What is the change in enthalpy for the final reaction?

Eq. 1 $2 Z n \left(s\right) + {O}_{2} \left(g\right) \rightarrow 2 Z n O \left(s\right)$ $\Delta H = - 696.0$kJ/mol Eq. 2 ${O}_{2} \left(g\right) + 2 {H}_{2} \left(g\right) \rightarrow 2 {H}_{2} O \left(l\right)$ $\Delta H = - 571.6$kJ/mol Eq. 3 $Z n \left(s\right) + 2 H C l \left(g\right) \rightarrow Z n C {l}_{2} \left(s\right) + {H}_{2} \left(g\right)$ $\Delta H = - 231.29$kJ/mol $Z n O \left(s\right) + 2 H C l \left(g\right) \rightarrow Z n C {l}_{2} \left(s\right) + {H}_{2} O \left(l\right)$

Oct 5, 2017

$\Delta {H}_{\text{target" = - "169.1 kJ mol}}^{- 1}$

Explanation:

Your goal here is to rearrange the thermochemical equations given to you in order to find a way to get to the target reaction

${\text{ZnO"_ ((s)) + 2"HCl"_ ((g)) -> "ZnCl"_ (2(s)) + "H"_ 2"O}}_{\left(l\right)}$

You know that you have

$2 \text{Zn"_ ((s)) + "O"_ (2(g)) -> 2"ZnO"_ ((s))" " DeltaH = - "696.0 kJ mol"^(-1)" } \textcolor{b l u e}{\left(1\right)}$

$\text{O"_ (2(g)) + 2"H"_ (2(g)) -> 2"H"_ 2"O"_ ((l))" "DeltaH = - "571.6 kJ mol"^(-1)" } \textcolor{b l u e}{\left(2\right)}$

$\text{Zn"_ ((s)) + 2"HCl"_ ((g)) -> "ZnCl"_ (2(s)) + "H"_ (2(g))" "DeltaH = - "231.29 kJ mol"^(-1)" } \textcolor{b l u e}{\left(3\right)}$

Now, the first thing to notice is that the target reaction has zinc oxide as a reactant, so reverse equation $\textcolor{b l u e}{\left(1\right)}$ to get

$2 \text{ZnO"_ ((s)) -> 2"Zn"_ ((s)) + "O"_ (2(g))" } \textcolor{b l u e}{\left({1}^{'}\right)}$

As you know, when you reverse a chemical reaction, you change the sign of its enthalpy change of reaction. This means that for equation $\textcolor{b l u e}{\left({1}^{'}\right)}$, you have

$\Delta {H}_{{1}^{'}} = + {\text{696.0 kJ mol}}^{- 1}$

Next, divide all the coefficients in reaction $\textcolor{b l u e}{\left({1}^{'}\right)}$ by $2$ to get

$\text{ZnO"_ ((s)) -> "Zn"_ ((s)) + 1/2"O"_ (2(g))" } \textcolor{b l u e}{\left({1}^{'} '\right)}$

After doing this, you need to divide the value of the enthalpy change of reaction by $2$ as well.

$\Delta {H}_{{1}^{'} '} = + {\text{348.0 kJ mol}}^{- 1}$

Next, divide all the coefficients in reaction $\textcolor{b l u e}{\left(2\right)}$ by $2$ to get

$\frac{1}{2} \text{O"_ (2(g)) + "H"_ (2(g)) -> "H"_ 2"O"_ ((l))" } \textcolor{b l u e}{\left({2}^{'}\right)}$

Remember to divide the enthalpy change of reaction by $2$ as well!

$\Delta {H}_{{2}^{'}} = - {\text{285.8 kJ mol}}^{- 1}$

You are now ready to add equations $\textcolor{b l u e}{\left({1}^{'} '\right)}$, $\textcolor{b l u e}{\left({2}^{'}\right)}$, and $\textcolor{b l u e}{\left(3\right)}$ to get your target equation.

$\textcolor{w h i t e}{a a a a a a a a a} \text{ZnO"_ ((s)) -> color(purple)(cancel(color(black)("Zn"_ ((s))))) + color(red)(cancel(color(black)(1/2"O"_ (2(g))))) " " " " " } +$
color(white)()color(red)(cancel(color(black)(1/2"O"_ (2(g))))) + color(green)(cancel(color(black)("H"_ (2(g))))) -> "H"_ 2"O"_ ((l))
$\textcolor{p u r p \le}{\cancel{\textcolor{b l a c k}{{\text{Zn"_ ((s))))) + 2"HCl"_ ((g)) -> "ZnCl"_ (2(s)) + color(green)(cancel(color(black)("H}}_{2 \left(g\right)}}}}$
$\frac{\textcolor{w h i t e}{a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a a}}{\textcolor{w h i t e}{a}}$

${\text{ZnO"_ ((s)) + 2"HCl"_ ((g)) -> "ZnCl"_ (2(s)) + "H"_ 2"O}}_{\left(l\right)}$

To find the enthalpy change of reaction, simply add the enthalpy changes of reaction that correspond to equations $\textcolor{b l u e}{\left({1}^{'} '\right)}$, $\textcolor{b l u e}{\left({2}^{'}\right)}$, and $\textcolor{b l u e}{\left(3\right)}$.

You will have

DeltaH_"target" = +"348.0 kJ mol"^(-1) + (-"285.8 kJ mol"^(-1)) + (-"231.29 kJ mol"^(-1))

DeltaH_"target" = color(darkgreen)(ul(color(black)(-"169.1 kJ mol"^(-1))))

The answer is rounded to one decimal place.