[H_3PO_4]_2=-0.349)$.
$k=-\text{slope}=\frac{\ln[H_3PO_4]_2-\ln[H_3PO_4]_1}{t_1 - t_2}=\frac{- 0.349-0}{0 - 10}=0.035\ s^{-1}$

Answer:

rate = $k[H_3PO_4]$
$k = 0.035\ s^{-1}$