Question

Chlorine dioxide (ClO2) is a disinfectant in municipal water-treatment plants. It dissolves in basic solution, producing ClO3- and ClO2- according to the redox reaction: 2ClO2(g) + 2OH-(aq) ?ClO3-(aq) + ClO2-(aq) +H2O. The following kinetic data were obtained at 298 K for the reaction:

Expt # [ClO2]0, M [OH-]0, M Initial rate (M/s)

1 0.060 0.030 0.0248

2 0.020 0.030 0.00827

3 0.020 0.090 0.0247

Determine the rate law, overall order of reaction, and the rate constant for this reaction at 298 K.

Answer 1

`\text{ rate = k[ClO$_(2)$][OH$^(-)$]; 2nd order; 138 M$^(-1)$s$^(-1)$}`

Step-by-step explanation:

We call this type of problem, "determining a rate law by the method of initial rates."

We want to know how much a change in the concentration of a reactant will change the rate of a reaction.

We must measure the reaction rate once. Then we repeat the experiment. We change the concentration of one reactant and leave everything else the same.

The equation for this reaction is

2ClO₂ + 2OH⁻ ⟶ ClO₃⁻ + ClO₂⁻ + H₂O

The general form of the rate law is

`\rm rate =k[ClO_(2)]^(m)[OH^(-)]^(n)`

Our task is to find the values of m and n.

Data:

`\begin{array}{cccl}\textbf{Expt} & \textbf{[ClO$_(2)]_(0)$/M} & \textbf{[OH$^(-)]_(0)$/M} & \textbf{rate/(M/s)}\\1 & 0.060 & 0.030 & 0.248\\2 & 0.020 & 0.030 & 0.00827\\3 & 0.020 & 0.090 & 0.0247\\\end{array}`

1. Calculate m

Consider experiments 1 and 2 .

`\rm ([ClO_(2)]_(1) )/([ClO_(2)]_(2)) = (0.60)/(0.00 ) = 3.0 \approx 3\\\\\rm (rate_(1))/(rate_(2)) = (0.0248)/(0.00827) = 3.00 \approx 3`

In going from Experiment 2 to Experiment 1, we tripled the concentration and the rate tripled.

If tripling the concentration triples the rate, the reaction is 1st order.

m = 1, so the rate law becomes

`\rm rate =k[ClO_(2)][OH^(-)]^(n)`

2. Calculate n

Consider experiments 2 and 3 .

`\rm ([ OH^(-)]_(3) )/([ OH^(-)]_(2)) = (0.0900)/(0.030 ) = 3.0 \approx 3\\\\\rm (rate_(3) )/(rate_(2)) = (0.0247)/(0.00827) = 2.99 \approx 3`

In going from Experiment 3 to Experiment 3, we tripled the concentration and the rate tripled.

n = 1, so the rate law becomes

`\large \boxed{\textbf{ rate = k[ClO$_(2)$][OH$^(-)$]}}`

3. Write the overall order

Overall order = m + n = 1 + 1 = 2

The overall reaction order is 2.

4. Calculate k

From Experiment 1,

`\begin{array}{rcl}\text{rate} &=& \text{[ClO$_(2)$][OH$^(-)]$}\\0.248 & = & k \rm * 0.060 * 0.030\\& = & k * 1.8 * 10^(-3)\\k & = & (0.248)/(1.8 * 10^(-3))\\\\ &=& \textbf{138 M$^{\mathbf{-1}}$s$^{\mathbf{-1}}$}\\\end{array}`