Kinetics: The Oxidation of Iodide by Hydrogen Peroxide

Results

Name q,

Part 1

Write the differential rate law for the reaction between iodide and hydrogen peroxide.

For most purposes, if a small percentage of the reactants are used (less than 10%), ?the concentration change may be ignored when comparing the initial rate to the average rate. Show, by calculation, that for the present experiments, the number of moles of peroxide used in each run and then calculate what percent of the original hydrogen peroxide was used.

Part 2

What are the values of the activation energy and the Arrhenius pre-exponential factor for the reaction between iodide and hydrogen peroxide? Be sure to include the correct units and number of significant figures.

A q,

Part 3

Compare the rate of the catalyzed reaction to the uncatalyzed reaction. Would you say the iron(II) ?ions are a good catalyst for this reaction? Why?

The Correct Answer and Explanation is :

To address the kinetics of the oxidation of iodide by hydrogen peroxide, here’s the detailed breakdown:

Part 1: Differential Rate Law

The reaction between iodide (( I^- )) and hydrogen peroxide (( H_2O_2 )) proceeds as follows:
[ 2 I^- + H_2O_2 + 2 H^+ \rightarrow I_2 + 2 H_2O ]

The differential rate law can be expressed as:
[ \text{Rate} = k[I^-]^m[H_2O_2]^n[H^+]^p ]
Where:

( k ) is the rate constant,
( m, n, ) and ( p ) are the reaction orders with respect to ( I^- ), ( H_2O_2 ), and ( H^+ ), respectively.
These orders can be determined experimentally by observing how the rate changes with varying concentrations of reactants.

Part 1: Percent of Original Hydrogen Peroxide Used

To calculate the percentage of hydrogen peroxide used:

Determine the moles of ( H_2O_2 ) initially:
[ \text{moles of } H_2O_2 = \text{concentration (M)} \times \text{volume (L)} ]
Determine the moles of ( H_2O_2 ) consumed in the reaction:
Use stoichiometry and the experimental data (e.g., iodine formed).
Calculate the percentage used:
[ \text{Percent used} = \left( \frac{\text{moles consumed}}{\text{moles initial}} \right) \times 100 ]

For these experiments, if less than 10% of the original hydrogen peroxide is consumed, the concentration change is negligible.

Part 2: Activation Energy and Arrhenius Pre-Exponential Factor

The Arrhenius equation is:
[ k = A e^{-\frac{E_a}{RT}} ]
Where:

( k ) is the rate constant,
( A ) is the pre-exponential factor,
( E_a ) is the activation energy (J/mol or kJ/mol),
( R ) is the gas constant (( 8.314 \, \text{J mol}^{-1} \text{K}^{-1} )),
( T ) is the temperature in Kelvin.

To find ( E_a ) and ( A ), you need experimental rate constants (( k )) at different temperatures. Using the natural logarithmic form:
[ \ln k = \ln A – \frac{E_a}{R} \cdot \frac{1}{T} ]

Plot ( \ln k ) vs ( \frac{1}{T} ).
The slope of the line (( -\frac{E_a}{R} )) gives ( E_a ), and the y-intercept gives ( \ln A ).

Ensure values are reported to appropriate significant figures based on the data.

Part 3: Catalyzed vs. Uncatalyzed Reaction

Catalyzed Reaction
The reaction can be catalyzed by iron(II) ions (( Fe^{2+} )):
[ 2 I^- + H_2O_2 + 2 H^+ \xrightarrow{Fe^{2+}} I_2 + 2 H_2O ]

Rate Comparison
The catalyzed reaction proceeds significantly faster because ( Fe^{2+} ) lowers the activation energy by providing an alternative pathway.

Example Calculation:

Uncatalyzed rate constant: ( k_1 )
Catalyzed rate constant: ( k_2 )

The ratio ( \frac{k_2}{k_1} ) demonstrates the increase in rate. A large ratio (( >10^2 )) confirms the catalyst’s effectiveness.

Explanation (300 words)

The catalytic effect of ( Fe^{2+} ) ions is due to their ability to act as intermediaries, reducing the activation energy of the reaction. In the uncatalyzed reaction, the energy required for reactants to reach the transition state is significantly higher. Adding ( Fe^{2+} ), the reaction is broken into smaller steps, each with a lower activation energy. For example, ( Fe^{2+} ) can alternate between ( Fe^{2+} ) and ( Fe^{3+} ), facilitating electron transfer between iodide and hydrogen peroxide.

Experimental data shows a dramatic increase in reaction rate upon adding the catalyst. This suggests ( Fe^{2+} ) effectively promotes the reaction, justifying its use. Furthermore, the catalyst is not consumed in the reaction, maintaining its efficiency over time.

This result highlights the importance of catalysts in chemical processes, especially in industrial and biological systems where reaction speed and energy efficiency are critical.