Question #98e63

1 Answer
Feb 27, 2017

Answer:

#DeltaH_"rxn" = + "74 kJ mol"^(-1)#

Explanation:

You know that

#"A " -> " B" " " " "DeltaH_1 = -"210 kJ mol"^(-1)#

#"A " -> " C" " " " "DeltaH_2 = -"50 kJ mol"^(-1)#

#"C " -> " D" " " " "DeltaH_3 = -"86 kJ mol"^(-1)#

Now, your goal here is to figure out the enthalpy change associated with the reaction

#"B " -> " D" " " " " DeltaH_"rxn" = ?#

The first thing to do here is to make sure that you gert #"B"# as a reactant by reversing the first reaction. You will have

#"B " -> " A"" " " "DeltaH_"1 rev"#

Keep in mind that when you reverse a reaction, you must change the sign of the enthalpy of reaction. In this case, you will have

#DeltaH_"1 rev" = - (-"210 kJ mol"^(-1))#

#DeltaH_"1 rev" = + "210 kJ mol"^(-1)#

Notice what happens when you add the first reaction reverse and the second reaction

#"B " -> " A"" " " "DeltaH_"1 rev" = + "210 kJ mol"^(-1)#

#"A " -> " C" " " " "DeltaH_2 = -"50 kJ mol"^(-1)#
#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#

#"B" + color(red)(cancel(color(black)("A"))) -> color(red)(cancel(color(black)("A"))) + "C"#

which is equivalent to

#"B " -> " C"#

The enthalpy change for this reaction will be the sum of the enthalpy changes of the two reactions we've added to get to this point

#DeltaH_"1 rev + 2" = DeltaH_"1 rev" + DeltaH_2#

#DeltaH_"1 rev + 2" = + "210 kJ mol"^(-1) + (-"50 kJ mol"^(-1))#

#DeltaH_"1 rev + 2" = + "160 kJ mol"^(-1)#

Finally, to get from #"B"# to #"D"#, add the above reaction and the third reaction

#"B " -> " C"" " " "DeltaH_"1 rev + 2" = + "160 kJ mol"^(-1)#

#"C " -> " D" " " " "DeltaH_2 = -"86 kJ mol"^(-1)#
#color(white)(aaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaaa)/color(white)(a)#

#"B" + color(red)(cancel(color(black)("C"))) -> color(red)(cancel(color(black)("C"))) + "D"#

which is equivalent to

#"B " -> " D"#

Once again, the enthalpy change for this reaction will be equal to the sum of the enthalpy changes associated with the two reactions we've added to get to this point

#DeltaH_"rxn" = DeltaH_"1 rev + 2" + DeltaH_3#

#DeltaH_"rxn" = + "160 kJ mol"^(-1) + (-"86 kJ mol"^(-1))#

#color(darkgreen)(ul(color(black)(DeltaH_"rxn" = + "74 kJ mol"^(-1))))#

This goes to show that the enthalpy change of a reaction is independent of the number of steps needed to make said reaction. #-># think Hess' Law here.

In this case, the enthalpy change of the reaction

#"B " -> " D"" " " "DeltaH_"rxn" = + "74 kJ mol"^(-1)#

is the same regardless if the reaction takes place in a single step or in the three steps we've used to get to it.