Question

The rate constant, `k`, for a particular reaction is `1.3xx10^(-4)"M"^(-1)"s"^(-1)` at `100^@"C"` and `1.5xx10^(-3)"M"^(-1)"s"^(-1)` at `150^@"C"`. What is the `E_a` (in kJ) for this reaction?

Answer 1

`"64 kJ mol"^(-1)`

Explanation:

Your tool of choice here will be the Arrhenius equation, which looks like this

`color(blue)(|bar(ul(color(white)(a/a)k = A * "exp"(-E_a/(RT))color(white)(a/a)|)))" "`, where

`k` - the rate constant for a given reaction
`A` - the pre-exponential factor, specific to a given reaction
`E_a` - the activation energy of the reaction
`R` - the universal gas constant, useful here as `8.314"J mol"^(-1)"K"^(-1)`
`T` - the absolute temperature at which the reaction takes place

As you can see, the Arrhenius equation establishes a relationship between the rate constant and the absolute temperature at which the reaction takes place.

In other words, this equation allows you to determine how a change in temperature affects the rate of the reaction.

Now, convert the two temperatures from degrees Celsius to Kelvin by using the conversion factor

`color(purple)(|bar(ul(color(white)(a/a)color(black)(T["K"] = t[""^@"C"] + 273.15)color(white)(a/a)|)))`

You will thus have

`T_1 = 100^@"C" + 273.15 = "373.15 K"`

`T_2 = 150^@"C" + 273.15 = "423.15 K"`

So, you know that at `T_1`, the rate constant for your reaction takes the form

`k_1 = A * "exp"( -E_a/(R * T_1))`

Likewise, at `T_2` the rate constant takes the form

`k_2 = A * "exp"( -E_a/(R * T_2))`

Divide these two equations to get rid of the pre-exponential factor, `A`

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

This will be equivalent to - using the properties of exponents

`k_1/k_2 = "exp"[E_a/R * (1/T_2 - 1/T_1)]`

Take the natural log of both sides of the equation to get

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

Rearrange to solve for `E_a`, the activation energy of the reaction

`color(purple)(|bar(ul(color(white)(a/a)color(black)(E_a = R * ln(k_1/k_2)/(1/T_2 - 1/T_1))color(white)(a/a)|)))`

Finally, plug in your values to get

`E_a = "8.314 J mol"^(-1) color(red)(cancel(color(black)("K"^(-1)))) * ln( (1.3 * 10^(-4)color(red)(cancel(color(black)("M"^(-1))))color(red)(cancel(color(black)("s"^(-1)))))/(1.5 * 10^(-3)color(red)(cancel(color(black)("M"^(-1))))color(red)(cancel(color(black)("s"^(-1))))))/((1/423.15 - 1/373.15)color(red)(cancel(color(black)("K"^(-1)))))`

`E_a = "64,212.3 J mol"^(-1)`

Expressed in kilojoules per mole, the answer will be

`E_a = color(green)(|bar(ul(color(white)(a/a)"64 kJ mol"^(-1)color(white)(a/a)|)))`

I'll leave the answer rounded to two sig figs.