The rate constant of a first-order reaction is 4.25 x 10^-4 s^-1 at 350 Celsius. If the activation energy is 103 kJ/mol, calculate the temperature at which its rate constant is 8.75 x 10^-4 s^-1.

Use the Arrhenius equation.

I got 17 Kelvin in the end. Does that mean my final answer will be 290 C (converted C to K)?

I don't think so.

273 + C = K
273 + C = 17
C = ? not 290??
Second think I notice is that your calculated temperature decreased. k2 is about double k1 so wouldn't you think T would be higher. Did you convert kJ/mol to J/mol? That's a common error.

In fact, a rule of thumb in the kinetics business is that reaction rate can be doubled for every 10 degrees T. Since this ratio k2/k1 = about 2.05 I would guess T2 would be close to 365 or something like that.

ah you are right DrBob222. I converted kJ to J.

Just to be sure we're on the same page my post could be misinterpreted. You are SUPPOSED to convert kJ/mol to J/mol. Using kJ/mol won't work.

To solve this problem, we can use the Arrhenius equation, which relates the rate constant of a chemical reaction to the temperature and the activation energy.

The Arrhenius equation is given by:

k = Ae^(-Ea/RT)

where:
- k is the rate constant
- A is the pre-exponential factor (or frequency factor)
- Ea is the activation energy
- R is the ideal gas constant (8.314 J/(mol·K))
- T is the temperature in Kelvin

We are given the rate constant at one temperature (350 degrees Celsius) and we need to find the temperature at which the rate constant is 8.75 x 10^-4 s^-1.

Let's start by rearranging the Arrhenius equation to solve for the temperature:

ln(k) = ln(A) - (Ea / (R * T))

To find the temperature T, we need to isolate it on one side of the equation. Let's rewrite the equation as:

ln(k) = ln(A) - (Ea / (R * T))

Rearrange and isolate T:

(Ea / (R * T)) = ln(A) - ln(k)

Now, substitute the known values into the equation:

(Ea / (8.314 J/(mol·K) * T)) = ln(A) - ln(k)

(Ea / (8.314 J/(mol·K))) = ln(A) - ln(k) * T

Solve for T:

T = Ea / (8.314 J/(mol·K) * (ln(A) - ln(k)))

Now we can plug in the values to find the temperature:

Ea = 103 kJ/mol = 103 * 10^3 J/mol
k1 = 4.25 x 10^-4 s^-1
k2 = 8.75 x 10^-4 s^-1

Convert Ea to J/mol:
Ea = 103 * 10^3 J/mol

Plug the values into the equation:

T = (103 * 10^3 J/mol) / (8.314 J/(mol·K) * (ln(A) - ln(k)))

Calculate ln(A) - ln(k):

ln(A) = ln(4.25 x 10^-4 s^-1)
ln(k) = ln(8.75 x 10^-4 s^-1)

Calculate ln(A) - ln(k):

ln(A) - ln(k) = ln(4.25 x 10^-4 s^-1) - ln(8.75 x 10^-4 s^-1)

T = (103 * 10^3 J/mol) / (8.314 J/(mol·K) * (ln(4.25 x 10^-4 s^-1) - ln(8.75 x 10^-4 s^-1)))

Simplify the equation to get the numerical value of T.