Question

From the diagram

\( \Delta H \) for the reaction C → A is:

• A. 35 j
• B. −15J
• C. −35J
• D. 15 J

Hint:

Reverse enthalpy changes are equal in magnitude but opposite in sign to forward changes.

Answer 1

The Correct Option is C

The problem involves determining the enthalpy change (\( \Delta H \)) for the reaction from C to A using the given reaction diagram.

The diagram shows:

• \( A \to 2B \) with \( \Delta H_1 = +10 \, \text{J} \)
• \( 2B \to C \) with \( \Delta H_2 = +25 \, \text{J} \)

Using Hess's Law, the total enthalpy change for the series of reactions is the sum of the enthalpy changes for each step.

Reversing the reactions to find \( C \to A \):

• For \( 2B \to A \), reverse of \( A \to 2B \), the \( \Delta H \) will be \(-10 \, \text{J} \).
• For \( C \to 2B \), reverse of \( 2B \to C \), the \( \Delta H \) will be \(-25 \, \text{J} \).

Add the reversed enthalpy changes together:

\(\Delta H (C \to A) = (-10) + (-25) = -35  \, \text{J}\)

Thus, the enthalpy change \( \Delta H \) for the reaction \( C \to A \) is -35 J.