Answer to Question #109463 in Chemistry for Jake

Question

For the reaction Q → W + X, the following data were obtained at 30°C. [Q] is M, Rate is in M/s
[Q]: 0.170 0.212. 0.357
Rate: 6.68 x 10^-3 1.04 x 10^-2. 2.94 x 10^-2
Determine the rate constant of this reaction and state its units?

Answer 1

Solution:

Q → W + X

The balanced chemical equation for the decomposition of Q is given above. Since there is only one reactant, the rate law for this reaction has the general form:

Rate = k * [Q]^m

In order to determine the rate constant of this reaction, we need, firstly, to determine the value of the exponent m.

Rate₁ = 6.68*10^-3 = k * [0.170]^m;

Rate₂ = 1.04*10^-2 = k * [0.212]^m.

We can set up a ratio of the first rate to the second rate:

[Rate₁ / Rate₂] = (6.68*10^-3 / 1.04*10^-2) = (0.170^m / 0.212^m);

[Rate₁ / Rate₂] = 06423 = 0.8019^m;

ln(0.6423) = m * ln(0.8019);

-0.4427 = m * -0.2208;

m = 2.005 = 2.

Since m=2, the decomposition is a second-order reaction.

Therefore,

Rate = k * [Q]².

Checking:

Rate₂ = 1.04*10^-2 = k * [0.212]^m;

Rate₃ = 2.94*10^-2 = k * [0.357]^m.

[Rate₂ / Rate₃] = (1.04*10^-2 / 2.94*10^-2) = (0.212^m / 0.357^m);

[Rate₂ / Rate₃] = 0.3537 = 0.5938^m;

ln(0.3537) = m * ln(0.5938);

-1.0393 = m * -0.5212;

m = 1.994 = 2. (Correct).

Units:

Rate = k * [Q]²;

[ M/s ] = [k] * [ M ]²;

[ k ] = [ M/s ] / [ M² ] = [ M^-1 s^-1 ];

[ k ] = M^-1 s^-1

Once we have determined the order of the reaction, we can go back and plug in one set of our initial values and solve for k (rate constant). We find that:

Rate = k * [Q]².

Substituting in our first set of values, we have

Rate₁ = 6.68*10^-3 = k * [0.170]²;

k = (6.68*10^-3) / (0.170²) = 0.231

k = 0.231 M^-1 s^-1.

Checking:

1) Rate₂ = 1.04*10^-2 = k * [0.212]²;

k = (1.04*10^-2) / (0.212²) = 0.231 M^-1 s^-1 (Correct).

2) Rate₃ = 2.94*10^-2 = k * [0.357]²;

k = (2.94*10^-2) / (0.357²) = 0.231 M^-1 s^-1 (Correct).

Answer:

The rate constant = k = 0.231 M^-1 s^-1.

[ k ] = [ M^-1 s^-1 ].