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 ΔE for the reaction \( A (g) + 2B(g) \rightarrow 2C(g) + 3D(g) \) with a given change of enthalpy (ΔH) of 19 kcal, we can use the relationship between ΔH, ΔE, and the change in the number of moles of gas (ΔnG).

### Step-by-Step Solution:

1. **Identify the given data**:
- ΔH = 19 kcal
- Temperature (T) = 27°C = 300 K (converted to Kelvin by adding 273)

2. **Calculate ΔnG**:
- ΔnG = (Number of moles of gaseous products) - (Number of moles of gaseous reactants)
- For the reaction:
- Products: 2C + 3D = 5 moles of gas
- Reactants: A + 2B = 3 moles of gas
- Therefore, ΔnG = 5 - 3 = 2

3. **Use the relationship between ΔH and ΔE**:
- The relationship is given by the equation:
\[ ΔH = ΔE + ΔnG \cdot R \cdot T \]
- Where R is the universal gas constant. In kcal, R = 0.001987 kcal/(mol·K) or approximately 2 cal/(mol·K).

4. **Substitute the known values into the equation**:
- Rearranging the equation gives:
\[ ΔE = ΔH - ΔnG \cdot R \cdot T \]
- Substituting the values:
\[ ΔE = 19 \text{ kcal} - (2) \cdot (0.001987 \text{ kcal/(mol·K)}) \cdot (300 \text{ K}) \]

5. **Calculate ΔE**:
- First, calculate the term \( ΔnG \cdot R \cdot T \):
\[ ΔnG \cdot R \cdot T = 2 \cdot 0.001987 \cdot 300 = 1.1922 \text{ kcal} \]
- Now substitute this back into the equation for ΔE:
\[ ΔE = 19 - 1.1922 = 17.8078 \text{ kcal} \]
- Rounding to three significant figures, we get:
\[ ΔE \approx 17.8 \text{ kcal} \]

### Final Answer:
ΔE = 17.8 kcal