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 $CO_{2}$.
$FeO(s) + CO(g) \longleftrightarrow Fe(s) + CO_{2}(g);$ $K_{p} = 0.265$ at $1050 \, K$
What are the equilibrium partial pressures of $CO$ and $CO_{2}$ at $1050 \, K$ if the initial partial pressures are: $p_{CO} = 1.4 \, atm$ and $p_{CO_{2}} = 0.80 \, atm$?
$FeO(s) + CO(g) \longleftrightarrow Fe(s) + CO_{2}(g);$ $K_{p} = 0.265$ at $1050 \, K$
What are the equilibrium partial pressures of $CO$ and $CO_{2}$ at $1050 \, K$ if the initial partial pressures are: $p_{CO} = 1.4 \, atm$ and $p_{CO_{2}} = 0.80 \, atm$?
(N/A) For the given reaction,
$FeO(s) + CO(g) \longleftrightarrow Fe(s) + CO_{2}(g)$
Initially: $p_{CO} = 1.4 \, atm$,$p_{CO_{2}} = 0.80 \, atm$
$Q_{p} = \frac{p_{CO_{2}}}{p_{CO}} = \frac{0.80}{1.4} \approx 0.571$
Since $Q_{p} > K_{p}$ $(0.571 > 0.265)$,the reaction proceeds in the backward direction.
Let $x$ be the change in pressure at equilibrium.
$K_{p} = \frac{p_{CO_{2}}}{p_{CO}} = \frac{0.80 - x}{1.4 + x} = 0.265$
$0.80 - x = 0.265(1.4 + x)$
$0.80 - x = 0.371 + 0.265x$
$1.265x = 0.429$
$x = 0.339 \, atm$
Equilibrium partial pressures:
$p_{CO_{2}} = 0.80 - 0.339 = 0.461 \, atm$
$p_{CO} = 1.4 + 0.339 = 1.739 \, atm$
$FeO(s) + CO(g) \longleftrightarrow Fe(s) + CO_{2}(g)$
Initially: $p_{CO} = 1.4 \, atm$,$p_{CO_{2}} = 0.80 \, atm$
$Q_{p} = \frac{p_{CO_{2}}}{p_{CO}} = \frac{0.80}{1.4} \approx 0.571$
Since $Q_{p} > K_{p}$ $(0.571 > 0.265)$,the reaction proceeds in the backward direction.
Let $x$ be the change in pressure at equilibrium.
$K_{p} = \frac{p_{CO_{2}}}{p_{CO}} = \frac{0.80 - x}{1.4 + x} = 0.265$
$0.80 - x = 0.265(1.4 + x)$
$0.80 - x = 0.371 + 0.265x$
$1.265x = 0.429$
$x = 0.339 \, atm$
Equilibrium partial pressures:
$p_{CO_{2}} = 0.80 - 0.339 = 0.461 \, atm$
$p_{CO} = 1.4 + 0.339 = 1.739 \, atm$
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The values of pressure equilibrium constant recorded at different temperatures for the following equilibrium reaction have been given below: $A(g) \rightleftharpoons B(g) + C(g)$.
$1/T \text{ (K}^{-1})$ $\log_{10} K_p$ $0.05$ $3.5$ $0.06$ $2.5$ $0.07$ $1.5$
The magnitude of $\frac{\Delta H^\circ}{R}$ calculated from the above data is . . . . . . . (Note: The slope $m = -\frac{\Delta H^\circ}{2.303 R}$)
| $1/T \text{ (K}^{-1})$ | $\log_{10} K_p$ |
| $0.05$ | $3.5$ |
| $0.06$ | $2.5$ |
| $0.07$ | $1.5$ |
The magnitude of $\frac{\Delta H^\circ}{R}$ calculated from the above data is . . . . . . . (Note: The slope $m = -\frac{\Delta H^\circ}{2.303 R}$)
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$(1)$ $A_{2(g)} + 3B_{2(g)} \rightleftharpoons 2AB_{3(g)}$ $(i)$ $(RT)^{-2}$ $(2)$ $A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$ $(ii)$ $(RT)^0$ $(3)$ $A_{(s)} + 1.5B_{2(g)} \rightleftharpoons AB_{3(g)}$ $(iii)$ $(RT)^{1/2}$ $(4)$ $AB_{2(g)} \rightleftharpoons AB_{(g)} + 0.5B_{2(g)}$ $(iv)$ $(RT)^{-1/2}$
| $(1)$ $A_{2(g)} + 3B_{2(g)} \rightleftharpoons 2AB_{3(g)}$ | $(i)$ $(RT)^{-2}$ |
| $(2)$ $A_{2(g)} + B_{2(g)} \rightleftharpoons 2AB_{(g)}$ | $(ii)$ $(RT)^0$ |
| $(3)$ $A_{(s)} + 1.5B_{2(g)} \rightleftharpoons AB_{3(g)}$ | $(iii)$ $(RT)^{1/2}$ |
| $(4)$ $AB_{2(g)} \rightleftharpoons AB_{(g)} + 0.5B_{2(g)}$ | $(iv)$ $(RT)^{-1/2}$ |
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