The reduction of iron(III) oxide to iron during steel-making can be summarized by this sequence of reactions: 2C(s) +02 (g) > 2CO (g) K1Fe2O3 (s) + 3CO(lg) > 2Fe(l)+3CO2 (g) K2The net reaction is: 2Fe2O3(s) + 6C(s) + 3O2(g) ---> 4Fe(I) + 6CO2 (g) Kwrite an equation that gives the overall equilibrium constant K in terms of the equilibrium constants K1 and K2· If you need to include any physical constants be sure you use their standard symbols, which you'll find in the ALEKS Calculator.

K = [Fe]4×[CO2] / [Fe2O3]2 × [C]6 × [O2]3

Explanation: the over all reaction is

2Fe2O3(s) + 6C(s) + 3O2(g) ---> 4Fe(I) + 6CO2 (g)

So equilibrium constant equals, product concentration/ reactant concentration.

K = [Fe]4×[CO2] / [Fe2O3]2 × [C]6 × [O2]3

answer: the enthalpy of the overall chemical reaction is +87.88 kj or +88 kj.

according to hess's law, the overall enthalpy of a reaction is the difference between the sum of enthalpy values of all the products and the sum of enthalpy values of all the reactants.

δh(reaction) = ∑nδh(products) - ∑nδh(reactants)

δh(reaction) = [δh +δh] - [δh]

δh(reaction) = [1(-287.02) + 1(0) ] - [1(-374.9)] kj = +87.88 kj

here, δh (pcl3) = -287.02 kj, δh(cl2) = 0, δh(pcl5) = -374.9 kj

therefore, enthalpy of the overall reaction is +87.88 kj or +88 kj.