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 solve the problem regarding the reaction \( 2X_3 \rightleftharpoons 3X_2 \) and to find the rate of formation of \( X_2 \), we can follow these steps:

### Step 1: Write the rate expressions for the reaction
For the reaction \( 2X_3 \rightleftharpoons 3X_2 \), we can express the rates of change of concentrations of the reactants and products. The rate of disappearance of \( X_3 \) and the rate of formation of \( X_2 \) can be defined as follows:

\[ \text{Rate of formation of } X_2 = \frac{1}{3} \frac{d[X_2]}{dt} \]

\[ \text{Rate of disappearance of } X_3 = -\frac{1}{2} \frac{d[X_3]}{dt} \]

### Step 2: Relate the rates of formation and disappearance
According to the stoichiometry of the reaction, the rates are related by their coefficients in the balanced equation. Thus, we can equate the rates as follows:

\[ \frac{1}{3} \frac{d[X_2]}{dt} = -\frac{1}{2} \frac{d[X_3]}{dt} \]

### Step 3: Solve for the rate of formation of \( X_2 \)
From the above equation, we can express the rate of formation of \( X_2 \) in terms of the rate of disappearance of \( X_3 \):

\[ \frac{d[X_2]}{dt} = -\frac{3}{2} \frac{d[X_3]}{dt} \]

This means that for every 2 moles of \( X_3 \) that react, 3 moles of \( X_2 \) are formed.

### Step 4: Conclusion
Thus, the rate of formation of \( X_2 \) is given by:

\[ \text{Rate of formation of } X_2 = \frac{3}{2} \times \text{Rate of disappearance of } X_3 \]

### Final Answer
The rate of formation of \( X_2 \) is \( \frac{3}{2} \) times the rate of disappearance of \( X_3 \).

---