# What is the change in enthalpy for the final reaction?

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**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)#

**Eq. 1**

**Eq. 2**

**Eq. 3**

##### 1 Answer

#### Answer:

#### 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

#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

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

Next, **divide** all the coefficients in reaction

#"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

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

Next, **divide** all the coefficients in reaction

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

Remember to divide the enthalpy change of reaction by

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

You are now ready to **add** equations

#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

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*.