Question

rate of a certain reaction is given by the following rate law: rate Use this information to answer the questions below. What is the reaction order in H2? What is the reaction order in I2? What is overall reaction order? At a certain concentration of H2 and I2, the initial rate of reaction is 2.0 x 104 M / s. What would the initial rate of the reaction be if the concentration of H2 were doubled? Round your answer to significant digits. The rate of the reaction is measured to be 52.0 M / s when [H2] = 1.8 M and [I2] = 0.82 M. Calculate the value of the rate constant. Round your answer to significant digits.

Answer 1

Complete Question

The  rate of a certain reaction is given by the following rate law:

            $rate = k [H_2][I_2]$

rate Use this information to answer the questions below.

What is the reaction order in $H_2$?

What is the reaction order in $I_2$?

What is overall reaction order?

At a certain concentration of H2 and I2, the initial rate of reaction is 2.0 x 104 M / s. What would the initial rate of the reaction be if the concentration of H2 were doubled? Round your answer to significant digits. The rate of the reaction is measured to be 52.0 M / s when [H2] = 1.8 M and [I2] = 0.82 M. Calculate the value of the rate constant. Round your answer to significant digits.

Answer:

The reaction order in $H_2$ is  n =  1

The reaction order in $I_2$ is  m = 1

The  overall reaction order z =  2

When the hydrogen is double the the initial rate is   $rate_n = 4.0*10^{-4} M/s$

The rate constant is   $k = 35.23 \ M^{-1} s^{-1}$

Explanation:

From the question we are told that

   The rate law is  $rate = k [H_2][I_2]$

   The rate of reaction is $rate = 2.0 *10^{4} M /s$

Let the reaction order for $H_2$ be  n and for $I_2$  be  m

From the given rate law the concentration of $H_2$ is raised to the power of 1 and this is same with $I_2$ so their reaction order is  n=m=1

   The overall reaction order is  

               $z = n +m$

               $z =1 +1$

               $z =2$

At  $rate = 2.0 *10^{4} M /s$

        $2.0*10^{4} = k [H_2] [I_2] ---(1)$

= >    $k = \frac{2.0*10^{4}}{[H_2] [I_2] }$

given that the concentration of hydrogen is doubled we have that

            $rate = k [2H_2] [I_2] ----(2)$

=>      $k = \frac{rate_n }{ [2H_2] [I_2]}$

 So equating the two k

           $\frac{2.0*10^{4}}{[H_2] [I_2] } = \frac{rate_n }{ [2H_2] [I_2]}$

    =>    $rate_n = 4.0*10^{-4} M/s$

So when

      $rate_x = 52.0 M/s$

        $[H_2] = 1.8 M$

         $[I_2] = 0.82 \ M$

We have

      $52 .0 = k(1.8)* (0.82)$

     $k = \frac{52 .0}{(1.8)* (0.82)}$

      $k = 35.23 M^{-2} s^{-1}$