For the reaction, `2X_(3)hArr 3X_(2)`, the rate of formation of `X_(2)` is

Options

A. `3(-(d[X_(3)])/(dt))`
B. `(1)/(2)(-(d[X_(3)])/(dt))`
C. `(1)/(3)(-(d[X_(3)])/(dt))`
D. `(3)/(2)(-(d[X_(3)])/(dt))`

Answer

To find the rate of formation of \( X_2 \) in the reaction \( 2X_3 \rightleftharpoons 3X_2 \), we can follow these steps:

### Step 1: Write the balanced chemical equation
The balanced equation for the reaction is:
\[ 2X_3 \rightleftharpoons 3X_2 \]

### Step 2: Identify the stoichiometric coefficients
From the balanced equation, we can see that:
- The stoichiometric coefficient of \( X_3 \) is 2.
- The stoichiometric coefficient of \( X_2 \) is 3.

### Step 3: Write the rate of change expressions
For the reaction, the rate of change of concentration can be expressed as:
- For \( X_3 \) (which is being consumed):
\[ -\frac{d[X_3]}{dt} = \frac{1}{2} \frac{d[X_2]}{dt} \]
- For \( X_2 \) (which is being formed):
\[ \frac{d[X_2]}{dt} = \frac{3}{2} \left(-\frac{d[X_3]}{dt}\right) \]

### Step 4: Relate the rates
From the stoichiometry of the reaction, we can relate the rate of formation of \( X_2 \) to the rate of consumption of \( X_3 \):
\[ \frac{d[X_2]}{dt} = \frac{3}{2} \left(-\frac{d[X_3]}{dt}\right) \]

### Step 5: Solve for the rate of formation of \( X_2 \)
Thus, the rate of formation of \( X_2 \) can be expressed as:
\[ \frac{d[X_2]}{dt} = \frac{3}{2} \left(-\frac{d[X_3]}{dt}\right) \]

### Conclusion
The rate of formation of \( X_2 \) is given by:
\[ \frac{d[X_2]}{dt} = \frac{3}{2} \left(-\frac{d[X_3]}{dt}\right) \]

### Final Answer
The rate of formation of \( X_2 \) is \( \frac{3}{2} \left(-\frac{d[X_3]}{dt}\right) \).

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