Question

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, ${C_{(s)} +CO2_{(g)} <=> 2CO_{(g)}}$ $K_c$ for this reaction at the above temperature is

• A. $1.53$
• B. $0.153$
• C. $0.53$
• D. $0.76$

Answer 1

Answer: (B)

$0.153$

Answer 2

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, ${C_{(s)} +CO_{2(g)} <=> 2CO_{(g)}}$ $K_{p}=\frac{p^{2}_{CO}}{p_{CO_2}}$ $=\frac{\left(0.938\right)^{2}}{0.062}=14.19$ Now, $\Delta n_{g}=2-1=1$, $K_{p}=K_{c}\left(RT\right)^{\Delta n}g$ or $K_{c}=\frac{K_{p}}{RT}=\frac{14.19}{0.0821\times1127}=0.153$