Question

The rate constant for a reaction at 25.0 degrees C is 0.010 `s^-1` and its activation energy is 35.8 KJ. How do you find the rate constant at 50.0 degrees C?

Answer 1

You compare them at two temperatures and determine the Arrhenius plot.

`color(white)(=>)k_2 = k_1e^((E_a)/R[1/(T_1) - 1/(T_2)])`

Be sure to use `8.314472xx10^(-3) "kJ/mol"cdot"K"` if `E_a` is in `"kJ"`, and to convert `T_1` and `T_2` into `"K"`.

You should get `"0.031 s"^(-1)`.

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The Arrhenius equations are:

`k_1 = Ae^(-E_a"/"RT_1)`
`k_2 = Ae^(-E_a"/"RT_2)`

The same reaction has the same activation energy `E_a` and the same universal gas constant `R`. But at different temperatures, the rate constant `k` differs.

Now, divide these equations.

`k_1/k_2 = e^(-E_a"/"RT_1)/e^(-E_a"/"RT_2)`

taking the natural logarithm allows you to separate the right side.

`ln(k_1/k_2) = ln(e^(-E_a"/"RT_1)/e^(-E_a"/"RT_2))`

`= ln(e^(-E_a"/"RT_1)) - ln(e^(-E_a"/"RT_2))`

`= -E_a/(RT_1) - (-E_a)/(RT_2)`

`= -(E_a)/R[1/(T_1) - 1/(T_2)]`

So your final equation is:

`bb(ln(k_1/k_2) = -(E_a)/R[1/(T_1) - 1/(T_2)])`

To find the rate constant at a new temperature, try switching the sign on the left and right:

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

`ln((k_1/k_2)^(-1)) = (E_a)/R[1/(T_1) - 1/(T_2)]`

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

There, now it's easier to solve for `k_2`:

`k_2/k_1 = e^((E_a)/R[1/(T_1) - 1/(T_2)])`

`=> color(blue)(k_2 = k_1e^((E_a)/R[1/(T_1) - 1/(T_2)]))`