Question

A student sets up two reactions. Reaction 1 uses 0.290 mol/L of reactant, and Reaction 2 uses 0.530 mol/L of reactant. How many times faster is Reaction 2 compared to Reaction 1?Express your answer as a multiple of the rate for Reaction 1 to three significant figures. Note that Reaction 1 is already written for you, so just enter the number.Hint Setting up the ratioSince 0.530/0.290=1.83 , the second reaction contains 1.83 times the amount of reactant compared to Reaction 1. The reaction rate is proportional to the square of the amount of reactant, so the rate will increase by a factor of the square of 1.83.

Answer 1

According to the given information, the reaction rate is proportional to the initial concentration of the reactants.

To compare how many times faster is reaction 2, we need to find the ratio of the initial concentrations of both reactions.

`(0.530)/(0.290)=1.83`

Now, let's state the rate law for both reactions:

`\begin{gathered} r1=k\lbrack A1]^2 \\ r2=k\lbrack A2]^2 \end{gathered}`

Remember that A2=1.83A1:

`\begin{gathered} r1=k\lbrack A1]^2 \\ r2=k\lbrack1.83A1]^2=k\cdot3.35\lbrack A1]^2 \end{gathered}`

Find the ratio of r2 and r1:

`\begin{gathered} (r2)/(r1)=(k\cdot3.35\lbrack A]^2)/(k\cdot\lbrack A]^2) \\ (r2)/(r1)=3.35 \\ r2=3.35r1 \end{gathered}`

It means that Reaction 2 is 3.35 times faster than Reaction 1.