When 20 g of CaCO_3 are subjected to decompo | vedclass

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Question

(A) The decomposition reaction is: $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$Initial moles of $CaCO_3$ = $\frac{20 \ g}{100 \ g/mol} = 0.2

Vedclass · Accepted Answer

$0.41$

Vedclass · Answer

(A) The decomposition reaction is: $CaCO_{3(s)} ightleftharpoons CaO_{(s)} + CO_{2(g)}$Initial moles of $CaCO_3$ = $\frac{20 \ g}{100 \ g/mol} = 0.2 \ mol$.At equilibrium,$50 \%$ of $CaCO_3$ remains unreacted,so $50 \%$ has decomposed:$n_{CaCO_3, 	ext{reacted}} = 0.2 	imes 0.5 = 0.1 \ mol$.According to the stoichiometry,$1 \ mol$ of $CaCO_3$ produces $1 \ mol$ of $CO_2$. Therefore,$n_{CO_2} = 0.1 \ mol$.Using the ideal gas law $PV = nRT$ for $CO_2$:$P_{CO_2} = \frac{n_{CO_2}RT}{V} = \frac{0.1 	imes 0.0821 	imes (227 + 273)}{10} = \frac{0.1 	imes 0.0821 	imes 500}{10} = 0.41 \ atm$.Since $K_P = P_{CO_2}$ for this reaction,$K_P = 0.41 \ atm$.

When $20 \ g$ of $CaCO_3$ are subjected to decomposition at $227 \ ^oC$ in a closed container of $10 \ L$ capacity,$50 \%$ of $CaCO_3$ remained unreacted at equilibrium. Calculate $K_P$ for $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$ in $atm$.

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