Question

Calculate the ΔH reaction?

CO(g) + 1/2O2(g) -----> CO2(g) ∆H = -282 kJ

C(s) + O2(g) -----> CO2(g) ∆H = -393 kJ

CO(g) ---> C(s) + 1/2O2(g) ∆H = ? kJ

I got an answer of -111 KJ but the answer is +111.. could somebody explain why? thanks

Answer 1

See below:

Explanation:

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

We can set up a Hess Cycle:

Applying Hess' Law you can see that, in terms of entalpy, the red route must equal the green route. This is because the arrows start and finish in the same place.

This gives us:

`sf(DeltaH-282=-393)`

`:.` `sf(DeltaH=-393+282=-111color(white)(x)kJ)`

However this gives us `sf(DeltaH)` for this reaction:

`sf(C+1/2O_2rarrCO)`

We want `sf(DeltaH)` for the reverse reaction i.e.

`sf(COrarrC+1/2O_2)`

This means we must reverse the sign of `sf(DeltaH)` :

`sf(DeltaH_(react)=-DeltaH=-(-111)=+111color(white)(x)kJ)`