For the reaction P `to` 2Q, the rate of formation of Q is 0.50 moles per litre-hour. What is the rate of disappearance of P?

Options

A. `0.50 mol//l-hr`
B. `0.25 mol//l-hr`
C. `1.0 mol//l-hr`
D. `0.75 mol//l-hr`

Answer

To solve the problem, we need to determine the rate of disappearance of reactant P in the reaction where P converts to 2Q. The rate of formation of Q is given as 0.50 moles per liter-hour.

### Step-by-Step Solution:

1. **Write the balanced chemical equation**:
The reaction is given as:
\[ P \rightarrow 2Q \]

2. **Identify the relationship between the rates**:
From the stoichiometry of the reaction, we can establish the relationship between the rate of disappearance of P and the rate of formation of Q. The rate of disappearance of P is related to the rate of formation of Q as follows:
\[ -\frac{d[P]}{dt} = \frac{1}{2} \frac{d[Q]}{dt} \]
Here, \(-\frac{d[P]}{dt}\) is the rate of disappearance of P, and \(\frac{d[Q]}{dt}\) is the rate of formation of Q.

3. **Substitute the known value**:
We know that the rate of formation of Q is given as:
\[ \frac{d[Q]}{dt} = 0.50 \text{ moles per liter per hour} \]
Now, substituting this value into the equation:
\[ -\frac{d[P]}{dt} = \frac{1}{2} \times 0.50 \]

4. **Calculate the rate of disappearance of P**:
Performing the calculation:
\[ -\frac{d[P]}{dt} = \frac{0.50}{2} = 0.25 \text{ moles per liter per hour} \]
Therefore, the rate of disappearance of P is:
\[ \frac{d[P]}{dt} = 0.25 \text{ moles per liter per hour} \]

### Final Answer:
The rate of disappearance of P is **0.25 moles per liter per hour**.

---