1. Tardigrade
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  3. Chemistry
  4. At 1127 K and 1 atm pressure, a gaseous mixture of CO and CO2 in equilibrium with solid carbon has 90.55 % CO by mass, C(s) +CO2(g) leftharpoons 2CO(g) Kc for this reaction at the above temperature is

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Q. At $1127\,K$ and $1 \,atm$ pressure, a gaseous mixture of $CO$ and $CO_2$ in equilibrium with solid carbon has $90.55\%\, CO$ by mass,
$C_{(s)} +CO_{2(g)} \rightleftharpoons 2CO_{(g)}$
$K_c$ for this reaction at the above temperature is

Equilibrium

Solution:

Let the total mass of the mixture of $CO$ and $CO_2$ is $100\,g$, then $CO = 90.55\,g$ and $CO_2 = 100 - 90.55 = 9.45 \,g$
Moles of $CO=\frac{90.55}{28}=3.234$ ;
Moles of $CO_{2}=\frac{9.45}{44}=0.215$
Mole fraction of $CO =\frac{3.234}{3.234+0.215}$
$=0.938$
Mole fraction of $CO_{2}=\frac{0.215}{3.234+0.215}$
$=0.062$
$\therefore p_{CO}=$ mole fraction $\times$ total pressure
$=0.938\times1\,atm=0.938\,atm$
$p_{CO_{2}}=0.062\times1\,atm=0.062\,atm$
$K_p$ for the reaction, C_{(s)} +CO_{2(g)} \rightleftharpoons 2CO_{(g)}$
$K_{p}=\frac{p^{2}_{CO}}{p_{CO_2}}$$=\frac{\left(0.938\right)^{2}}{0.062}=14.19$
Now, $\Delta n_{g}=2-1=1$,
$K_{p}=K_{c}\left(RT\right)^{\Delta n}g$
or $K_{c}=\frac{K_{p}}{RT}=\frac{14.19}{0.0821\times1127}=0.153$