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 find the rate of disappearance of X for the reaction \(2X + Y \rightarrow Z\), we can follow these steps:

### Step 1: Understand the relationship between the rates of reaction
For the reaction \(2X + Y \rightarrow Z\), the stoichiometry tells us that:
- 2 moles of X disappear for every 1 mole of Z that appears.

### Step 2: Write the rate expressions
The rate of disappearance of X (\(-\frac{d[X]}{dt}\)) and the rate of appearance of Z (\(\frac{d[Z]}{dt}\)) are related by their stoichiometric coefficients. We can express this relationship as:
\[ -\frac{1}{2} \frac{d[X]}{dt} = \frac{d[Z]}{dt} \]
This means that the rate of disappearance of X is twice the rate of appearance of Z.

### Step 3: Substitute the given rate of appearance of Z
We are given that the rate of appearance of Z is:
\[ \frac{d[Z]}{dt} = 0.05 \, \text{mol L}^{-1} \text{min}^{-1} \]

### Step 4: Calculate the rate of disappearance of X
Using the relationship established in Step 2, we can express the rate of disappearance of X as:
\[ -\frac{d[X]}{dt} = 2 \times \frac{d[Z]}{dt} \]
Substituting the value of \(\frac{d[Z]}{dt}\):
\[ -\frac{d[X]}{dt} = 2 \times 0.05 \, \text{mol L}^{-1} \text{min}^{-1} = 0.1 \, \text{mol L}^{-1} \text{min}^{-1} \]

### Step 5: Write the final answer
Thus, the rate of disappearance of X is:
\[ \frac{d[X]}{dt} = 0.1 \, \text{mol L}^{-1} \text{min}^{-1} \]

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

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