Question

A first order reaction has an Arrhenius constant for the reaction is `9.9 xx 10^10 sec^ -1`, and an activation energy of `"62.0 kJ"`. At what temperature (in Celsius) is the rate constant equal to `0.434 sec^-1`?

Answer 1

`12^@"C"`

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

Notice that he problem provides you with the pre-exponential factor, the activation energy, and the rate constant, and asks for the temperature in degrees Celsius at which the reaction has that particular rate constant.

Your strategy here will be to use the Arrhenius equation to solve for `T`, the absolute temperature at which the reaction takes place, then use the conversion factor

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

to go from Kelvin to degrees Celsius.

So, rearrange the Arrhenius equation to isolate the exponential term on one side of the equation

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

Take the natural log of both sides of the equation

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

This will be equivalent to

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

Rearrange to isolate `T` on one side of the equation

`T = -E_a/R * 1/ln(k/A)`

Plug in your values to get - do not forget to convert the activation energy from kilojoules to joules

`T = - (62.0 * 10^3color(red)(cancel(color(black)("J"))))/(8.314color(red)(cancel(color(black)("J")))color(red)(cancel(color(black)("mol"^(-1))))"K"^(-1)) * 1/(ln( (0.434 color(red)(cancel(color(black)("s"^(-1)))))/(9.9 * 10^(10)color(red)(cancel(color(black)("s"^(-1))))))`

`T = "285.14 K"`

Expressed in degrees Celsius, the temperature will be

`t[""^@"C"] = "285.14 K" - 273.15 = color(green)(|bar(ul(color(white)(a/a)12^@"C"color(white)(a/a)|)))`

The answer is rounded to two sig figs, the number of sig figs you have for the pre-exponential factor.