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Assuring an elementary reaction H2O2+3I^...

Assuring an elementary reaction `H_2O_2+3I^(-)+2H^(+)rarr2H_2O+I_3^(-)` . The effect on the rate of this reaction brought about by doubling the concentration of `I^(-)` without changing the order

A

The rate would increase by a factor of 3

B

The rate would increase by a factor of 8

C

The rate would decrease by a factor of 1/3

D

The rate would increase by a factor of 9

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The correct Answer is:
To solve the problem, we need to analyze the given elementary reaction and determine how the rate of the reaction changes when the concentration of \( I^- \) is doubled. ### Step-by-Step Solution: 1. **Identify the Reaction and Rate Law**: The reaction given is: \[ H_2O_2 + 3I^- + 2H^+ \rightarrow 2H_2O + I_3^- \] For an elementary reaction, the rate law can be expressed as: \[ \text{Rate} = k [H_2O_2]^m [I^-]^n [H^+]^p \] where \( m, n, \) and \( p \) are the stoichiometric coefficients of the reactants in the balanced equation. 2. **Determine the Order of the Reaction**: From the balanced equation: - The order with respect to \( H_2O_2 \) is 1 (coefficient of 1). - The order with respect to \( I^- \) is 3 (coefficient of 3). - The order with respect to \( H^+ \) is 2 (coefficient of 2). Therefore, the overall rate law becomes: \[ \text{Rate} = k [H_2O_2]^1 [I^-]^3 [H^+]^2 \] 3. **Express the Initial Rate**: Let’s denote the initial rate of the reaction as \( R \): \[ R = k [H_2O_2]^1 [I^-]^3 [H^+]^2 \] 4. **Calculate the New Rate When \( [I^-] \) is Doubled**: If we double the concentration of \( I^- \): \[ [I^-] \rightarrow 2[I^-] \] The new rate \( R' \) will be: \[ R' = k [H_2O_2]^1 (2[I^-])^3 [H^+]^2 \] Simplifying this gives: \[ R' = k [H_2O_2]^1 \cdot 8[I^-]^3 [H^+]^2 \] 5. **Relate the New Rate to the Original Rate**: Since \( R = k [H_2O_2]^1 [I^-]^3 [H^+]^2 \), we can express \( R' \) in terms of \( R \): \[ R' = 8 \cdot R \] 6. **Conclusion**: Therefore, when the concentration of \( I^- \) is doubled, the rate of the reaction increases by a factor of 8. ### Final Answer: The rate of the reaction increases by a factor of 8 when the concentration of \( I^- \) is doubled. ---
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