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)#

1 Answer
Oct 5, 2017

Answer:

#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.