Consider the following chemical reaction and the corresponding kinetic data showing the initial reaction rate as a function of the initial concentrations of the reactions :
`H_3AsO_4 (aq)+2H_3O^(+)(aq)+3l^(-)(aq)toHAsO_2(aq)+l_3^(-)(aq)+4H_2O(liq)`
`{:("Initial Rate"xx10^(-5)(M//sec),[H_3AsO_4],[H_3O^(+)],[I]),(3.7,0.001,0.01,0.10),(7.4,0.001,0.01,0.20),(7.4,0.002,0.01,0.10),(3.7,0.002,0.005,0.20):}`
Using the data, establish the correct reaction composite order.

Explanation

To determine the composite order of the given reaction using the provided kinetic data, we will follow these steps:

### Step 1: Write the rate law expression
The rate law for the reaction can be expressed as:
\[ \text{Rate} = k [H_3AsO_4]^\alpha [H_3O^+]^\beta [I^-]^\gamma \]
where \( \alpha \), \( \beta \), and \( \gamma \) are the orders of the reaction with respect to each reactant.

### Step 2: Analyze the provided data
We have the following data:
1. \( R_1 = 3.7 \times 10^{-5} \) M/s, \( [H_3AsO_4] = 0.001 \) M, \( [H_3O^+] = 0.01 \) M, \( [I^-] = 0.10 \) M
2. \( R_2 = 7.4 \times 10^{-5} \) M/s, \( [H_3AsO_4] = 0.001 \) M, \( [H_3O^+] = 0.01 \) M, \( [I^-] = 0.20 \) M
3. \( R_3 = 7.4 \times 10^{-5} \) M/s, \( [H_3AsO_4] = 0.002 \) M, \( [H_3O^+] = 0.01 \) M, \( [I^-] = 0.10 \) M
4. \( R_4 = 3.7 \times 10^{-5} \) M/s, \( [H_3AsO_4] = 0.002 \) M, \( [H_3O^+] = 0.005 \) M, \( [I^-] = 0.20 \) M

### Step 3: Determine the order with respect to \( [I^-] \)
Compare \( R_1 \) and \( R_2 \):
- \( R_1 = 3.7 \times 10^{-5} \) M/s with \( [I^-] = 0.10 \) M
- \( R_2 = 7.4 \times 10^{-5} \) M/s with \( [I^-] = 0.20 \) M

Using the rate law:
\[ \frac{R_2}{R_1} = \frac{k [H_3AsO_4]^\alpha [H_3O^+]^\beta [I^-]^{\gamma_2}}{k [H_3AsO_4]^\alpha [H_3O^+]^\beta [I^-]^{\gamma_1}} \]
Substituting the values:
\[ \frac{7.4 \times 10^{-5}}{3.7 \times 10^{-5}} = \frac{[0.001]^\alpha [0.01]^\beta [0.20]^\gamma}{[0.001]^\alpha [0.01]^\beta [0.10]^\gamma} \]
This simplifies to:
\[ 2 = \left( \frac{0.20}{0.10} \right)^\gamma \Rightarrow 2 = 2^\gamma \Rightarrow \gamma = 1 \]

### Step 4: Determine the order with respect to \( [H_3AsO_4] \)
Now compare \( R_1 \) and \( R_3 \):
- \( R_1 = 3.7 \times 10^{-5} \) M/s with \( [H_3AsO_4] = 0.001 \) M
- \( R_3 = 7.4 \times 10^{-5} \) M/s with \( [H_3AsO_4] = 0.002 \) M

Using the rate law:
\[ \frac{R_3}{R_1} = \frac{k [H_3AsO_4]^{\alpha_2} [H_3O^+]^\beta [I^-]^\gamma}{k [H_3AsO_4]^{\alpha_1} [H_3O^+]^\beta [I^-]^\gamma} \]
Substituting the values:
\[ \frac{7.4 \times 10^{-5}}{3.7 \times 10^{-5}} = \frac{[0.002]^\alpha [0.01]^\beta [0.10]^\gamma}{[0.001]^\alpha [0.01]^\beta [0.10]^\gamma} \]
This simplifies to:
\[ 2 = \left( \frac{0.002}{0.001} \right)^\alpha \Rightarrow 2 = 2^\alpha \Rightarrow \alpha = 1 \]

### Step 5: Determine the order with respect to \( [H_3O^+] \)
Now compare \( R_3 \) and \( R_4 \):
- \( R_3 = 7.4 \times 10^{-5} \) M/s with \( [H_3O^+] = 0.01 \) M
- \( R_4 = 3.7 \times 10^{-5} \) M/s with \( [H_3O^+] = 0.005 \) M

Using the rate law:
\[ \frac{R_4}{R_3} = \frac{k [H_3AsO_4]^\alpha [H_3O^+]^{\beta_2} [I^-]^\gamma}{k [H_3AsO_4]^\alpha [H_3O^+]^{\beta_1} [I^-]^\gamma} \]
Substituting the values:
\[ \frac{3.7 \times 10^{-5}}{7.4 \times 10^{-5}} = \frac{[0.002]^\alpha [0.005]^\beta [0.10]^\gamma}{[0.002]^\alpha [0.01]^\beta [0.10]^\gamma} \]
This simplifies to:
\[ \frac{1}{2} = \left( \frac{0.005}{0.01} \right)^\beta \Rightarrow \frac{1}{2} = \left( \frac{1}{2} \right)^\beta \Rightarrow \beta = 1 \]

### Step 6: Calculate the overall order
Now, we can sum the individual orders:
\[ \text{Overall order} = \alpha + \beta + \gamma = 1 + 1 + 1 = 3 \]

### Final Answer
The composite order of the reaction is 3.

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