Question

Question #47c93

Answer 1

Here's what I got.

Explanation:

Your tool of choice for this problem will be the Arrhenius equation, which establishes a relationship between the rate constant of a reaction, `k`, and the temperature at which it occurs, `T`

`color(blue)(k = A * "exp"(-E_a/(RT))" " " "color(purple)((1))" "`, where

`A` - a pre-exponential factor
`E_a` - the activation energy of the reaction
`R` - the universal gas constant, in this context given as `8.3145"J"/("mol K")`

Now, the Arrhenius parameters are `A` and `E_a`.

Since you're basically dealing with one equation and two unknowns, you will have to use the information given to find a way to eliminate one of the two Arrhenius parameters first.

You can do that by writing the Arrhenius equation in non-exponential form

`ln(k) = ln[A * "exp"(-E_a/(RT))]`

This is equivalent to

`ln(k) = ln(A) + ln["exp"(-E_a/(RT))]`

`ln(k) = ln(A) - E_a/(RT)`

Now, notice that the problem provides you with information about the value of the rate constant at two different temperatures

`T_1 = 273.15 + 24 = "297.15 K"`

`T_2 = 273.15 + 37 = "310.15 K"`

This means that you can write

`ln(k_1) = ln(A) - E_a/(R * T_1) ->` for `k_1` and `T_1`

`ln(k_2) = ln(A) - E_a/(R * T_2) ->` for `k_2` and `T_2`

Notice that you can get rid of the `ln(A)` term by subtracting the second equation for the first one.

`ln(k_1) - ln(k_2) = color(red)(cancel(color(black)(ln(A)))) - color(red)(cancel(color(black)(ln(A)))) - E_a/R * (1/T_1 - 1/T_2)`

This will get you

`color(blue)(ln(k_1/k_2) = - E_a/R * (1/T_1 - 1/T_2))`

You can now find the activation energy of the reaction - keep in mind that activation energy must carry a positive sign, since it expresses energy required

`E_a = - (R * ln(k_1/k_2))/(1/T_1 - 1/T_2)`

`E_a = - (8.3145 "J"/("mol" * color(red)(cancel(color(black)("K")))) * ln( (1.70 * 10^(-2) color(red)(cancel(color(black)("L mol"^(-1)"s"^(-1)))))/(2.01 * 10^(-2) color(red)(cancel(color(black)("L mol"^(-1)"s"^(-1)))))))/((1/297.15 - 1/310.15) 1/color(red)(cancel(color(black)("K"))))`

`E_a = "9873.5 J/mol"`

To get the value of the pre-exponential factor, pick a value for `k` and `T` and plug in the activation energy into equation `color(purple)((1))`.

For `k_1` and `T_1`, you will get

`k_1 = A * "exp"(-E_a/(RT_1))`

This will get you

`A = k_1/("exp"(-E_a/(RT)))`

Notice that since the exponential has to be unitless, `A` will have the same units as `k`. Plug in your values to get

`A = (1.70 * 10^(-2)"L mol"^(-1)"s"^(-1))/("exp"(- (9873.5 color(red)(cancel(color(black)("J")))/color(red)(cancel(color(black)("mol"))))/(8.3145color(red)(cancel(color(black)("J")))/(color(red)(cancel(color(black)("mol"))) * color(red)(cancel(color(black)("K")))) * 297.15color(red)(cancel(color(black)("K"))))))`

`A = "0.92475 L mol"^(-1)"s"^(-1)`

I'll leave the answers rounded to three sig figs.

`E_a = color(green)("9.87 kJ/mol") ->` expressed in kilojoules per mole

`A = color(green)("0.925 L mol"^(-1)"s"^(-1))`