Question

given the standard enthalpy changes for the following two reactions: (1) 2ni(s) + o₂(g)→2nio(s) δh°=-479.4 kj (2) 2hg(l) + o₂(g)→2hgo(s) δh°=-181.6 kj what is the standard enthalpy change for the reaction: (3) nio(s) + hg(l)→ni(s) + hgo(s) δh°=?

Explanation:

Step1: Reverse reaction (1)

The reverse of $2Ni(s)+O_2(g)
ightarrow2NiO(s)$ with $\Delta H^o=- 479.4\ kJ$ is $2NiO(s)
ightarrow2Ni(s)+O_2(g)$ and $\Delta H^o = 479.4\ kJ$.

Step2: Keep reaction (2) as it is

$2Hg(l)+O_2(g)
ightarrow2HgO(s)$ with $\Delta H^o=-181.6\ kJ$.

Step3: Add the two - modified reactions

$(2NiO(s)
ightarrow2Ni(s)+O_2(g))+(2Hg(l)+O_2(g)
ightarrow2HgO(s))$ gives $2NiO(s)+2Hg(l)
ightarrow2Ni(s)+2HgO(s)$.
The $\Delta H^o$ for the overall reaction is the sum of the $\Delta H^o$ values of the two modified reactions. So, $\Delta H^o=479.4+( - 181.6)$.

Step4: Calculate the $\Delta H^o$

$\Delta H^o=479.4 - 181.6=297.8\ kJ$.
For the reaction $NiO(s)+Hg(l)
ightarrow Ni(s)+HgO(s)$, we divide the above - calculated $\Delta H^o$ by 2. So, $\Delta H^o=\frac{297.8}{2}=148.9\ kJ$.

Answer:

$148.9\ kJ$