Question

At $1127K$ and $1atm$ pressure , a gaseous mixture of $CO$and $C{O_2}$in equilibrium with solid carbon $90.55\%$ $CO$ by mass;
$C(s) + C{O_2}(g) \rightleftarrows 2CO(g)$
Calculate ${K_c}$ for the reaction at the above temperature.

Answer 1

We know that ${K_c}$ is the equilibrium constant of a chemical reaction in terms of the molar concentrations. The active mass of the solids in a chemical equation is taken $1$.The molar concentration of the solid is taken $1$. The partial pressure is also taken as $1$ of solids.

Complete step by step answer:
As in the question we are given the temperature $T = 1127K$ .The total pressure given is ${P_T} = 1atm$. Let us consider the total mass of the gaseous mixture to be $100g$. Then the mass of $CO$ will be $90.55g$ and the mass of $C{O_2}$will be $\Rightarrow 100 - 90.55 = 9.45g$. The molecular mass of $CO$ is $28$ and the molecular mass of $C{O_2}$ will be $44$. The moles of $CO$ will be $\Rightarrow {n_{CO}} = \dfrac{{90.55}}{{28}} = 3.234mol$. The moles of $\Rightarrow {n_{C{O_2}}} = \dfrac{{9.45}}{{44}} = 0.215mol$. The partial pressure of $CO$ will be
$\Rightarrow {P_{CO}} = \dfrac{{{n_{CO}}}}{{{n_{CO}} + {n_{C{O_2}}}}} \times {P_T} \\\ \Rightarrow {P_{CO}} = \dfrac{{3.234}}{{3.234 + 0.215}} \times 1 \\\ \Rightarrow {P_{CO}} = 0.938atm \\\$
The partial pressure of $C{O_2}$ will be
$\Rightarrow {P_{C{O_2}}} = \dfrac{{{n_{C{O_2}}}}}{{{n_{CO}} + {n_{C{O_2}}}}} \times {P_T} \\\ \Rightarrow {P_{C{O_2}}} = \dfrac{{0.215}}{{3.234 + 0.215}} \times 1 \\\ \Rightarrow {P_{C{O_2}}} = 0.062atm \\\$
Now the ${K_p}$ of the equation is given by ,
${K_P} = \dfrac{{{{[{K_{CO}}]}^2}}}{{[C{O_2}]}} = \dfrac{{{{(0.938)}^2}}}{{0.062}} = 14.19$
The relation between the ${K_P}$ and the ${K_C}$ of a chemical equation is given by the formula
$\Rightarrow {K_P} = {K_C}{(RT)^{\Delta n}}$. Here $R$ is the ideal gas constant and $\Delta n$ is the change in mole of the chemical equation. Now putting the values we can find the ${K_c}$ for the reaction.
$\Rightarrow {K_P} = {K_C}{(RT)^{\Delta n}} \\\ \Rightarrow {K_C} = \dfrac{{{K_P}}}{{{{(RT)}^{\Delta n}}}} \\\ \Rightarrow {K_C} = \dfrac{{14.19}}{{(0.082 \times 1127)}} \\\ \Rightarrow {K_C} = 0.154 \\\$
So from the above explanation and calculation it is clear to us that the correct answer of the given question is $\Rightarrow {K_C} = 0.154$.

Additional information: The ideal gas equation is given by the formula $\Rightarrow PV = nRT$. $P$ is the pressure of the gas , $V$ is the volume of the gas , $R$ is the ideal gas constant, $n$ is the number of moles.

Note: Always remember that the relation between ${K_P}$ and the ${K_C}$ of a chemical equation is given by the formula $\Rightarrow {K_P} = {K_C}{(RT)^{\Delta n}}$.${K_c}$ is the equilibrium constant of a chemical reaction in terms of the molar concentrations and ${K_P}$ is the equilibrium constant of the reaction in terms of the partial pressure of the products and the reactants. Always avoid calculation errors while solving the numerical.