Question

Consider a reaction with an activation energy of 12779.4 J/mol run at 298.15 K and at 2225 K.

How many times larger is the rate constant, E. for the reaction at 2225 K than at 298 K?

Answer 1

The rate constant at 2225 K is approximately 2.25 x 10^-17 times larger than the rate constant at 298 K.

Step-by-step explanation:

The rate constant for a reaction can be determined using the Arrhenius equation, which relates the rate constant (k) to the activation energy (Ea) and the temperature (T):

k = A*e^(-Ea/RT)

Where:

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

To determine how many times larger the rate constant is at 2225 K compared to 298 K, we can calculate the ratio of the rate constants at the two temperatures.

Let's first find the rate constant at 298 K using the given activation energy and the rate constant at 2225 K:

k_298 = A*e^(-Ea/(R*298))

k_2225 = A*e^(-Ea/(R*2225))

The ratio of the rate constants is:

ratio = k_2225 / k_298

Substituting the expressions for k_2225 and k_298:

ratio = (A*e^(-Ea/(R*2225))) / (A*e^(-Ea/(R*298)))

The A factor cancels out, and we can simplify the expression:

ratio = e^(-Ea/(R*2225) + Ea/(R*298))

Now we can substitute the given activation energy and calculate the ratio:

ratio = e^((-12779.4 J/mol)/(8.314 J/mol·K)*(1/2225 K - 1/298 K))

ratio = e^(-39.16)

ratio ≈ 2.25 x 10^-17

Therefore, the rate constant at 2225 K is approximately 2.25 x 10^-17 times larger than at 298 K.