Question #f93f3

1 Answer
Sep 3, 2015

Answer:

The answer is (d) #DeltaH_3 = -DeltaH_1 - DeltaH_2#

Explanation:

The idea here is that you can play around with your reactions to try and find a relationship between the first one and the other two.

Notice that you have

#"A" + "B" -> "C" + "D"# #" "color(blue)((1))#

but that the two other reactions have

#"C" + "X" -> "A" + "Y"# #" "color(blue)((2))#

#"D" + "Y" -> "B" + "X"# #" "color(blue)((3))#

If you add reactions #color(blue)((2))# and #color(blue)((3))# you will get

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

#"C" + "D" -> "A" + "B"# #" "(DeltaH_2 + DeltaH_3)#

This is none other than the reverse reaction for reaction #color(blue)((1))#.

For a general chemical reaction that has the enthalpy change of reaction #DeltaH_"forward"#, the enthalpy change of reaction for the reverse reaction will be

#DeltaH_"reverse" = - DeltaH_"forward"#

In this case, you have

#"A" + "B" -> "C" + "D"#, #" "DeltaH_1#

and

#"C" + "D" -> "A" + "B"# #" "(DeltaH_2 + DeltaH_3)#

This means that you have

#DeltaH_1 = -(DeltaH_2 + DeltaH_3)#

Rearrange this to get

#DeltaH_1 = - DeltaH_2 - DeltaH_3 implies color(green)(DeltaH_3 = -DeltaH_1 - DeltaH_2)#