MCQ
At $1000\, K$ and $2$ $atm$ pressure , a gaseous mixture of $CO$ and $CO_2$ in equilibrium with solid carbon has $84\%$ $CO_{(g)}$ by mass. Calculate $K_p$ for the reaction ; $C_{(s)} + CO_{2(g)}  \rightleftharpoons  2CO_{(g)}$ at this temperature .....$atm$
  • $14.89$
  • B
    $8.76$
  • C
    $22.34$
  • D
    $3.34$

Answer

Correct option: A.
$14.89$
a
Source :-

$NCERT$ Class $XI^{th}$ Page $220 .$ Question $7.23$ Let total mass of gaseous mixture be $100\, \mathrm{gm}$. mass of $\mathrm{CO}_{(\mathrm{g})}=84 \, \mathrm{gm}$ and mass of

$\mathrm{CO}_{2(\mathrm{s})}=(100-84) \mathrm{gm}=16\, \mathrm{gm}$

Moles of $\mathrm{CO}_{(\mathrm{g})}=\frac{84}{28}=3$

mole of $\mathrm{CO}_{2}=\frac{16}{44}=0.36$

Partial pressure of $\mathrm{CO}=\frac{3}{3.36} \times 2=1.785\, \mathrm{atm}$

Partial pressure of $\mathrm{CO}_{2}=\frac{0.36}{3.36} \times 2=0.214\, \mathrm{atm}$

$\mathrm{K}_{\mathrm{p}}=\frac{\left(\mathrm{P}_{\mathrm{CO}}\right)^{2}}{\left(\mathrm{P}_{\mathrm{CO}_{2}}\right)}=\frac{(1.785)^{2}}{(0.214)}=14.89\, \mathrm{atm}$

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