A) Give the name of H2. b) Determine the value of ΔH3 based on the provided energy cycle and Hess's Law.

Hess's Law states that the total enthalpy change for a reaction is independent of the pathway taken.
From the diagram, the direct path from $\text{H}_2(\text{g}) + \text{O}_2(\text{g})$ to $\text{H}_2\text{O}_2(\text{l})$ is $\Delta H_1$.
The indirect path is from $\text{H}_2(\text{g}) + \text{O}_2(\text{g})$ to $\text{H}_2\text{O}(\text{l}) + \frac{1}{2}\text{O}_2(\text{g})$ (which is $\Delta H_2$), and then from $\text{H}_2\text{O}(\text{l}) + \frac{1}{2}\text{O}_2(\text{g})$ to $\text{H}_2\text{O}_2(\text{l})$ (which is $\Delta H_3$).
Therefore, according to Hess's Law:

$$\Delta H_1 = \Delta H_2 + \Delta H_3$$

Given $\Delta H_1 = -188 \text{ kJ mol}^{-1}$ and $\Delta H_2 = -286 \text{ kJ mol}^{-1}$.

$$-188 \text{ kJ mol}^{-1} = -286 \text{ kJ mol}^{-1} + \Delta H_3$$ $$\Delta H_3 = -188 \text{ kJ mol}^{-1} - (-286 \text{ kJ mol}^{-1})$$ $$\Delta H_3 = -188 \text{ kJ mol}^{-1} + 286 \text{ kJ mol}^{-1}$$ $$\Delta H_3 = \boxed{98 \text{ kJ mol}^{-1}}$$