In a reaction, `2X + Y rarrX_2Y` , the X disappears at

Options

A. Half the rate as that of disappearance of Y
B. The same rate as that of disappearance of Y
C. The same rate as that of appearance of `X_2Y`
D. Twice the rate as that of appearance of `X_2Y`

Answer

To solve the problem, we need to analyze the reaction given:

**Reaction:**
\[ 2X + Y \rightarrow X_2Y \]

### Step-by-Step Solution:

1. **Identify the Reaction Components:**
- In this reaction, \(2X\) and \(Y\) are the reactants, and \(X_2Y\) is the product.

2. **Write the Rate of Disappearance:**
- The rate of disappearance of a reactant is expressed in terms of its change in concentration over time. For reactant \(X\), the rate of disappearance can be written as:
\[ -\frac{1}{2} \frac{d[X]}{dt} \]
The factor of \(\frac{1}{2}\) is due to the stoichiometry of the reaction, where 2 moles of \(X\) are consumed.

3. **Write the Rate of Disappearance for \(Y\):**
- Similarly, for reactant \(Y\), the rate of disappearance is:
\[ -\frac{d[Y]}{dt} \]

4. **Write the Rate of Appearance for the Product \(X_2Y\):**
- The rate of appearance of the product \(X_2Y\) can be expressed as:
\[ \frac{d[X_2Y]}{dt} \]

5. **Relate the Rates:**
- According to the stoichiometry of the reaction:
\[ -\frac{1}{2} \frac{d[X]}{dt} = -\frac{1}{1} \frac{d[Y]}{dt} = \frac{1}{1} \frac{d[X_2Y]}{dt} \]

6. **Determine the Relationship:**
- From the equations, we can derive that:
\[ \frac{d[X]}{dt} = -2 \frac{d[Y]}{dt} \]
- This indicates that the rate at which \(X\) disappears is twice the rate at which \(Y\) disappears.

7. **Conclusion:**
- Therefore, the rate of disappearance of \(X\) is twice the rate of disappearance of \(Y\) and can be expressed as:
\[ -\frac{d[X]}{dt} = 2 \left(-\frac{d[Y]}{dt}\right) \]

### Final Answer:
The rate at which \(X\) disappears is twice the rate at which \(Y\) disappears.

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