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Equilibrium — Chemistry STD 11 Science — Question

Gujarat BoardEnglish MediumSTD 11 ScienceChemistryEquilibrium5 Marks
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 $CO_2$.
$\text{FeO (S) + CO (g)}\rightleftharpoons\text{Fe (S) + CO}_2\text{ (g)};\text{ K}_{\text{p}}=0.265\text{atm at}1050\text{K}$
What are the equilibrium partial pressures of $CO$ and $CO_2$ at 1050 K if the initial partial pressures are: $p_{CO}= 1.4atm$ and = $0.80atm$?
✓

Answer

For the given reaction,
$\text{FeO}_{\text{(g)}}$ $+$ $\text{CO}_{\text{(g)}}$ $\leftrightarrow$ $\text{Fe}_{\text{(s)}}$ $+$ $\text{CO}_{2\text{(g)}}$
Initialy,   $1.4\text{atm}$       $0.80\text{atm}$
$\text{Q}_{\text{p}}=\frac{\text{p}_{\text{co}_2}}{\text{p}_{\text{co}}}$
$=\frac{0.80}{1.4}$
$=0.571$ It is given that $\text{K}_{\text{p}}=0.265.$ Since $\text{Q}_{\text{p}}>\text{K}_{\text{p}},$ the reaction will proceed in the backward direction. Therefore, we can say that the pressure of CO will increase while the pressure of $CO_2$_ will decrease. Now, let the increase in pressure of CO = decrease in pressure of $CO_2$ be p. Then, we can write, $\text{K}_{\text{p}}=\frac{\text{p}_{\text{co}_2}}{\text{p}_{\text{co}}}$
$\Rightarrow0.265=\frac{0.80-\text{p}}{1.4+\text{p}}$
$\Rightarrow0.371+0.265\text{p}=0.80-\text{p}$
$\Rightarrow1.265\text{p}=0.429$
$\Rightarrow\text{p}=0.339\text{atm}$ Therefore, equilibrium partial of $\text{CO}_2,\text{ p}_{\text{co}_2}=0.80-0.339=0.461\text{atm.}$ And, equilibrium partial pressure of $\text{CO},\text{ p}_{\text{co}_2}=1.4-0.339=1.739\text{atm.}$

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  1. Define standard enthalpy of formation. Explain why the enthalpy changes for the reaction given below are not enthalpies of formation of $CaCO_3$ and HBr:
  1. $\text{CaO(s)}+\text{CO}_2(\text{g})\overrightarrow{\ \ \ \ \ \ \ \ }\ \text{CaCO}_3(\text{s});$ $\Delta_\text{r}\text{H}^\circ=-178.3\text{kJ mol}^{-1}$
  2. $\text{H}_2(\text{g})+\text{Br}_2(\text{g})\overrightarrow{\ \ \ \ \ \ \ }\ 2\text{HBr(g)};$ $\Delta_\text{r}\text{H}^\circ=-72.8\text{kJ mol}^{-1}$
  1. Use the bond enthalpies listed below to determine the enthalpy of reaction:
$\ \ \ \ \ \ \ \ \text{H}\\ \ \ \ \ \ \ \ \ \ |\\\text{H}-\text{C}-\text{H(g)}+2\text{O}=\text{O(g)}\overrightarrow{\ \ \ \ \ }\ \text{O}=\text{C}=\text{O(g)}\\ \ \ \ \ \ \ \ \ \ |\\ \ \ \ \ \ \ \ \ \text{H}\\+2\text{H}-\text{O}-\text{H(g)}$

Bond enthalpy $(\Delta\text{H}^\circ)/\text{kJ mol}^{-1}$ of C=O = 741; C-H = 414; H-O = 464; O=O = 498.
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