Question

For the reaction at $27^\circ C$
$\left(MgCO\right)_{3}\left(s\right)\rightleftharpoonsMgO\left(s\right)+\left(CO\right)_{2};K_{P}=100atm$
If $42g$ of $MgCO_{3}$ is taken in $20L$ vessel then calculate percentage dissociation of $MgCO_{3}$ .
[Take : $R=0.08Latmmol^{- 1}K^{- 1}$ ]

• A. $50\%$
• B. $100\%$
• C. $80\%$
• D. $25\%$

Answer 1

Answer: (B)

$n_{CO_{2}}=\frac{100 \times 20}{0 . 08 \times 300}mol=83.33mol$ at equilibrium if it is established
But, $n_{MgCO_{3}}=\frac{42}{84}mol=0.5mol$
$\therefore n_{CO_{2}}$ (at most after complete dissociation) $=0.5mol$
$\therefore $ Even after $100\%$ dissociation, equilibrium will not be established.
$\therefore $ There will be $100\%$ dissociation.