calculate the standard...?

Question

standard enthalpy change of : 2C(s) + O2 -> 2CO(g)

given : C(s) + O2(g)->CO2(g) -394kJ
CO(g) + 1/2 O2 (g) -> CO2(g) -284kJ

Answer 1

Lancenigo di Villorba (TV), Italy

As you know, enthalpy belongs to a particular class of thermodynamics properties, e.g. the "state's functions".
The latters are CLOSELY dependents only on physico-chemicals variables. Talking about any process, it is possible individuate one INITIAL amount and one FINAL "thermodynamics state" of the considerate system. Now, it is possible assume that any changes of "state's functions" depends only on parameters value in initial and final states.
These properties of ENTHALPY are on the basis of "Hess's law". Executing an "Enthalpy's balancement", you can determine the "Enthalpy's change" related to a chemical reaction if you know the "enthalpy's changes" associated to other reactions very closely similar to the former reaction.

CALCULUS
Since you know the "Enthalpy's changes" related to the following reactions :

b) C(s) + O2(g) ---> CO2(g)
c) CO(g) + 1/2 O2(g) ---> CO2(g)

so you can combine them in order to re-obtain the chemical equation as a)

a) 2 C(s) + O2(g) ---> 2 CO(g)

In effect, you retrieve a = 2 * (b - c)
so the calculus is

DeltaH(a) = 2 * (DeltaH(b) - DeltaH(c)) = 2 * (-394 - (-284)) =
= - 220 kJ

I hope this helps you.

Answer 2

c+O2 ---> Co2 -394kj
Co2----->Co+1/2O2 284 kj
c+O2---->Co+1/2 O2 -394+284 kj
C+1/2O2---->co -110kj
2c+o2----> 2co -220 kj