the data in the table concern the lactonization of hydroxyvaleric acid at 25 degrees celsius. they give the…

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.\nfind the average rate of reaction for the following time intervals: (a) 2 ≤ t ≤ 6, (b) 2 ≤ t ≤ 4, (c) 0 ≤ t ≤ 2\nround your answer to four decimal places.\n(a) (moles/l)/min\n(b) (moles/l)/min\n(c) (moles/l)/min

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.\nfind the average rate of reaction for the following time intervals: (a) 2 ≤ t ≤ 6, (b) 2 ≤ t ≤ 4, (c) 0 ≤ t ≤ 2\nround your answer to four decimal places.\n(a) (moles/l)/min\n(b) (moles/l)/min\n(c) (moles/l)/min

Answer

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) $-0.0115$