Question

which equation represents the standard enthalpy of formation, \\( delta h_{f}^{circ} \\), of \\( hno_{2}(aq) \\) at 25 \\( { }^{circ} c \\)? \\( hno_{2}(aq) \to \frac{1}{2} h_{2}(g)+\frac{1}{2} n_{2}(g)+o_{2}(g) \\) \\( hno_{2}(aq) \to h(g)+n(g)+2 o(g) \\) \\( \frac{1}{2} h_{2}(g)+\frac{1}{2} n_{2}(g)+o_{2}(g) \to hno_{2}(aq) \\) \\( h(g)+n(g)+o_{2}(g) \to hno_{2}(aq) \\) \\( h_{2}(g)+n_{2}(g)+o_{3}(g) \to hno_{3}(aq) \\)

Explanation:

The standard enthalpy of formation ($\Delta H_f^{\circ}$) is the change in enthalpy when one mole of a compound is formed from its elements in their standard states.

• For the first option: It shows the decomposition of $HNO_2(aq)$ instead of its formation.
• For the second option: $H$, $N$, and $O$ are not in their standard states (standard state of $H$ is $H_2(g)$, of $N$ is $N_2(g)$, and of $O$ is $O_2(g)$).
• For the third option: $\frac{1}{2}H_2(g)$ (standard state of $H$), $\frac{1}{2}N_2(g)$ (standard state of $N$), and $O_2(g)$ (standard state of $O$) combine to form $HNO_2(aq)$. This is in line with the definition of standard enthalpy of formation.
• For the fourth option: $H$ and $N$ are not in their standard states.
• For the fifth option: It forms $HNO_3(aq)$ instead of $HNO_2(aq)$ and $O_3$ is not the standard state of oxygen (standard state is $O_2$).

Answer:

$\frac{1}{2}H_2(g)+\frac{1}{2}N_2(g)+O_2(g)\to HNO_2(aq)$