For a chemical reaction,
`2A_(2)(g)+5B_(2)(g)to2A_(2)B_(2)(g)`, at `27^(@)C` the difference between `DeltaHandDeltaE` is X.
Then the ratio X/R -

Options

A. Zero
B. Unity
C. `-5xx10^(0)`
D. `-1.5xx10^(3)`

Answer

To solve the problem, we need to find the ratio \( \frac{X}{R} \) where \( X \) is the difference between the change in enthalpy (\( \Delta H \)) and the change in internal energy (\( \Delta E \)) for the given reaction.

### Step-by-Step Solution:

1. **Write the Reaction:**
The given reaction is:
\[ 2A_{2}(g) + 5B_{2}(g) \rightarrow 2A_{2}B_{2}(g) \]

2. **Convert Temperature to Kelvin:**
The temperature given is \( 27^\circ C \). To convert this to Kelvin:
\[ T = 27 + 273 = 300 \, K \]

3. **Use the Relation Between \( \Delta H \) and \( \Delta E \):**
The relationship between change in enthalpy and change in internal energy is given by:
\[ \Delta H = \Delta E + \Delta N_g RT \]
where \( \Delta N_g \) is the change in the number of moles of gas.

4. **Calculate \( \Delta N_g \):**
- Moles of gaseous products = 2 (from \( 2A_{2}B_{2} \))
- Moles of gaseous reactants = \( 2 + 5 = 7 \)
- Therefore,
\[ \Delta N_g = \text{Moles of products} - \text{Moles of reactants} = 2 - 7 = -5 \]

5. **Substitute \( \Delta N_g \) into the Equation:**
Substitute \( \Delta N_g \) into the equation:
\[ \Delta H - \Delta E = \Delta N_g RT \]
This implies:
\[ X = \Delta H - \Delta E = (-5)RT \]

6. **Calculate \( X \):**
Using \( R = 8.314 \, J/(mol \cdot K) \) and \( T = 300 \, K \):
\[ X = -5 \times 8.314 \times 300 \]
Calculating this gives:
\[ X = -12471 \, J \]

7. **Find the Ratio \( \frac{X}{R} \):**
Now, we need to find the ratio \( \frac{X}{R} \):
\[ \frac{X}{R} = \frac{-12471}{8.314} \]
Performing the division:
\[ \frac{X}{R} \approx -1500 \]

8. **Express in Scientific Notation:**
The final answer can be expressed in scientific notation:
\[ \frac{X}{R} \approx -1.5 \times 10^{3} \]

### Final Answer:
The ratio \( \frac{X}{R} \) is approximately:
\[ -1.5 \times 10^{3} \]