1) Write a basic form of the rate law for the following reaction. CIO4 + 8Br + 8H* → Cl + 4B12 + 4H2O

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**Chemical Kinetics: Reaction Rate and Activation Energy**

1) **Rate Law Determination**

   **Problem**: Write a basic form of the rate law for the following reaction.

   \[ \text{ClO}_4^- + 8\text{Br}^- + 8\text{H}^+ \rightarrow \text{Cl}^- + 4\text{Br}_2 + 4\text{H}_2\text{O} \]

   **Solution Approach**: The rate law is generally expressed as:

   \[ \text{Rate} = k[\text{ClO}_4^-]^m[\text{Br}^-]^n[\text{H}^+]^p \]

   where \( k \) is the rate constant, and \( m \), \( n \), and \( p \) are the orders of the reaction with respect to each species.

2) **Order of Reaction**

   **Problem**: If doubling the concentration of perchlorate ion in the reaction above causes the rate of reaction to double, what is the order of reaction with respect to perchlorate ion?

   **Solution Approach**: If doubling [ClO₄⁻] results in a doubling of the rate, then the reaction is first order with respect to perchlorate ion. Mathematically, this implies \( m = 1 \).

3) **Activation Energy Calculation**

   **Problem**: A plot of ln(Rate) vs. \( 1/T \) for the above reaction gave a straight line with a slope of \(-5021\). What is the activation energy for this reaction?

   **Solution Approach**: The relationship between the activation energy (Ea) and the slope of the Arrhenius plot is given by:

   \[ \text{slope} = -\frac{E_a}{R} \]

   where \( R \) is the gas constant (8.314 J/mol·K). Rearranging the equation for \( E_a \) gives:

   \[ E_a = -\text{slope} \times R = -(-5021) \times 8.314 \]

   Calculating \( E_a \) gives the activation energy for the reaction.

---

This exercise helps to understand the relationship between reaction rate, reaction order, and activation energy in chemical kinetics.
Transcribed Image Text:**Chemical Kinetics: Reaction Rate and Activation Energy** 1) **Rate Law Determination** **Problem**: Write a basic form of the rate law for the following reaction. \[ \text{ClO}_4^- + 8\text{Br}^- + 8\text{H}^+ \rightarrow \text{Cl}^- + 4\text{Br}_2 + 4\text{H}_2\text{O} \] **Solution Approach**: The rate law is generally expressed as: \[ \text{Rate} = k[\text{ClO}_4^-]^m[\text{Br}^-]^n[\text{H}^+]^p \] where \( k \) is the rate constant, and \( m \), \( n \), and \( p \) are the orders of the reaction with respect to each species. 2) **Order of Reaction** **Problem**: If doubling the concentration of perchlorate ion in the reaction above causes the rate of reaction to double, what is the order of reaction with respect to perchlorate ion? **Solution Approach**: If doubling [ClO₄⁻] results in a doubling of the rate, then the reaction is first order with respect to perchlorate ion. Mathematically, this implies \( m = 1 \). 3) **Activation Energy Calculation** **Problem**: A plot of ln(Rate) vs. \( 1/T \) for the above reaction gave a straight line with a slope of \(-5021\). What is the activation energy for this reaction? **Solution Approach**: The relationship between the activation energy (Ea) and the slope of the Arrhenius plot is given by: \[ \text{slope} = -\frac{E_a}{R} \] where \( R \) is the gas constant (8.314 J/mol·K). Rearranging the equation for \( E_a \) gives: \[ E_a = -\text{slope} \times R = -(-5021) \times 8.314 \] Calculating \( E_a \) gives the activation energy for the reaction. --- This exercise helps to understand the relationship between reaction rate, reaction order, and activation energy in chemical kinetics.
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