Be sure to answer all parts. By what factor does the rate shown below change if each of the following changes occurs: Rate = k[BrO; ][Br ][#*] If BrO, is doubled, the rate (select) by a factor of If Br is halved, the rate (select) by a factor of If H* | is quadrupled, the rate (select) by a factor of

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**Kinetics Question: Understanding Rate Changes**

Be sure to answer all parts.

By what factor does the rate shown below change if each of the following changes occurs:

\[ \text{Rate} = k[ \text{BrO}_3^- ][ \text{Br}^- ] [ \text{H}^+ ]^2 \]

1. **If** \([ \text{BrO}_3^- ]\) **is doubled, the rate** 
   \[ \text{(select) by a factor of} \ \_\_ .\]
   
2. **If** \([ \text{Br}^- ]\) **is halved, the rate** 
   \[ \text{(select) by a factor of} \ \_\_ .\]
   
3. **If** \([ \text{H}^+ ]\) **is quadrupled, the rate** 
   \[ \text{(select) by a factor of} \ \_\_ .\]

**Explanation of the Rate Law Equation Components:**

- \(\text{k}\) is the rate constant.
- \([ \text{BrO}_3^- ]\) represents the concentration of bromate ions.
- \([ \text{Br}^- ]\) represents the concentration of bromide ions.
- \([ \text{H}^+ ]^2\) represents the square of the concentration of hydrogen ions.

**Analysis of Each Change:**

1. Doubling \([ \text{BrO}_3^- ]\): Since \([ \text{BrO}_3^- ]\) is directly proportional to the rate, doubling this concentration will double the rate.
2. Halving \([ \text{Br}^- ]\): Since \([ \text{Br}^- ]\) is directly proportional to the rate, halving this concentration will halve the rate.
3. Quadrupling \([ \text{H}^+ ]\): Since the rate depends on \([ \text{H}^+ ]\) squared, quadrupling \([ \text{H}^+ ]\) will increase the rate by a factor of \(4^2 = 16\).

Please ensure your answers match the proportional changes for each scenario based on the provided rate law.
Transcribed Image Text:**Kinetics Question: Understanding Rate Changes** Be sure to answer all parts. By what factor does the rate shown below change if each of the following changes occurs: \[ \text{Rate} = k[ \text{BrO}_3^- ][ \text{Br}^- ] [ \text{H}^+ ]^2 \] 1. **If** \([ \text{BrO}_3^- ]\) **is doubled, the rate** \[ \text{(select) by a factor of} \ \_\_ .\] 2. **If** \([ \text{Br}^- ]\) **is halved, the rate** \[ \text{(select) by a factor of} \ \_\_ .\] 3. **If** \([ \text{H}^+ ]\) **is quadrupled, the rate** \[ \text{(select) by a factor of} \ \_\_ .\] **Explanation of the Rate Law Equation Components:** - \(\text{k}\) is the rate constant. - \([ \text{BrO}_3^- ]\) represents the concentration of bromate ions. - \([ \text{Br}^- ]\) represents the concentration of bromide ions. - \([ \text{H}^+ ]^2\) represents the square of the concentration of hydrogen ions. **Analysis of Each Change:** 1. Doubling \([ \text{BrO}_3^- ]\): Since \([ \text{BrO}_3^- ]\) is directly proportional to the rate, doubling this concentration will double the rate. 2. Halving \([ \text{Br}^- ]\): Since \([ \text{Br}^- ]\) is directly proportional to the rate, halving this concentration will halve the rate. 3. Quadrupling \([ \text{H}^+ ]\): Since the rate depends on \([ \text{H}^+ ]\) squared, quadrupling \([ \text{H}^+ ]\) will increase the rate by a factor of \(4^2 = 16\). Please ensure your answers match the proportional changes for each scenario based on the provided rate law.
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