For a hypothetical reaction,
`A + 3B rarr P" "DeltaH=-2x` kJ/mole of A and
`Mrarr2Q+R" "Delta=+x` kJ/mole of M
If these reaction are carried simultaneously in a reactor such that temperature is not changing. If rate of disappearance of B is y M `sec^(-1)` then rate of formatiom (in M `sec^(-1)`) of Q is :

Options

A. `(2)/(3)y`
B. `(3)/(2)y`
C. `(4)/(3)y`
D. `(3)/(4)y`

Answer

To solve the problem, we need to analyze the two reactions given and their respective enthalpy changes. Let's break down the solution step by step.

### Step 1: Understand the Reactions and Their Enthalpy Changes
We have two reactions:
1. \( A + 3B \rightarrow P \) with \( \Delta H = -2x \) kJ/mole of A
2. \( M \rightarrow 2Q + R \) with \( \Delta H = +x \) kJ/mole of M

### Step 2: Analyze the Simultaneous Reactions
Since these reactions are occurring simultaneously in a reactor at constant temperature, the heat released by the first reaction (exothermic) will be utilized by the second reaction (endothermic).

### Step 3: Adjust the Enthalpy of the Second Reaction
To balance the heat exchange, we need to multiply the second reaction by 2 so that its enthalpy change becomes \( +2x \) kJ/mole of M. This gives us:
- \( 2M \rightarrow 4Q + 2R \) with \( \Delta H = +2x \) kJ/mole of 2M

### Step 4: Set Up the Rate of Reactions
Now, we can relate the rates of the two reactions. The rate of the first reaction can be expressed in terms of the disappearance of B:
\[ \text{Rate of reaction 1} = -\frac{dB}{dt} \cdot \frac{1}{3} \]
Thus, we have:
\[ \text{Rate of reaction 1} = -\frac{1}{3} \frac{dB}{dt} \]

For the second reaction, the rate of formation of Q can be expressed as:
\[ \text{Rate of reaction 2} = \frac{dQ}{dt} \cdot \frac{1}{2} \]
Thus, we have:
\[ \text{Rate of reaction 2} = \frac{1}{2} \frac{dQ}{dt} \]

### Step 5: Equate the Rates of the Two Reactions
Since the heat released by the first reaction is equal to the heat absorbed by the second reaction, we can equate the rates:
\[ -\frac{1}{3} \frac{dB}{dt} = \frac{1}{2} \frac{dQ}{dt} \]

### Step 6: Solve for the Rate of Formation of Q
We know that the rate of disappearance of B is given as \( y \) M/sec:
\[ -\frac{dB}{dt} = y \quad \Rightarrow \quad \frac{dB}{dt} = -y \]
Substituting this into the equation:
\[ -\frac{1}{3}(-y) = \frac{1}{2} \frac{dQ}{dt} \]
This simplifies to:
\[ \frac{y}{3} = \frac{1}{2} \frac{dQ}{dt} \]

### Step 7: Rearranging to Find \( \frac{dQ}{dt} \)
Now, we can solve for \( \frac{dQ}{dt} \):
\[ \frac{dQ}{dt} = \frac{2y}{3} \]

### Final Step: Conclusion
Thus, the rate of formation of Q is:
\[ \frac{dQ}{dt} = \frac{2y}{3} \text{ M/sec} \]