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Salt of Weak Acid and Strong Base

One of the re...

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One of the reaction that takes place in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide to give iron metal and $C O_{2}$ . $F e O (s) + C O (g) ⇌ F e (s) + C O_{2} (g)$ ; $K_{P} = 0.265 a t m$ at $1050 K$ . What are the equilibrium partial pressures of CO and $C O_{2}$ respectively at 1050K if the initial partial pressures are: $p_{c o} = 1.4 a t m$ and $p_{c o_{2}} = 0.80 a t m$ ?

A

[P_{C O}] = 1.739 a t m

and

P_{C O_{2}}] = 0.461 a t m

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B

[P_{C O}] = 17.39 a t m

and

P_{C O_{2}} = 0.461 a t m

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C

[P_{C O}] = 1.79 a t m

and

P_{C O_{2}}] = 0.46 a t m

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D

[P_{C O}] = 2.739 a t m

and

P_{C O_{2}}] = 1.461 a t m

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Solution

The correct option is A $[P_{C O}] = 1.739 a t m$ and $P_{C O_{2}}] = 0.461 a t m$
The equilibrium reaction is $F e O (s) + C O (g) ⇌ F e (s) + C O_{2} (g)$ .
The initial partial pressures of $C O$ and $C O_{2}$ are 1.4 atm and 0.8 atm respectively.
The expression for the reaction quotient is $Q = \frac{0.8}{1.4} = \frac{4}{7} = 0.55$ .
The value of the reaction quotient is greater than the value of the equilibrium constant. $(Q > K)$ , the reaction will shift in the backward direction.
The equilibrium partial pressures of $C O$ and $C O_{2}$ are $1.4 + x$ atm and $0.8 - x$ atm respectively.
The expression for the equilibrium constant is $K_{P} = \frac{P_{C O_{2}}}{P_{C O}}$
Substitute values in the above expression.
$0.265 = \frac{0.8 - x}{1.4 + x}$
Thus, $x = 0.339$ .
The equilibrium partial pressure of $C O$ is $[P_{C O}] = 1.739 a t m$ and the equilibrium partial pressure of $C O_{2}$ is $[P_{C O_{2}}] = 0.461 a t m$ .
Thus option A is the correct answer.

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Similar questions

Q. One of the reaction that takes place in producing steel from iron ore is the reduction of iron(II) oxide by carbon monoxide 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

atm at 1050K. What are the equilibrium partial pressures of CO and

C O_{2}

respectively at 1050 K if the initial partial pressures are:

P_{C O} = 1.4

p_{C O_{2}} = 0.80 a t m ?

Q. 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

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if the initial partial pressures are:

p_{C O} = 1.4 a t m

p_{{C O}_{2}} = 0.80 a t m

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?

Q. One of the reaction that takes place in producing steel from iron ore is the reduction of iron (II) oxide by carbon monoxide 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

atm at 1050 K.
What are the equilibrium partial pressures of

C O

C O_{2}

respectively at 1050 K if the initial partial pressures are:

P_{C O} = 1.4

P_{C O_{2}} = 0.80

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