Question

Using the initial rate date to determine the activation energy (in kJ/mol) for the reaction A + B (arrow sign) C:

Experiment          Temperature (K)                 Initial A                    Initial B                 Initial Reactive Rate (m/s)

1                                700                                 0.20                         0.10                         1.8 x 10^-5

2                                 700                                 0.40                        0.10                         3.6 x 10^-5

3                                700                                   0.10                       0.20                          3.6 x 10^-5

4                                 600                                  0.50                       0.50                          4.3 x 10^-5

What is the reaction order of HI?____________________

What is the energy for the reaction?_________________ kJ/mole (2 significant figures)

What is the value of the reaction constant at 700K?___________________(mole*second) (2 significant figures)

What is the value of the reaction constant at 650K?____________________ (mole*second) (2 significant figures)

What is the initial concentration of HI at 650K_____________________ mole/Liter (2 significant figures)

****Please put your answers on the lines when you reply or at least number the answers 1 thru 5, so that I can better understand what the answers are. Thank you.

Answer 1

1. Here, HI is not mentioned in the given reaction. So, The order of reaction w.r. t both the reactants A and B is mentioned below for the convenience.

Let the order of reaction w.r.t. A = m and that w.r.t. B = n

According to the given data, the rate increase 2 times when the concentration of A increases 2 times and concentration of B is constant.

i.e. 2^m * 1ⁿ = 2¹, i.e. 2^m = 2¹

Therefore, m = 1

According to the given data, the rate increase 2 times when the concentration of A decreases to half and concentration of B increases 2 times.

i.e. (1/2)^m * 2ⁿ = 2¹, i.e. 2^-m * 2ⁿ = 2¹

i.e. 2^n-m = 2¹, i.e. n-m = 1

i.e. n-1 = 1, therefore n =2

3. According to the rate law, we can write as shown below.

The rate of reaction = k * [A]¹ * [B]², where k = rate constant

For the experiment 1

1.8*10^-5 = k * (0.2)¹ * (0.1)²

i.e. (k)_{at 700K} = 9*10^-3 M²s^-1 .

2. Since rate of reaction becomes twice for every 10 ^oC rise in temperature

c(9*10^-3)/1024 ~ 9*10^-6 M²s^-1

ln((k)_{at 700K}/(k)_{at 600K}) = E/R (1/600 - 1/700), where E = energy of the reaction

You can calculate E by substituting the corresponding values.

4. Since rate of reaction becomes twice for every 10 ^oC rise in temperature

(k)_{at 700K} = 9*10^-3 M²s^-1 .

(k)_{at 650K} = (9*10^-3) / 32, i.e. 2.8*10^-4 M²s^-1 .