Consider the following processes :-
`{:(,DeltaH(kJ//mol)),((1)/(2)A rarr B,+150),(3B rarr2C+D,-125),(E+A rarr 2D,+350),("For "B+D rarr E+2C",",Delta H" will be"):}`

Options

A. 325 kJ / mol
B. 525 kJ / mol
C. `-175` kJ / mol
D. `-325` kJ / mol

Explanation

To find the value of ΔH for the reaction \( B + D \rightarrow E + 2C \), we will manipulate the given reactions and their associated enthalpy changes (ΔH).

### Given Reactions:
1. \( \frac{1}{2}A \rightarrow B \) (ΔH = +150 kJ/mol)
2. \( 3B \rightarrow 2C + D \) (ΔH = -125 kJ/mol)
3. \( E + A \rightarrow 2D \) (ΔH = +350 kJ/mol)

### Step-by-step Solution:

**Step 1: Adjust the first reaction to isolate B.**
- We need B on the left side for our target reaction. We can multiply the first reaction by 2 to get:
\[ A \rightarrow 2B \quad (\Delta H = 2 \times 150 = +300 \text{ kJ/mol}) \]

**Step 2: Adjust the second reaction.**
- The second reaction already has B on the left side, but we need to reverse it to get B on the left side:
\[ 2C + D \rightarrow 3B \quad (\Delta H = +125 \text{ kJ/mol}) \]

**Step 3: Combine the adjusted first and second reactions.**
- Now we combine the adjusted first reaction and the reversed second reaction:
\[ A + 2C + D \rightarrow 2B + 3B \]
- This simplifies to:
\[ A + 2C + D \rightarrow 5B \quad (\Delta H = +300 + 125 = +425 \text{ kJ/mol}) \]

**Step 4: Adjust the third reaction.**
- We need to isolate A and D, so we can reverse the third reaction:
\[ 2D \rightarrow E + A \quad (\Delta H = -350 \text{ kJ/mol}) \]

**Step 5: Combine the reactions to form the target reaction.**
- Now we combine the results from Steps 3 and 4:
\[ A + 2C + D + 2D \rightarrow 5B + E + A \]
- This simplifies to:
\[ B + D \rightarrow E + 2C \quad (\Delta H = 425 - 350 = +75 \text{ kJ/mol}) \]

### Final Calculation of ΔH:
- Thus, the ΔH for the reaction \( B + D \rightarrow E + 2C \) is:
\[ \Delta H = +75 \text{ kJ/mol} \]

### Conclusion:
The value of ΔH for the reaction \( B + D \rightarrow E + 2C \) is +75 kJ/mol.

---