Question

Question #f93f3

Answer 1

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)`