For a chemical reaction `2X + Y rarrZ`, the rate of appearance of `Z` is `0.05 mol L^(-1) min^(-1)`. The rate of diappearance of `X` will be

Options

A. `0.05 mol L^(-1) hr^(-1)`
B. `0.05 mol L^(-1) min^(-1)`
C. `0.1 mol L^(-1) min^(-1)`
D. `0.25 mol L^(-1) min^(-1)`

Explanation

To solve the problem, we need to relate the rate of appearance of product Z to the rate of disappearance of reactant X using stoichiometry. The given reaction is:

\[ 2X + Y \rightarrow Z \]

### Step-by-Step Solution:

1. **Identify the Stoichiometric Coefficients**:
In the reaction, the stoichiometric coefficients are:
- For \( X \): 2
- For \( Y \): 1
- For \( Z \): 1

2. **Write the Rate Expressions**:
The rate of a reaction can be expressed in terms of the change in concentration of the reactants and products. For the reaction:
\[ \text{Rate of disappearance of } X = -\frac{1}{2} \frac{d[X]}{dt} \]
\[ \text{Rate of disappearance of } Y = -\frac{1}{1} \frac{d[Y]}{dt} \]
\[ \text{Rate of appearance of } Z = \frac{d[Z]}{dt} \]

3. **Relate the Rates**:
From the stoichiometry of the reaction, we can relate the rates:
\[ -\frac{1}{2} \frac{d[X]}{dt} = \frac{d[Z]}{dt} \]
This means that for every 1 mole of Z produced, 2 moles of X are consumed.

4. **Substitute the Given Rate**:
We are given that the rate of appearance of Z is:
\[ \frac{d[Z]}{dt} = 0.05 \, \text{mol L}^{-1} \text{min}^{-1} \]
Now substituting this value into the rate expression:
\[ -\frac{1}{2} \frac{d[X]}{dt} = 0.05 \]

5. **Solve for the Rate of Disappearance of X**:
To find \(\frac{d[X]}{dt}\), we rearrange the equation:
\[ \frac{d[X]}{dt} = -2 \times 0.05 \]
\[ \frac{d[X]}{dt} = -0.1 \, \text{mol L}^{-1} \text{min}^{-1} \]

6. **Interpret the Result**:
The negative sign indicates that X is disappearing. Therefore, the rate of disappearance of X is:
\[ \frac{d[X]}{dt} = 0.1 \, \text{mol L}^{-1} \text{min}^{-1} \]

### Final Answer:
The rate of disappearance of \( X \) is \( 0.1 \, \text{mol L}^{-1} \text{min}^{-1} \).

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