For the reaction, `A (g) + 2B(g) rarr 2C(g) + 3D(g)` the change of enthalpy at `27^@C` is 19 kcal. The value of `Delta E` is:

Options

A. 21.2 kcal
B. 17.8 kcal
C. 18.4 kcal
D. 20.6 kcal

Explanation

To find the value of \(\Delta E\) for the reaction \(A (g) + 2B(g) \rightarrow 2C(g) + 3D(g)\) with a given change of enthalpy (\(\Delta H\)) of 19 kcal, we can follow these steps:

### Step 1: Identify the given data
We know:
- \(\Delta H = 19 \text{ kcal}\)
- The reaction is at \(27^\circ C\).

### Step 2: Convert \(\Delta H\) to calories
Since \(1 \text{ kcal} = 1000 \text{ calories}\):
\[ \Delta H = 19 \text{ kcal} = 19 \times 1000 \text{ calories} = 19000 \text{ calories} \]

### Step 3: Calculate \(\Delta N_G\)
\(\Delta N_G\) is the change in the number of moles of gaseous products minus the number of moles of gaseous reactants.
- Moles of gaseous products (\(n_P\)): \(2C + 3D = 5\)
- Moles of gaseous reactants (\(n_R\)): \(A + 2B = 3\)

Thus,
\[ \Delta N_G = n_P - n_R = 5 - 3 = 2 \]

### Step 4: Use the formula for \(\Delta E\)
The relationship between \(\Delta E\) and \(\Delta H\) is given by:
\[ \Delta E = \Delta H - \Delta N_G \cdot R \cdot T \]
Where:
- \(R\) is the universal gas constant, approximately \(2 \text{ cal/K/mol}\).
- \(T\) is the temperature in Kelvin. Convert \(27^\circ C\) to Kelvin:
\[ T = 27 + 273 = 300 \text{ K} \]

### Step 5: Calculate \(\Delta N_G \cdot R \cdot T\)
Now, calculate:
\[ \Delta N_G \cdot R \cdot T = 2 \cdot 2 \text{ cal/K/mol} \cdot 300 \text{ K} = 1200 \text{ calories} \]

### Step 6: Substitute values into the \(\Delta E\) equation
Now substitute the values into the equation:
\[ \Delta E = 19000 \text{ calories} - 1200 \text{ calories} = 17800 \text{ calories} \]

### Step 7: Convert \(\Delta E\) back to kilocalories
Convert calories back to kilocalories:
\[ \Delta E = \frac{17800 \text{ calories}}{1000} = 17.8 \text{ kcal} \]

### Final Answer
Thus, the value of \(\Delta E\) is:
\[ \Delta E = 17.8 \text{ kcal} \]

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