**Topic: Reaction Kinetics****Understanding the Order of a Reaction Relative to Its Stoichiometry**3. **Relationship Between Reaction Order and Stoichiometry:**If the order of a reaction is the same as its stoichiometry, the order of each reactant is the same number as the coefficient of that reactant in the balanced equation. When this is the case, it is possible (but not necessary) that the reaction takes place in one step. If the order of a reaction is different from its stoichiometry, the reaction must occur in steps. For the reaction used in this experiment, we will show how the rate law (equation 1 on page 1) would look if the reaction took place in one step:**Rate Law Representation:**\[ \text{rate} = k[\text{I}^−]^x[\text{BrO}_3^−]^y[\text{H}^+]^z \]Where:- $ k $ is the rate constant,- $ [\text{I}^−] $, $ [\text{BrO}_3^−] $, and $ [\text{H}^+] $ are the concentrations of the reactants,- $ x, y, $ and $ z $ are the orders of the reaction with respect to each reactant.**Balanced Chemical Equation:**\[ 6 \text{I}^− (\text{aq}) + \text{BrO}_3^− (\text{aq}) + 6 \text{H}^+ (\text{aq}) \rightarrow 3 \text{I}_2 (\text{aq}) + \text{Br}^− (\text{aq}) + 3 \text{H}_2\text{O} \]**Explanation of Symbols and Notations:**- $ \text{I}^− $ stands for the iodide ion.- $ \text{BrO}_3^− $ stands for the bromate ion.- $ \text{H}^+ $ represents the hydrogen ion (proton).- $ \text{I}_2 $ is molecular iodine.- $ \text{Br}^− $ is the bromide ion.- $ \text{H}_2\text{O} $ is water.The coefficients in the balanced equation indicate the

The rate law of a chemical reaction describes the relation between the…

3. If the order of a reaction is the same as its stoichiometry, the order of each reactant is the same number as the coef- ficient of that reactant in the balanced equation. When this is the case, it is possible (but not necessary) that the reac- tion takes place in one step. If the order of a reaction is different than its stoichiometry, then the reaction must take place in steps. For the reaction used in this experiment, show how the rate law (equation 1 on page 1) would look if the reaction took place in one step: = k[I¯ ]*[BrO;]¹[H* ]² rate= 61-(aq) + BrO3(aq) + 6H*(aq) → 3 1₂(aq) + Br¯(aq) + 3 H₂O

3. If the order of a reaction is the same as its stoichiometry, the order of each reactant is the same number as the coef- ficient of that reactant in the balanced equation. When this is the case, it is possible (but not necessary) that the reac- tion takes place in one step. If the order of a reaction is different than its stoichiometry, then the reaction must take place in steps. For the reaction used in this experiment, show how the rate law (equation 1 on page 1) would look if the reaction took place in one step: = k[I¯ ][BrO;]¹[H ]² rate= 61-(aq) + BrO3(aq) + 6H*(aq) → 3 1₂(aq) + Br¯(aq) + 3 H₂O

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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