A gaseous hypothetical chemical equation `2A hArr 4B + C` is carried out in a closed vessel. The concentration of B is found to increase by `5xx 10^(-4)mol.l^(-1) s^(-1)" in"`10 second. The rate of appearance of B is:

Options

A. `5 xx 10^(-4)mol.l^(-1) s^(-1) `
B. `5 xx 10^(-5) mol.l^(-1) s^(-1) `
C. `6 xx 10^(-5) mol.l^(-1) s^(-1)`
D. `4 xx 10^(-4) mol.l^(-1) s^(-1)`

Answer

To solve the problem, we need to determine the rate of appearance of substance B in the given chemical reaction. The reaction is:

\[ 2A \rightleftharpoons 4B + C \]

### Step-by-Step Solution:

1. **Identify the Given Information**:
- The concentration of B increases by \( 5 \times 10^{-4} \, \text{mol L}^{-1} \) over a time period of \( 10 \, \text{s} \).

2. **Understand the Rate of Appearance**:
- The rate of appearance of a substance in a reaction is defined as the change in concentration of that substance over the time taken for that change.

3. **Calculate the Rate of Appearance of B**:
- The formula for the rate of appearance of B (\( R_B \)) can be expressed as:
\[ R_B = \frac{\Delta [B]}{\Delta t} \]
where \( \Delta [B] \) is the change in concentration of B and \( \Delta t \) is the change in time.

4. **Substitute the Values**:
- From the problem, we have:
- \( \Delta [B] = 5 \times 10^{-4} \, \text{mol L}^{-1} \)
- \( \Delta t = 10 \, \text{s} \)
- Plugging in these values:
\[ R_B = \frac{5 \times 10^{-4} \, \text{mol L}^{-1}}{10 \, \text{s}} = 5 \times 10^{-5} \, \text{mol L}^{-1} \text{s}^{-1} \]

5. **Final Answer**:
- The rate of appearance of B is \( 5 \times 10^{-5} \, \text{mol L}^{-1} \text{s}^{-1} \).