The given reactions and their respective enthalpy changes are:
- \( A \to 2B \), \( \Delta H_1 = +10 \, \text{J} \)
- \( 2B \to C \), \( \Delta H_2 = +25 \, \text{J} \)
To find \( \Delta_r H \) for the reaction \( C \to A \):
According to Hess's Law, we reverse the given reactions:
- Reversing \( A \to 2B \) gives \( 2B \to A \), with \( \Delta H = -10 \, \text{J} \).
- Reversing \( 2B \to C \) gives \( C \to 2B \), with \( \Delta H = -25 \, \text{J} \).
Adding the reactions:
\[ \Delta H_{\text{total}} = -25 + (-10) = -35 \, \text{J} \]
The enthalpy change for the reaction \( C \to A \) is \( \Delta_r H = -35 \, \text{J} \), so the correct answer is (C) -35 J.