For a gaseous reaction, `A(g)+3B(g)to3C(g)+3D(g),DeltaU` is 17 kcal at `27^(@)C`. Assuming `R=2cal" "K^(-1)mol^(-1)`, the value of `DeltaH` for the above reaction is:

Options

A. 15.8kcal
B. 18.2 kcal
C. 20.0 kcal
D. 16.4 kcal

Answer

To solve the problem, we will use the relationship between the change in internal energy (ΔU) and the change in enthalpy (ΔH) for a gaseous reaction, which is given by the equation:

\[ \Delta H = \Delta U + \Delta n_g RT \]

Where:
- \(\Delta H\) = change in enthalpy
- \(\Delta U\) = change in internal energy
- \(\Delta n_g\) = change in the number of moles of gas
- \(R\) = universal gas constant
- \(T\) = temperature in Kelvin

### Step-by-Step Solution:

1. **Identify the Reaction and Given Data:**
The reaction is:
\[ A(g) + 3B(g) \rightarrow 3C(g) + 3D(g) \]
Given data:
- \(\Delta U = 17 \text{ kcal} = 17,000 \text{ cal}\)
- \(R = 2 \text{ cal K}^{-1} \text{ mol}^{-1}\)
- Temperature = \(27^\circ C = 27 + 273 = 300 \text{ K}\)

2. **Calculate \(\Delta n_g\):**
\(\Delta n_g\) is the change in the number of moles of gas, calculated as:
\[ \Delta n_g = \text{(moles of products)} - \text{(moles of reactants)} \]
Here, the moles of products = 3 (from 3C and 3D) and the moles of reactants = 4 (from A and 3B).
\[ \Delta n_g = (3 + 3) - (1 + 3) = 6 - 4 = 2 \]

3. **Substitute Values into the Enthalpy Equation:**
Now we substitute \(\Delta U\), \(\Delta n_g\), \(R\), and \(T\) into the equation:
\[ \Delta H = \Delta U + \Delta n_g RT \]
\[ \Delta H = 17,000 \text{ cal} + (2)(2 \text{ cal K}^{-1} \text{ mol}^{-1})(300 \text{ K}) \]

4. **Calculate the Second Term:**
Calculate \(2 \times 2 \times 300\):
\[ 2 \times 2 = 4 \]
\[ 4 \times 300 = 1200 \text{ cal} \]

5. **Final Calculation of \(\Delta H\):**
Now, add this to \(\Delta U\):
\[ \Delta H = 17,000 \text{ cal} + 1200 \text{ cal} = 18,200 \text{ cal} \]

6. **Convert to Kilocalories:**
Convert \(\Delta H\) from calories to kilocalories:
\[ \Delta H = \frac{18,200 \text{ cal}}{1000} = 18.2 \text{ kcal} \]

### Final Answer:
\[ \Delta H = 18.2 \text{ kcal} \]