Question

1. which of the following equations is an endothermic reaction?

• $ce{h_{2(g)} + 1/2 o_{2(g)} -> h_{2}o_{(l)}}$ $delta h = -285.8 \text{kj}$
• $ce{hg_{(l)} + 1/2 o_{2(g)} -> hgo_{(s)}}$ $delta h = -90.7 \text{kj}$
• $ce{4co_{2(g)} + 6h_{2}o_{(l)} -> 2c_{2}h_{6(g)} + 7o_{2(g)}}$ $delta h = +3120 \text{kj}$
• $ce{s_{(s)} + o_{2(g)} -> so_{2(g)}}$ $delta h = -295 \text{kj}$ (unanswered)

Explanation:

To determine an endothermic reaction, we use the sign of the enthalpy change ($\Delta H$). An endothermic reaction has a positive $\Delta H$ (heat is absorbed), while an exothermic reaction has a negative $\Delta H$ (heat is released).

• For $\ce{H_{2}(g) + \frac{1}{2}O_{2}(g) -> H_{2}O(l)}$ with $\Delta H = -285.8\ \text{kJ}$, the negative $\Delta H$ indicates exothermic.
• For $\ce{Hg(l) + \frac{1}{2}O_{2}(g) -> HgO(s)}$ with $\Delta H = -90.7\ \text{kJ}$, the negative $\Delta H$ indicates exothermic.
• For $\ce{4CO_{2}(g) + 6H_{2}O(l) -> 2C_{2}H_{6}(g) + 7O_{2}(g)}$ with $\Delta H = +3120\ \text{kJ}$, the positive $\Delta H$ indicates endothermic.
• For $\ce{S(s) + O_{2}(g) -> SO_{2}(g)}$ with $\Delta H = -295\ \text{kJ}$, the negative $\Delta H$ indicates exothermic.

Answer:

The equation for the endothermic reaction is $\boldsymbol{4\ce{CO_{2}(g)} + 6\ce{H_{2}O(l)}
ightarrow 2\ce{C_{2}H_{6}(g)} + 7\ce{O_{2}(g)}}$ with $\Delta H = +3120\ \text{kJ}$.