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NCERT Class XI Chemistry Equilibrium Solutions

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Question : 20 of 73
Marks: +1, -0
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\mathrm{CO}_2.
FeO(s)+CO(g)\mathrm{FeO}_{(s)} + \mathrm{CO}_{(g)} ⇌ Fe(s)+CO2(g)\mathrm{Fe}_{(s)} + \mathrm{CO}_{2(g)} ; KpK_p = 0.265 atm at 1050 K
What are the equilibrium partial pressures of CO and CO2\mathrm{CO}_2 at 1050 K if the partial pressures are : pCOp_{\mathrm{CO}} = 1.4 atm and pCO2p_{\mathrm{CO}_2} = 0.80 atm?
Solution:  
FeO(s)+CO(g)\mathrm{FeO}_{(s)} + \mathrm{CO}_{(g)} ⇌ Fe(s)+CO2(g)\mathrm{Fe}_{(s)} + \mathrm{CO}_{2(g)}
Initial pressures1.4 atm0.80 atmPressure at equilibrium1.4−p0.80+p\begin{array}{lcc} \text{Initial pressures} & 1.4\ \text{atm} & 0.80\ \text{atm} \\ \text{Pressure at equilibrium} & 1.4 - p & 0.80 + p \end{array}
∴ equilibrium, pCO2p_{\mathrm{CO}_2} = (0.80 + p) atm, pCOp_{\mathrm{CO}} = (1.4 – p) atm
KpK_p = pCO2pCO\frac{p_{\mathrm{CO}_2}}{p_{\mathrm{CO}}} ⇒ 0.265 = 0.80+p1.4−p\frac{0.80+p}{1.4-p}
or, 0.265 (1.4 – p) = 0.80 + p
or, 0.371 – 0.265p = 0.80 + p
or, 1.265p = – 0.429 or, p = –0.339 atm
∴ (pCO)eq(p_{\mathrm{CO}})_{\text{eq}} = 1.4 + 0.339 atm = 1.739 atm
(pCO2)eq(p_{\mathrm{CO}_2})_{\text{eq}} = 0.80 – 0.339 atm = 0.461 atm
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