Question

the data in the table concern the lactonization of hydroxyvaleric acid at 25 degrees celsius. they give the of this acid in moles per liter after t minutes.
find the average rate of reaction for the following time intervals: (a) 2 ≤ t ≤ 6, (b) 2 ≤ t ≤ 4, (c) 0 ≤ t ≤ 2
round your answer to four decimal places.
(a) (moles/l)/min
(b) (moles/l)/min
(c) (moles/l)/min

Explanation:

Step1: Recall average - rate formula

The average rate of reaction over the interval $[a,b]$ is $\frac{C(b)-C(a)}{b - a}$, where $C(t)$ is the concentration at time $t$.

Step2: Solve for (a) $2\leq t\leq6$

Here, $a = 2$, $b = 6$, $C(2)=0.0570$ and $C(6)=0.0295$.
$\frac{C(6)-C(2)}{6 - 2}=\frac{0.0295 - 0.0570}{4}=\frac{- 0.0275}{4}=-0.0069$

Step3: Solve for (b) $2\leq t\leq4$

Here, $a = 2$, $b = 4$, $C(2)=0.0570$ and $C(4)=0.0408$.
$\frac{C(4)-C(2)}{4 - 2}=\frac{0.0408 - 0.0570}{2}=\frac{-0.0162}{2}=-0.0081$

Step4: Solve for (c) $0\leq t\leq2$

Here, $a = 0$, $b = 2$, $C(0)=0.0800$ and $C(2)=0.0570$.
$\frac{C(2)-C(0)}{2 - 0}=\frac{0.0570 - 0.0800}{2}=\frac{-0.023}{2}=-0.0115$

Answer:

(a) $-0.0069$
(b) $-0.0081$
(c) $-0.0115$