Question

In: Chemistry

For the reaction A + B + 2C = 3D at 25 Celsius, the following initial...

For the reaction A + B + 2C = 3D at 25 Celsius, the following initial rate data were obtained:

Experiment #1: [A] = 0.1M, [B] = 0.1M, [C] = 0.2M, Initial Rate of Formation for D = 4 e -4 M/min

Experiment #2: [A] = 0.3M, [B] = 0.2M, [C] = 0.2M, Initial Rate of Formation for D = 1.2 e -3 M/min

Experiment #3: [A] = 0.1M, [B] = 0.3M, [C] = 0.2M, Initial Rate of Formation for D = 4 e -4 M/min

Experiment #4: [A] = 0.3M, [B] = 0.4M, [C] = 0.6M, Initial Rate of Formation for D = 3.6 e -3 M/min

If Experiment #2 is repeated at 50 Celsius and the initial rate is measured to be 1.6 e -3 M/min, what is the activation energy?

PLEASE PROVIDE ALL STEPS! :)

Solutions

Expert Solution

Let the orders of the given reaction w.r.to A, B and C are respectively x, y and z.

For experiment-1: r1 = 4*10-4 M/min = k*[0.1M]x * [0.1M]y * [0.2M]z   ------- (1)

For experiment-2: r2 = 1.2*10-3 M/min = k*[0.3M]x * [0.2M]y * [0.2M]z   ------- (2)

For experiment-3: r3 = 4*10-4 M/min = k*[0.1M]x * [0.3M]y * [0.2M]z   ------- (3)

For experiment-4: r4 = 3.6*10-3 M/min = k*[0.3M]x * [0.4M]y * [0.6M]z   ------- (4)

Dividing equation(1) and (3) we get

4*10-4 M/min / 4*10-4 M/min = [0.1 M / 0.3M]y

=> 1 = (1/3)y

=> y = 0

Dividing eqn(1) by (2)

4*10-4 M/min / 1.2*10-3 M/min =  [0.1 M / 0.3M]x

=> 0.333 = (0.333)x

=> x = 1

Dividing eqn(3) by (4)

4*10-4 M/min / 3.6*10-3 M/min =  [0.2 M / 0.6M]x

=> 0.111 = (0.333)z

=> z = 2

Hence the rate equation for the reaction is rate = k * [A]1 * [B]0 * [C]2 = k * [A]*[C]2

From Experiment - 2: at T1 = 25 C = 298 K

r2 = 1.2*10-3 M/min = k1*[0.3M]x * [0.2M]y * [0.2M]z = k*[0.3M]*[0.2M]2

=> k1 = 1.2*10-3 M/min / [0.3M]*[0.2M]2 = 0.1 M-2min-1  

At temperature, T2 = 50 C = 273 + 50 = 323 K

r = 1.6*10-3 M/min = k2*[0.3M]x * [0.2M]y * [0.2M]z = k*[0.3M]*[0.2M]2

=> k2 = 1.6*10-3 M/min / [0.3M]*[0.2M]2 = 0.1333 M-2min-1  

Now applying Arrhenius equation

ln(k2 / k1) = (Ea/R) * [1/T1 - 1/T2] = (Ea/R) * [1/298K - 1/323K]

=> ln(0.1333 M-2min-1 / 0.1 M-2min-1) = 0.2874 = Ea * 25K / 8.314 Jmol-1K-1 * 298K * 323K

=> Ea = 9200 J/mol = 9.2 KJ/mol (answer)

queen_honey_blossom answered 2 years ago
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