Question

1. the following data was obtained for the reaction:

a_{(g)} + b_{(l)} --> c_{(g)}

a) write the experimental rate law and state the order of the reaction.
b) calculate the specific rate constant (including units if you wish to try).
c) what is the rate of this reaction when a and b are each 0.010
mol/l.

Explanation:

Step1: Determine the order with respect to A

Compare runs 1 and 2 (where [B] is constant).
When [A] is doubled (\(\frac{[A]_2}{[A]_1}=\frac{0.2}{0.1} = 2\)), the rate is quadrupled (\(\frac{\text{Rate}_2}{\text{Rate}_1}=\frac{0.048}{0.012}=4\)).
Since \(\text{Rate}\propto[A]^n\), and \(4 = 2^n\), then \(n = 2\) (order with respect to A).

Step2: Determine the order with respect to B

Compare runs 1 and 4 (where [A] is constant).
When [B] is halved (\(\frac{[B]_4}{[B]_1}=\frac{0.3}{0.6}=0.5\)), the rate is halved (\(\frac{\text{Rate}_4}{\text{Rate}_1}=\frac{0.006}{0.012}=0.5\)).
Since \(\text{Rate}\propto[B]^m\), and \(0.5=0.5^m\), then \(m = 1\) (order with respect to B).

Step3: Write the rate law

The rate law is \(\text{Rate}=k[A]^2[B]\).
The overall order of the reaction is \(n + m=2 + 1=3\).

Step4: Calculate the rate constant \(k\)

Using data from run 1: \(\text{Rate}=k[A]^2[B]\)
\(k=\frac{\text{Rate}}{[A]^2[B]}\)
Substitute \(\text{Rate}=0.012\space mol/L\cdot s\), \([A]=0.1\space mol/L\), \([B]=0.6\space mol/L\)
\(k=\frac{0.012}{(0.1)^2\times0.6}\)
\(k=\frac{0.012}{0.006}= 2\space L^2/mol^2\cdot s\)

Step5: Calculate the rate when \([A]=[B]=0.010\space mol/L\)

Using \(\text{Rate}=k[A]^2[B]\) and \(k = 2\space L^2/mol^2\cdot s\)
\(\text{Rate}=2\times(0.010)^2\times0.010\)
\(\text{Rate}=2\times10^{-6}\space mol/L\cdot s\)

Answer:

a) The experimental rate law is \(\text{Rate}=k[A]^2[B]\) and the overall order of the reaction is \(3\).
b) The specific rate constant \(k = 2\space L^2/mol^2\cdot s\).
c) The rate of the reaction is \(2\times 10^{-6}\space mol/L\cdot s\).