One of the reaction that takes place to producing steel from iron ore is the reduction of iron(III) oxide by carbon monoxde to give iron metal and ${C O}_{2}$
$F e O (s) + C O (g) ⇌ F e (s) + {C O}_{2} (g); K_{p} = 0.265$
What are the equilibrium partial pressures of $C O$ and ${C O}_{2}$ at $1050 K$ if the initial partial pressures are: $p_{C O} = 1.4 a t m$ and $p_{{C O}_{2}} = 0.80 a t m$ ?

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Solution

$F e O (s) + C O (g) ⇄ F e (s) + C O_{2} (g)$
at $t = 0$ $1.4$ $a t m$ $0.8$ $a t m$
$K_{p} = 0.265$
$Q_{p} = \frac{P_{{C O}_{2}}}{P_{C O}}$
$Q_{p} = \frac{0.8}{1.4} = 0.571$
$K_{p} < Q_{p},$ the reaction proceeds in reverse direction
$⟹ C O_{2}$ pressure decreases and that of $C O$ increases.
$P_{{C O}_{2}} = (0.8 - P) K_{p} = \frac{P_{{C O}_{2}}}{P_{C O}} 0.265 = \frac{(0.8 - P)}{(1.4 + P)}$
$⟹ P = 0.339$ $a t m$
At eqb.
$P_{{C O}_{2}} = 0.8 - 0.339 = 0.461 a t m P_{C O} = 1.40 + 0.339 = 1.739 a t m$

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One of the reactions that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO₂.

FeO (s) + CO (g) Fe (s) + CO₂ (g); K_p= 0.265 at 1050 K.

What are the equilibrium partial pressures of CO and CO₂ at 1050 K if the initial partial pressures are: p_CO = 1.4 atm and = 0.80 atm?

One of the reactions that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide to give iron metal and CO₂. FeO (s) + CO (g) Fe (s) + CO₂ (g); K_p= 0.265 at 1050 K. What are the equilibrium partial pressures of CO and CO₂ at 1050 K if the initial partial pressures are: p_CO = 1.4 atm and CO₂ = 0.80 atm?