For the raction
`B_(2)H_(6)(g)+3O_(2)(g)rarrB_(2)O_(2)(s)+3H_(2)O(l)`
`DeltaE=-2143.2KJ`
Calculate `DeltaH` for the reaction at `25^(@)C`

Options

A. `-2148.2KJmol^(-1)`
B. `-2138.6KJmol^(-1)`
C. `-2133.2KJmol^(-1)`
D. `-2143.2KJmol^(-1)`

Explanation

To calculate the enthalpy change (ΔH) for the reaction:

\[ B_2H_6(g) + 3O_2(g) \rightarrow B_2O_2(s) + 3H_2O(l) \]

given that ΔE = -2143.2 kJ, we will follow these steps:

### Step 1: Convert the temperature to Kelvin
The temperature is given as 25°C. To convert this to Kelvin, we use the formula:

\[ T(K) = T(°C) + 273.15 \]

So,

\[ T = 25 + 273.15 = 298.15 \, K \approx 298 \, K \]

### Step 2: Calculate ΔNG (Change in the number of moles of gas)
Next, we need to determine ΔNG, which is the change in the number of moles of gaseous reactants and products.

- **Reactants**:
- 1 mole of \( B_2H_6(g) \)
- 3 moles of \( O_2(g) \)

Total moles of gaseous reactants = 1 + 3 = 4 moles

- **Products**:
- 0 moles of \( B_2O_2(s) \) (solid)
- 3 moles of \( H_2O(l) \) (liquid)

Total moles of gaseous products = 0 moles

Thus,

\[ \Delta N_G = \text{Moles of gaseous products} - \text{Moles of gaseous reactants} = 0 - 4 = -4 \]

### Step 3: Use the first law of thermodynamics to calculate ΔH
According to the first law of thermodynamics, the relationship between ΔH and ΔE is given by:

\[ \Delta H = \Delta E + \Delta N_G \cdot R \cdot T \]

Where:
- \( R = 8.314 \, \text{J/mol·K} = 0.008314 \, \text{kJ/mol·K} \) (since we need ΔH in kJ)
- \( T = 298 \, K \)

Now substituting the values into the equation:

\[ \Delta H = -2143.2 \, \text{kJ} + (-4) \cdot (0.008314 \, \text{kJ/mol·K}) \cdot (298 \, K) \]

Calculating the second term:

\[ -4 \cdot 0.008314 \cdot 298 = -9.93 \, \text{kJ} \]

Now substituting this back into the ΔH equation:

\[ \Delta H = -2143.2 \, \text{kJ} - 9.93 \, \text{kJ} = -2153.13 \, \text{kJ} \]

### Final Result
Thus, the value of ΔH for the reaction at 25°C is approximately:

\[ \Delta H \approx -2153.13 \, \text{kJ} \]