Which of the following statements is/are correct?

Options

A. The rate of the reaction involving the converison of ortho-hydrogen to parahydrogen is `-(d[H_(2)])/(dt) = k[H_(2)]^(3//2)`.
B. The rate of the reaction involving the thermal decompoistion of acetaldehyde is `k[CH_(3)CHO]^(3//2)`.
C. In the formation of phosgene gas form `CO` and `Cl`, the rate of the reaction is `k[CO][Cl_(2)]^(1//2)`.
D. In the decompoistion of `H_(2)O`, the rate of the reaction is `k[H_(2)O_(2)]`.

Answer

To solve the problem, we need to analyze the equilibrium reaction between AgCl and bromide ions (Br⁻) and determine the concentration of bromide ions at equilibrium. Here’s a step-by-step breakdown of the solution:

### Step 1: Understand the Reaction
The reaction between AgCl and bromide ions can be represented as follows:
\[ \text{AgCl (s)} + \text{Br}^- (aq) \rightleftharpoons \text{AgBr (s)} + \text{Cl}^- (aq) \]
In this equilibrium, AgCl is in excess, and we are interested in the changes in concentrations of Br⁻ and Cl⁻ ions.

### Step 2: Set Up Initial Concentrations
Initially, we have:
- Concentration of Br⁻ = 0.1 M
- Concentration of Cl⁻ = 0 M (since it is produced in the reaction)

At equilibrium:
- Concentration of Br⁻ = \( 0.1 - x \)
- Concentration of Cl⁻ = \( x \)

### Step 3: Calculate the Cell Potential
The cell potential (E_cell) can be calculated using the reduction potentials provided:
\[ E_{\text{cell}} = E_{\text{cathode}} - E_{\text{anode}} \]
Where:
- \( E_{\text{cathode}} \) (AgCl reduction potential) = 0.222 V
- \( E_{\text{anode}} \) (AgBr reduction potential) = 0.095 V

Calculating E_cell:
\[ E_{\text{cell}} = 0.222 - 0.095 = 0.127 \, \text{V} \]

### Step 4: Relate E_cell to Concentrations
Using the Nernst equation:
\[ E_{\text{cell}} = E^\circ - \frac{0.059}{n} \log Q \]
Where:
- \( Q = \frac{[\text{Cl}^-]}{[\text{Br}^-]} = \frac{x}{0.1 - x} \)
- \( n = 1 \) (since one electron is transferred)

Substituting the values:
\[ 0.127 = 0.059 \log \left( \frac{x}{0.1 - x} \right) \]

### Step 5: Solve for Q
Rearranging gives:
\[ \log \left( \frac{x}{0.1 - x} \right) = \frac{0.127}{0.059} \]
Calculating the right side:
\[ \frac{0.127}{0.059} \approx 2.152 \]

Thus:
\[ \frac{x}{0.1 - x} = 10^{2.152} \approx 142 \]

### Step 6: Solve for x
From the equation:
\[ x = 142(0.1 - x) \]
Expanding and rearranging:
\[ x + 142x = 14.2 \]
\[ 143x = 14.2 \]
\[ x = \frac{14.2}{143} \approx 0.09929 \, \text{M} \]

### Step 7: Find Concentration of Br⁻
Now, substituting x back to find the concentration of Br⁻:
\[ [\text{Br}^-] = 0.1 - x = 0.1 - 0.09929 \approx 0.00071 \, \text{M} \]

### Conclusion
The concentration of bromide ions at equilibrium is approximately 0.00071 M.

### Final Answer
The concentration of bromide ions is **0.00071 M**.

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