Question

example problem: the following initial rates were measured for the reaction x + y → products. run x₀ (m) y₀ (m) rate (m s⁻¹) 1 0.20 0.50 0.040 2 0.40 0.50 0.080 3 0.20 1.00 0.080 (a) write the experimental rate law and state the overall order of reaction. (b) evaluate k. (c) what is the initial rate of this reaction if x₀ = y₀ = 0.30 m? (d) the integrated rate law for this reaction is x = x₀ - kt. find the half - life of x for x₀ = 0.05 m.

Explanation:

Step1: Determine reaction order with respect to X

Compare runs 1 and 2. [Y] is constant, [X] doubles from 0.20 M to 0.40 M, and the rate doubles from 0.040 M s⁻¹ to 0.080 M s⁻¹. So the reaction is first - order with respect to X ($n_X = 1$).

Step2: Determine reaction order with respect to Y

Compare runs 1 and 3. [X] is constant, [Y] doubles from 0.50 M to 1.00 M, and the rate doubles from 0.040 M s⁻¹ to 0.080 M s⁻¹. So the reaction is first - order with respect to Y ($n_Y = 1$).

Step3: Write the rate law

The rate law is rate = $k[X][Y]$. The overall order of the reaction is $n=n_X + n_Y=1 + 1 = 2$.

Step4: Calculate the rate constant k

Using data from run 1, rate = $k[X][Y]$. Substitute rate = 0.040 M s⁻¹, [X]=0.20 M and [Y]=0.50 M. Then $k=\frac{\text{rate}}{[X][Y]}=\frac{0.040\ M\ s^{-1}}{0.20\ M\times0.50\ M}=0.40\ M^{-1}s^{-1}$.

Step5: Calculate the initial rate for [X]_0 = [Y]_0 = 0.30 M

Substitute [X]=0.30 M, [Y]=0.30 M and $k = 0.40\ M^{-1}s^{-1}$ into the rate law. rate = $k[X][Y]=0.40\ M^{-1}s^{-1}\times0.30\ M\times0.30\ M = 0.036\ M\ s^{-1}$.

Step6: Calculate the half - life of X

For a first - order reaction with respect to X (from the integrated rate law $[X]=[X]_0-kt$), the half - life formula for a first - order reaction is $t_{1/2}=\frac{[X]_0}{2k}$. Substitute $[X]_0 = 0.05\ M$ and $k = 0.40\ M^{-1}s^{-1}$. $t_{1/2}=\frac{0.05\ M}{2\times0.40\ M^{-1}s^{-1}}=0.0625\ s$.

Answer:

(a) Rate law: rate = $k[X][Y]$, Overall order = 2
(b) $k = 0.40\ M^{-1}s^{-1}$
(c) $0.036\ M\ s^{-1}$
(d) $0.0625\ s$