Question

Consider the following processes \[ \Delta H (\text{kJ/mol}) = \frac{1}{2} B + 150 \] For \( B + D \rightarrow E + 2C \), \( \Delta H \) will be

• A. \( 525 \, \text{kJ/mol} \)
• B. \( -175 \, \text{kJ/mol} \)
• C. \( -325 \, \text{kJ/mol} \)
• D. \( 325 \, \text{kJ/mol} \)

Hint:

Use stoichiometric relations and given enthalpy changes to determine the total enthalpy change for a reaction.

Answer 1

The Correct Option is B

Using the given relationship and the stoichiometry of the reaction, we can apply the thermodynamic calculations to find the change in enthalpy for \( B + D \rightarrow E + 2C \). Substituting into the equation, we get \( \Delta H = -175 \, \text{kJ/mol} \).

Step 2: Conclusion.
The enthalpy change for the reaction is \( -175 \, \text{kJ/mol} \), corresponding to option (b).