### Chemical Reaction Kinetics: Initial Rates MethodThe following reaction was studied using the method of initial rates:\[ 5 \, \text{Br}^-(\text{aq}) + \text{BrO}_3^-(\text{aq}) + \text{H}^+(\text{aq}) \rightarrow 3 \, \text{Br}_2(\text{aq}) + 3 \, \text{H}_2\text{O}(l) \]The table below shows the experimental results:| Experiment | [Br\(^-\)]\(_0\) (M) | [BrO\(_3^-\)] (M) | [H\(^+\)] (M) | \((\Delta[\text{BrO}_3^-]/\Delta t)_0\) \((\text{M s}^{-1})\) ||:----------:|:------------------:|:----------------:|:-------------:|:----------------------------------:|| I | 0.10 | 0.10 | 0.10 | \(6.8 \times 10^{-4}\) || II | 0.15 | 0.10 | 0.10 | \(1.0 \times 10^{-3}\) || III | 0.10 | 0.20 | 0.10 | \(1.4 \times 10^{-3}\) || IV | 0.10 | 0.10 | 0.25 | \(4.3 \times 10^{-3}\) |### Questions:1. **What is the rate law for the reaction?** - Example input: `rate = k . [A]^2 . [B]`.2. **What is the value of the rate constant at the temperature of the experiments?** - Input box provided for numerical answer.3. **What are the units of the rate constant?** - Options: - \( \text{M}^{-2}\text{s}^{-2} \) - \( \text{M}^{-2} \text{s}^{-1} \) - \( \text{M}^{-3} \text{s}^{-1} \) - \( \text

Given :- The given reaction is 5Br-(aq) + BrO3-(aq) + H+(aq) → 3Br2(aq) + 3H2O (l) --------------------------------------Explanation :- Suppose the rate law is rate = k[Br-]x[BrO3-]y[H+]z..... (A) To write rate law, the values of the exponents x, y and z are to be calculated From table we have, r1 = 6.8×10-4 (M/s) = k(0.10 M)x×(0.10 M)y×(0.10 M)z ......(1) r2 = 1.0×10-3 (M/s) = k(0.15 M)x×(0.10 M)y×(0.10 M)z ......(2) r3 = 1.4×10-3 (M/s) = k(0.10 M)x×(0.20 M)y×(0.10 M)z......(3) r4 = 4.3×10-3 (M/s) = k(0.10 M)x×(0.10 M)y×(0.25 M)z......(4) -----------------------------------------(1) Determination of x Now, divide equation (2) by equation (1) we get, r2r1 i.e. 1.0×10-3 (M/s)6.8×10-4 (M/s) = k(0.15 M)x×(0.10 M)y×(0.10 M)zk(0.10 M)x×(0.10 M)y×(0.10 M)z i.e. 1.5 = (0.15 M)x(0.10 M)x i.e. 1.5 = (0.15 M0.10 M)x i.e. 1.5 = (1.5)x i.e. 1.5 = (1.5)1 = (1.5)x i.e. x = 1 -----------------------------------------(2) Determination of y Now, divide equation (3) by equation (1) we get, r3r1 i.e. 1.4×10-3 (M/s)6.8×10-4 (M/s) = k(0.10 M)x×(0.20 M)y×(0.10 M)zk(0.10 M)x×(0.10 M)y×(0.10 M)z i.e. 2 = (0.20 M)y(0.10 M)y i.e. 2 = (0.20 M0.10 M)y i.e. 2 = (2)y i.e. 2 = 21 = (2)y i.e. y = 1 -----------------------------------------(3) Determination of z Now, divide equation (4) by equation (1) we get, r4r1 i.e. 4.3×10-3 (M/s)6.8×10-4 (M/s) = k(0.10 M)x×(0.10 M)y×(0.25 M)zk(0.10 M)x×(0.10 M)y×(0.10 M)z i.e. 6.3 = (0.25 M)z(0.10 M)z i.e. 6.3 = (0.25 M0.10 M)z i.e. 6.3 = (2.5)z i.e. 6.3 = 2.52 = (2.5)z i.e. z = 2 -----------------------------------------(4) Complete rate law for reaction The rate law is rate = k[Br-]x[BrO3-]y[H+]z..... (A) Since, x = 1 , y = 1 and z = 2 Therefore, From equation (A) The rate law for given reaction is rate = k[Br-][BrO3-][H+]2 .... (B) -----------------------------------------(5) Determination of rate constant k The rate law for given reaction is rate = k[Br-][BrO3-][H+]2 From equation (1) we have, 6.8×10-4 (M/s) = k(0.10 M)1×(0.10 M)1×(0.10 M)2 i.e. 6.8×10-4 (M/s) = k×0.10 M×0.10 M×0.010 M2 i.e. k = 6.8×10-4 (M/s) 0.10 M×0.10 M×0.010 M2 i.e. k = 6.8(M-3.s-1) -----------------------------------------The unit of rate constant Since we have, k = 6.8(M-3.s-1) Therefore, the unit of rate constant is M-3.s-1 -----------------------------------------Answer :- 1) rate = k[Br-][BrO3-][H+]2 2) Rate constant = k = 6.8(M-3.s-1) 3) The unit of rate constant = M-3.s-1 *************************************

