Question

At $1127$K and $1$ atm pressure, a gaseous mixture of CO and $C{O}_2$ in equilibrium with solid carbon has $90.55\%$ CO by mass,
$C_{\left(s\right)}+C{O}_{2\left(g\right)}\rightleftharpoons 2C{O}_{\left(g\right)}$
$K_c$ for this reaction at the above temperature is:

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

Answer 1

Answer: (B)

$0.153$

The given reaction is :-

$C\left(s\right)+{CO}_2\left(g\right)\rightleftharpoons 2CO\left(g\right)$

Let, total mass of mixture containing $CO<{CO}_2=100g$

Now, $9.55$% by mass of the mixture is $CO.$ so,

Mass of $CO=90.55$ g

Mass of ${CO}_2=(100-90.55)=9.45g$

Moles of $CO=\frac{90.55}{28}=3.234$

Moles of ${CO}_2=\frac{9.45}{44}=0.215$

Mole fraction of $CO$, $X_A=\frac{3.234}{3.234+0.215}=0.938$

Mole fraction of ${CO}_2$, $X_B=\frac{0.215}{0.215+3.234}=0.062$

so, $p{CO}_2=X_B\times P$

$\downarrow$ $\downarrow$ $\searrow$

partial pressure mole total pressure

of ${CO}_2$ fraction

of ${CO}_2$

$\Rightarrow p{CO}_2=0.062\times 1=0.062atm$

& $p{CO}_2=0.938\times 1=0.938atm$

Now, $K_P=\frac{{\left(pCO\right)}^2}{p{CO}_2}$

$=\frac{0.938\times 0.938}{0.062}=14.19$

we have, $K_P=K_C{\left(RT\right)}^{△n}$

$\Rightarrow K_C=\frac{K_P}{{\left(RT\right)}^{△n}}=\frac{14.19}{{\left(RT\right)}^{△}}=\frac{14.19}{0.0821\times 1127}=0.153$

{$△n=$ no. of moles of (gaseous product - gaseous reactant)}

$=2-1=1$