The rate of disappearance of A at two temperature is given by `A hArr B`
i. `(-d[A])/(dt)=2xx10^(-2)[A]-4xx10^(-3)[B]` at `300 K`
ii. `(-d[A])/(dt)=4xx10^(-2)[A]-16xx10^(-4)[B]` at `400 K`
From the given values of heat of reaction

Options

A. `3.86 kcal`
B. `6.93 kcal`
C. `1.68 kcal`
D. `1.68xx10^(-2) kcal`

Explanation

To solve the problem, we will follow these steps:

### Step 1: Write the rate equations for both temperatures.

At 300 K:
\[ -\frac{d[A]}{dt} = 2 \times 10^{-2}[A] - 4 \times 10^{-3}[B] \]

At 400 K:
\[ -\frac{d[A]}{dt} = 4 \times 10^{-2}[A] - 16 \times 10^{-4}[B] \]

### Step 2: Set the rates to zero at equilibrium.

At equilibrium, the rate of disappearance of A is zero. Therefore, we can set the equations to zero:

For 300 K:
\[ 2 \times 10^{-2}[A] = 4 \times 10^{-3}[B] \]

For 400 K:
\[ 4 \times 10^{-2}[A] = 16 \times 10^{-4}[B] \]

### Step 3: Solve for the equilibrium constant \( K_c \) at both temperatures.

For 300 K:
\[ \frac{B}{A} = \frac{2 \times 10^{-2}}{4 \times 10^{-3}} = 5 \]
Thus, \( K_{c, 300} = 5 \).

For 400 K:
\[ \frac{B}{A} = \frac{4 \times 10^{-2}}{16 \times 10^{-4}} = 25 \]
Thus, \( K_{c, 400} = 25 \).

### Step 4: Use the van 't Hoff equation to relate \( K_c \) and temperature.

The van 't Hoff equation is given by:
\[ \log K_c = -\frac{\Delta H}{2.303 R} \left( \frac{1}{T} \right) + \frac{\Delta S}{2.303 R} \]

We can write two equations based on the values of \( K_c \) at the two temperatures.

For 300 K:
\[ \log 5 = -\frac{\Delta H}{2.303 R} \left( \frac{1}{300} \right) + \frac{\Delta S}{2.303 R} \]

For 400 K:
\[ \log 25 = -\frac{\Delta H}{2.303 R} \left( \frac{1}{400} \right) + \frac{\Delta S}{2.303 R} \]

### Step 5: Subtract the two equations to eliminate \( \Delta S \).

Subtracting the first equation from the second gives:
\[ \log 25 - \log 5 = -\frac{\Delta H}{2.303 R} \left( \frac{1}{400} - \frac{1}{300} \right) \]

### Step 6: Simplify and solve for \( \Delta H \).

Calculating \( \log 25 - \log 5 \):
\[ \log 25 - \log 5 = \log \left( \frac{25}{5} \right) = \log 5 \]

So we have:
\[ \log 5 = -\frac{\Delta H}{2.303 R} \left( \frac{1}{400} - \frac{1}{300} \right) \]

Calculating \( \frac{1}{400} - \frac{1}{300} \):
\[ \frac{1}{400} - \frac{1}{300} = \frac{3 - 4}{1200} = -\frac{1}{1200} \]

Substituting this back into the equation:
\[ \log 5 = \frac{\Delta H}{2.303 R} \cdot \frac{1}{1200} \]

### Step 7: Calculate \( \Delta H \).

Rearranging gives:
\[ \Delta H = \log 5 \cdot 2.303 R \cdot 1200 \]

Using \( R = 8.314 \, \text{J/mol K} \):
\[ \Delta H = \log 5 \cdot 2.303 \cdot 8.314 \cdot 1200 \]

Calculating \( \log 5 \approx 0.6990 \):
\[ \Delta H \approx 0.6990 \cdot 2.303 \cdot 8.314 \cdot 1200 \]

Calculating gives:
\[ \Delta H \approx 3.8 \, \text{kcal} \]

### Final Answer:
The heat of the reaction \( \Delta H \) is approximately \( 3.8 \, \text{kcal} \).

---