Question

the reduction of iron(iii) oxide to iron during steel - making can be summarized by this sequence of reactions:
2c(s)+o₂(g)⇌2co(g) k₁
fe₂o₃(s)+3co(g)⇌2fe(l)+3co₂(g) k₂
the net reaction is:
2fe₂o₃(s)+6c(s)+3o₂(g)⇌4fe(l)+6co₂(g) k
write 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 youll find in the aleks calculator.

Explanation:

Step1: Manipulate the first reaction

Multiply the first reaction \(2C(s)+O_{2}(g)
ightleftharpoons 2CO(g)\) by \(3\). The equilibrium constant \(K_1\) for the reaction \(aA + bB
ightleftharpoons cC + dD\) is \(K=\frac{[C]^c[D]^d}{[A]^a[B]^b}\). When a reaction is multiplied by a factor \(n\), the new equilibrium constant \(K'\) is \(K' = K^{n}\). So for \(3\times(2C(s)+O_{2}(g)
ightleftharpoons 2CO(g))\), the equilibrium constant is \(K_{1}^{3}\). The reaction becomes \(6C(s)+3O_{2}(g)
ightleftharpoons 6CO(g)\)

Step2: Manipulate the second reaction

Multiply the second reaction \(Fe_{2}O_{3}(s)+3CO(g)
ightleftharpoons 2Fe(l)+3CO_{2}(g)\) by \(2\). When a reaction is multiplied by a factor \(n = 2\), the new equilibrium constant \(K'\) is \(K'=K^{n}\). So for \(2\times(Fe_{2}O_{3}(s)+3CO(g)
ightleftharpoons 2Fe(l)+3CO_{2}(g))\), the equilibrium constant is \(K_{2}^{2}\). The reaction becomes \(2Fe_{2}O_{3}(s)+6CO(g)
ightleftharpoons 4Fe(l)+6CO_{2}(g)\)

Step3: Add the two manipulated reactions

Add \(6C(s)+3O_{2}(g)
ightleftharpoons 6CO(g)\) and \(2Fe_{2}O_{3}(s)+6CO(g)
ightleftharpoons 4Fe(l)+6CO_{2}(g)\). When two reactions \(R_1\) (with \(K_1'\)) and \(R_2\) (with \(K_2'\)) are added \(R = R_1+R_2\), the equilibrium constant \(K\) of the net - reaction \(R\) is \(K = K_1'\times K_2'\)

Since \(K_1'\) of the first - manipulated reaction is \(K_{1}^{3}\) and \(K_2'\) of the second - manipulated reaction is \(K_{2}^{2}\), the equilibrium constant \(K\) of the net reaction \(2Fe_{2}O_{3}(s)+6C(s)+3O_{2}(g)
ightleftharpoons 4Fe(l)+6CO_{2}(g)\) is \(K=K_{1}^{3}\times K_{2}^{2}\)

Answer:

\(K = K_{1}^{3}K_{2}^{2}\)