Question

What is the change in enthalpy for the final reaction?

Eq. 1 `2Zn(s)+O_2(g)rarr2ZnO(s)` `DeltaH=-696.0`kJ/mol
Eq. 2 `O_2(g)+2H_2(g)rarr2H_2O(l)` `DeltaH=-571.6`kJ/mol
Eq. 3 `Zn(s)+2HCl(g)rarrZnCl_2(s)+H_2(g)` `DeltaH=-231.29`kJ/mol

`ZnO(s)+2HCl(g)rarrZnCl_2(s)+H_2O(l)`

Answer 1

`DeltaH_"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

`"ZnO"_ ((s)) + 2"HCl"_ ((g)) -> "ZnCl"_ (2(s)) + "H"_ 2"O"_ ((l))`

You know that you have

`2"Zn"_ ((s)) + "O"_ (2(g)) -> 2"ZnO"_ ((s))" " DeltaH = - "696.0 kJ mol"^(-1)" "color(blue)((1))`

`"O"_ (2(g)) + 2"H"_ (2(g)) -> 2"H"_ 2"O"_ ((l))" "DeltaH = - "571.6 kJ mol"^(-1)" "color(blue)((2))`

`"Zn"_ ((s)) + 2"HCl"_ ((g)) -> "ZnCl"_ (2(s)) + "H"_ (2(g))" "DeltaH = - "231.29 kJ mol"^(-1)" "color(blue)((3))`

Now, the first thing to notice is that the target reaction has zinc oxide as a reactant, so reverse equation `color(blue)((1))` to get

`2"ZnO"_ ((s)) -> 2"Zn"_ ((s)) + "O"_ (2(g))" "color(blue)((1^'))`

As you know, when you reverse a chemical reaction, you change the sign of its enthalpy change of reaction. This means that for equation `color(blue)((1^'))`, you have

`DeltaH_ (1^') = + "696.0 kJ mol"^(-1)`

Next, divide all the coefficients in reaction `color(blue)((1^'))` by `2` to get

`"ZnO"_ ((s)) -> "Zn"_ ((s)) + 1/2"O"_ (2(g))" "color(blue)((1^''))`

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

`DeltaH_ (1^'') = + "348.0 kJ mol"^(-1)`

Next, divide all the coefficients in reaction `color(blue)((2))` by `2` to get

`1/2"O"_ (2(g)) + "H"_ (2(g)) -> "H"_ 2"O"_ ((l))" "color(blue)((2^'))`

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

`DeltaH_ (2^') = - "285.8 kJ mol"^(-1)`

You are now ready to add equations `color(blue)((1^''))`, `color(blue)((2^'))`, and `color(blue)((3))` to get your target equation.

`color(white)(aaaaaaaaa)"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))`
`color(purple)(cancel(color(black)("Zn"_ ((s))))) + 2"HCl"_ ((g)) -> "ZnCl"_ (2(s)) + color(green)(cancel(color(black)("H"_ (2(g)))))`
`color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)`

`"ZnO"_ ((s)) + 2"HCl"_ ((g)) -> "ZnCl"_ (2(s)) + "H"_ 2"O"_ ((l))`

To find the enthalpy change of reaction, simply add the enthalpy changes of reaction that correspond to equations `color(blue)((1^''))`, `color(blue)((2^'))`, and `color(blue)((3))`.

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.