Question #295d0

`sf(A.)`

To do this type of question you need to construct a Hess Cycle.

Hess' Law states that the overall enthalpy change of a chemical reaction is independent of the route taken.

All the substances are made from the same elements so the elements can go on the bottom.

Then construct the cycle:

Note that I have divided the enthalpy of formation of `sf(CO)` by 2 since the value given refers to formation of 2 moles.

Also note that the enthalpy formation of the element carbon is zero.

Hess' Law tells us that the enthalpy change of the `sf(color(blue)(blue))` route is equal to the enthalpy change of the `sf(color(red)(red))` route.

This is because the arrows start and finish in the same place.

Applying the law `sf(rArr)`

`sf(DeltaH+0+106.8=-221.0/2)`

`:.``sf(DeltaH=-110.5-106.8=-217.3color(white)(x)kJ)`

This refers to `sf(Pb_((s))+1/2O_(2(g))rarrPbO_((s)))`

`sf(B).`

The `sf(A_r)` of Pb is 207.2

So 1 mole of Pb weighs 207.2 g.

From the equation we can say:

`sf(207.2color(white)(x)grarr-217.3color(white)(x)kJ)`

`:.``sf(1color(white)(x)grarr-217.3/207.2color(white)(x)kJ)`

`:.``sf(250.0color(white)(x)grarr-217.3/(207.2)xx250.0=-262.2color(white)(x)kJ)`

The minus sign tells us that `sf(262.2color(white)(x)kJ)` is released.