Question

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 ?

Answer 1

Answer:

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

Answer 2

Answer:

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)