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Salt of Weak Acid and Strong Base

One of the re...

Question

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

C O

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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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{C O}_{(2 (g)} + C_{(g)} ⇌ 2 {C O}_{(g)}

3.0

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