Question

A student sets up two reactions. reaction 1 uses 0.230 mol/l of reactant, and reaction 2 uses 0.570 mol/l of reactant. how many times faster is reaction 2 compared to reaction 1?

Answer 1

Assuming that the rate of reaction is of first order, so that the rate is only dependent on the concentration A:

rate = - k A

 

So to know how much faster is reaction 2 compared to reaction 1, simply take the ratio of concentrations:

0.570 / 0.230 = 2.48

So reaction 2 is about 2.48 times faster.

Answer 2

Answer: The rate of second reaction is $(2.47)^a$ times the rate of the first reaction.

Explanation:

$aA\rightarrow B$

For the first reaction where concentration o the reactant is [A]= 0.230 mol/L

$R_1=k\times [A]^a=k\times [0.230 mol/L]^a$..(1)

For the second reaction where concentration o the reactant is [A]= 0.570 mol/L

{tex]R_2=k\times [A]^a=k\times [0.570 mol/L]^a[/tex]...(2)

Dividing (2) and (1)

$\frac{R_2}{R_1}=\frac{k\times [0.570 mol/L]^a}{k\times [0.230 mol/L]^a}$

$R_2=R_1\times(2.47)^a$

The rate of second reaction is $(2.47)^a$ times the rate of the first reaction.