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 solve the problem, we need to determine the ΔH for the reaction \( B + D \rightarrow E + 2C \) using the provided reactions and their respective ΔH values. Here’s a step-by-step breakdown of the solution:

### Step 1: Write down the given reactions and their ΔH values.

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 2: Adjust the first reaction to have whole numbers.

The first reaction is given as \( \frac{1}{2} A \rightarrow B \). To convert it to whole numbers, we can multiply the entire equation by 2:

\[ A \rightarrow 2B \quad (\Delta H = 2 \times 150 = +300 \text{ kJ/mol}) \]

### Step 3: Reverse the third reaction.

The third reaction is \( E + A \rightarrow 2D \). To use this reaction in our calculations, we need to reverse it:

\[ 2D \rightarrow E + A \quad (\Delta H = -350 \text{ kJ/mol}) \]

### Step 4: Write the second reaction as it is.

The second reaction remains unchanged:

\[ 3B \rightarrow 2C + D \quad (\Delta H = -125 \text{ kJ/mol}) \]

### Step 5: Combine the adjusted reactions.

Now we have:

1. \( A \rightarrow 2B \) (ΔH = +300 kJ/mol)
2. \( 3B \rightarrow 2C + D \) (ΔH = -125 kJ/mol)
3. \( 2D \rightarrow E + A \) (ΔH = -350 kJ/mol)

Now, let's add these reactions together:

- From the first reaction, we get \( A \rightarrow 2B \).
- From the second reaction, we get \( 3B \rightarrow 2C + D \).
- From the reversed third reaction, we get \( 2D \rightarrow E + A \).

### Step 6: Cancel out species and write the overall reaction.

When we add these reactions, we can cancel out \( A \) and \( D \):

\[ B + D \rightarrow E + 2C \]

### Step 7: Calculate the overall ΔH.

Now we sum the ΔH values:

\[ \Delta H = (+300) + (-125) + (-350) = 300 - 125 - 350 = -175 \text{ kJ/mol} \]

### Final Answer:
The ΔH for the reaction \( B + D \rightarrow E + 2C \) is **-175 kJ/mol**.

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