Ncert Exercises & Solutions

7.20 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 CO2.

FeO (s) + CO (g) Fe (s) + CO2 (g); Kp = 0.265 atm at 1050K

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

For the given reaction,

{FeO}_{(g)} + {CO}_{(g)} \leftrightarrow {Fe}_{(g)} + {CO}_{2 (g)}

Initialy, 1.4 atm 0.80 atm

Q_{p} = \frac{{pco}_{2}}{p_{co}}

= \frac{0.80}{1.4}

= 0.571

It is given that

K_{p}

=0.265.

Since,

Q_{p} > K_{p}

, the reaction will proceed in the backward direction.

Therefore, we can say that the pressure of CO will increase while the pressure of

C O_{2}

will decrease.

Now, let the increase in pressure of CO = decrease in pressure of

C O_{2}

be p.

Then, we can write,

K_{p} = \frac{p_{{co}_{2}}}{p_{co}}

\Rightarrow 0.265 = \frac{0.80 - p}{1.4 + p}

\Rightarrow 0.371 + 0.265 p = 0.80 - p

\Rightarrow 1.265 p = 0.429

\Rightarrow p = 0.339 atm

Therefore, equilibrium partial of

C O_{2}, P_{c o_{2}} = 0.80 - 0.339 = 0.461 a t m .

And, equilibrium partial pressure of

C O, P_{c o} = 1.4 + 0.339 = 1.739 a t m .

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