Question

Consider the following processes : $\Delta H (kJ / mol)$ ½ A → B + 150 3B → 2C + D -125 E + A → 2D +350 For B + D → E + 2C, ΔH will be-

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

Answer 1

The Correct Option is C

To determine the enthalpy change \(\Delta H\) for the reaction \(B + D \rightarrow E + 2C\), we will use Hess's law. According to Hess's law, if a chemical reaction can be expressed as the sum of multiple reactions, then the enthalpy change for the overall process is the sum of the enthalpy changes for the individual steps.

Given reactions with their enthalpy changes: 

• \(\frac{1}{2} A \rightarrow B \quad \Delta H = +150 \text{ kJ/mol}\)
• \(3B \rightarrow 2C + D \quad \Delta H = -125 \text{ kJ/mol}\)
• \(E + A \rightarrow 2D \quad \Delta H = +350 \text{ kJ/mol}\)

We need the target reaction:

• \(B + D \rightarrow E + 2C\)

We'll manipulate the given reactions to obtain the target reaction. Let's rearrange and reverse the equations as necessary:

1. Start by reversing the second reaction to match our required format:

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

1. Use the third reaction as it includes \(E\):

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

1. Now, reverse the third reaction:

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

1. For consistency, add all reactions:
   • \(\frac{1}{2} A \rightarrow B \quad \Delta H = +150 \text{ kJ/mol}\)
   • \(2C + D \rightarrow 3B \quad \Delta H = +125 \text{ kJ/mol}\)
   • \(2D \rightarrow E + A \quad \Delta H = -350 \text{ kJ/mol}\)

Summing these reactions, we obtain:

• \(B + D \rightarrow E + 2C\)

Thus, the enthalpy change for the target reaction is:

\(\Delta H = 150 + 125 - 350 = -175 \text{ kJ/mol}\)

Hence, the correct answer is \(-175 \text{ kJ/mol}\), making the option '

-175 kJ/mol

' correct.