The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy. If the rate constant of this reaction is at, what will the rate constant be at?

K2 = 61.2 M^-1. S^-1

Explanation:

We complete the question fully:

The rate constant of a certain reaction is known to obey the Arrhenius equation, and to have an activation energy Ea = 71.0kJ/mol. If the rate constant of this reaction is 6.7M^ (-1) * s^ (-1) at 244.0 degrees Celsius, what will the rate constant be at 324.0 degrees Celsius?

Answer is as follows:

The question asks us to calculate the value of the rate constant at a certain temperature, given that it is at a particular value for a particular temperature. We solve the question as follows:

According to Arrhenius equation, the relationship between temperature and activation energy is as follows:

k = Ae^ - (Ea/RT)

where, k = rate constant

A = pre-exponential factor

Ea = activation energy

R = gas constant

T = temperature in kelvin

From the equation, the following was derived for a double temperature problem:

ln (k2/k1) = (-Ea/R) * (1/T1 - 1/T2)

We list out the parameters as follows:

T1 = (244 + 273.15) K = 517.15 K

T2 = (324 + 273.15) K = 597.15 K

K1 = 6.7, K2 = ?

R = 8.314 J/mol K

Ea = 71.0 kJ/mol = 71000 J/mol

Putting the given values into the above formula as follows:

ln (k2/6.7) = (-71000/8.314) * (1/517.15 - 1/597.15)

lnk2 - 1.902 = 8539.8 * 0.000259

lnK2 = 1.902 + 2.21

lnK2 = 4.114

K2 = e^ (4.114)

K2 = 61.2

Hence, K2 = 61.2 (M. S) ^-1

60.34 M^ (-1) * s^ (-1)

Explanation:

In order to solve this question we are going to be making use of the Arrhenius equation which is given below;

k = Ae^ - (Ea/RT). Where k = rate constant, Ea = activation energy = 71 * 10^3 J, R = gas constant, T = temperature and A = frequency factor.

So, we are dealing with Question that involves two rate constants. Therefore, we take a natural logarithm of the Arrhenius equation two times (remember we are dealing with 2 rate constant), then subtract the two equations which gives us;

ln (k1/k2) = Ea/R [ 1/T2 - 1/T1].

So, ln (6.7) / ln k2 = 71 * 10^3 / 8.31 * [ 1 / 597 - 1 / 517].

1.9 / ln k2 = 8543.923 * [ 0.001675 - 0.00193].

1.9 - ln k2 = - 2.18.

- ln k2 = - 2.18 - 1.9.

-ln k2 = - 4.08

ln k2 = 4.1.

k2 = e^4.1.

k2 = 60.34 M^ (-1) * s^ (-1)