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 B in the given reaction:

**Step 1: Understand the reaction and the information given.**
The reaction is:
\[ 2A \rightleftharpoons 4B + C \]
We are given that the concentration of B increases by \(5 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1}\) over a time period of 10 seconds.

**Step 2: Calculate the total increase in concentration of B over the time period.**
The increase in concentration of B over 10 seconds is:
\[ \Delta [B] = 5 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1} \times 10 \, \text{s} = 5 \times 10^{-3} \, \text{mol L}^{-1} \]

**Step 3: Calculate the rate of appearance of B.**
The rate of appearance of B is defined as the change in concentration per unit time. Therefore, we can calculate it as follows:
\[ \text{Rate of appearance of B} = \frac{\Delta [B]}{\Delta t} = \frac{5 \times 10^{-3} \, \text{mol L}^{-1}}{10 \, \text{s}} = 5 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1} \]

**Final Answer:**
The rate of appearance of B is:
\[ 5 \times 10^{-4} \, \text{mol L}^{-1} \, \text{s}^{-1} \]

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