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The following reaction was followed by the method of initial rates: 5 Br (aq) + BrO3¯(aq) + H*(aq) →→→ 3 Br₂(aq) + 3 H₂O(/) with the following results. Exp I II III IV [Br] [BrO3¯ ] 0.10 0.15 0.10 0.10 0.10 0.10 0.20 0.10 What are the units of the rate constant? O M-²5-2 O M-²5-1 O M-³5-1 OM-³5-2 H+ 0.10 0.10 0.10 0.25 (A[BrO3 ]/At)o M.S-1 6.8 x 10-4 1.0 x 10-3 1.4 x 10-3 4.3 x 10-3³ What is the rate law for the reaction? (Example input 'rate = k. [A]^2 . [B]'.) What is the value of the rate constant at the temperature of the experiments? 4.0

The following reaction was followed by the method of initial rates: 5 Br (aq) + BrO3¯(aq) + H*(aq) →→→ 3 Br₂(aq) + 3 H₂O(/) with the following results. Exp I II III IV [Br] [BrO3¯ ] 0.10 0.15 0.10 0.10 0.10 0.10 0.20 0.10 What are the units of the rate constant? O M-²5-2 O M-²5-1 O M-³5-1 OM-³5-2 H+ 0.10 0.10 0.10 0.25 (A[BrO3 ]/At)o M.S-1 6.8 x 10-4 1.0 x 10-3 1.4 x 10-3 4.3 x 10-3³ What is the rate law for the reaction? (Example input 'rate = k. [A]^2 . [B]'.) What is the value of the rate constant at the temperature of the experiments? 4.0

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Chapter1: Chemical Foundations

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### Chemical Reaction Kinetics: Initial Rates Method

The following reaction was studied using the method of initial rates:

\[
5 \, \text{Br}^-(\text{aq}) + \text{BrO}_3^-(\text{aq}) + \text{H}^+(\text{aq}) \rightarrow 3 \, \text{Br}_2(\text{aq}) + 3 \, \text{H}_2\text{O}(l)
\]

The table below shows the experimental results:

| Experiment | [Br\(^-\)]\(_0\) (M) | [BrO\(_3^-\)] (M) | [H\(^+\)] (M) | \((\Delta[\text{BrO}_3^-]/\Delta t)_0\) \((\text{M s}^{-1})\) |
|:----------:|:------------------:|:----------------:|:-------------:|:----------------------------------:|
| I | 0.10 | 0.10 | 0.10 | \(6.8 \times 10^{-4}\) |
| II | 0.15 | 0.10 | 0.10 | \(1.0 \times 10^{-3}\) |
| III | 0.10 | 0.20 | 0.10 | \(1.4 \times 10^{-3}\) |
| IV | 0.10 | 0.10 | 0.25 | \(4.3 \times 10^{-3}\) |

### Questions:

1. **What is the rate law for the reaction?**
- Example input: `rate = k . [A]^2 . [B]`.

2. **What is the value of the rate constant at the temperature of the experiments?**
- Input box provided for numerical answer.

3. **What are the units of the rate constant?**
- Options:
- \( \text{M}^{-2}\text{s}^{-2} \)
- \( \text{M}^{-2} \text{s}^{-1} \)
- \( \text{M}^{-3} \text{s}^{-1} \)
- \( \text

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