# Question e4ee3

Jun 12, 2015

The change in enthalpy for your reaction will be $\Delta H = \text{-78.5 kJ}$.

#### Explanation:

I'll start by saying that the equations you wrote make no sense. Not only did you mistype compounds, but you wrote the same incorrect reaction twice.

$H C l + N a N {O}_{2} \to H N {O}_{2} + N a C l$, " "DeltaH_"rxn"=?

The four reactions that you're supposed to use look like this:

$\textcolor{b l u e}{\left(1\right)}$ $2 N a C l + {H}_{2} O \to 2 H C l + N {a}_{2} O$, $\text{ "DeltaH_1 = "+507 kJ}$

$\textcolor{b l u e}{\left(2\right)}$ $N O + N {O}_{2} + N {a}_{2} O \to 2 N a N {O}_{2}$, $\text{ "DeltaH_2 = "-427 kJ}$

$\textcolor{b l u e}{\left(3\right)}$ $N O + N {O}_{2} \to {N}_{2} O + {O}_{2}$, $\text{ "DeltaH_3 = "-43 kJ}$

$\textcolor{b l u e}{\left(4\right)}$ $2 H N {O}_{2} \to {N}_{2} O + {O}_{2} + {H}_{2} O$, $\text{ "DeltaH_4 = "+34 kJ}$

Take a look at the four reactions you have. Notice that the target equation has sodium chloride on the products's side and hydrochloric acid on the reactants' side.

Reaction $\textcolor{b l u e}{\left(1\right)}$ has them the other way around. Not only that, but this reaction has 2 moles of each, instead of 1 mole of each. Don't worry about that right now, just flip reaction $\textcolor{b l u e}{\left(1\right)}$. This will become

$\textcolor{b l u e}{\left({1}^{'}\right)}$ $2 H C l + N {a}_{2} O \to 2 N a C l + {H}_{2} O$,

$\Delta {H}_{1}^{'} = \left(- \Delta {H}_{1}\right) = \text{-507.31 kJ" = "-507 kJ}$

Now focus on sodium nitrite. Once again, the target reaction has it on the reactants' side, but reaction $\textcolor{b l u e}{\left(2\right)}$ has it on the products's side.

Flip reaction $\textcolor{b l u e}{\left(2\right)}$ to get

$\textcolor{b l u e}{\left({2}^{'}\right)}$ $2 N a N {O}_{2} \to N O + N {O}_{2} + N {a}_{2} O$

$\Delta {H}_{2}^{'} = \left(- \Delta {H}_{2}\right) = \text{+ 427.14 kJ" = "+427 kJ}$

Finally, focus on nitrous acid. You need it on the products' side, but have it on the reactants' side in reaction $\textcolor{b l u e}{\left(4\right)}$. Once again, flip reaction $\textcolor{b l u e}{\left(4\right)}$ to get

$\textcolor{b l u e}{\left({4}^{'}\right)}$ ${N}_{2} O + {O}_{2} + {H}_{2} O \to 2 H N {O}_{2}$

$\Delta {H}_{4}^{'} = \left(- \Delta {H}_{4}\right) = \text{-34.25 kJ" = "-34 kJ}$

You have everything that you need for the target reaction, so leave reaction $\textcolor{b l u e}{\left(3\right)}$ unchanged.

Add all these reactions and eliminate compounds that are on both sides of the general reaction to get

$2 H C l + \cancel{N {a}_{2} O} + 2 N a N {O}_{2} + \cancel{N O} + \cancel{N {O}_{2}} + \cancel{N {a}_{2} O} + \cancel{{O}_{2}} + \cancel{{H}_{2} O} = 2 N a C l + \cancel{{H}_{2} O} + \cancel{N O} + \cancel{N {O}_{2}} + \cancel{N {a}_{2} O} + \cancel{{N}_{2} O} + \cancel{{O}_{2}} + 2 H N {O}_{2}$

This is reduced to the form

$2 H C l + 2 N a N {O}_{2} \to 2 N a C l + 2 H N {O}_{2}$

$\Delta {H}_{\text{rxn}} = \Delta {H}_{1}^{'} + \Delta {H}_{2}^{'} + \Delta {H}_{3} + \Delta {H}_{4}^{'}$

DeltaH_"rxn" = "-507" + 427 + ("-43") + ("-34 kJ")

$\Delta {H}_{\text{rxn" = "-157 kJ}}$

All you have to do now is divide this reaction by 2 to get the final form

$H C l + N a N {O}_{2} \to N a C l + H N {O}_{2}$

DeltaH_"rxn" = ("-157 kJ"/2) = color(green)("-78.5 kJ")#