Class 11Chemistry6-1.Equilibrium (Chemical Equilibrium)Mix Examples- 6-1.Equilibrium (Chemical Equilibrium)
MediumAt $1000 \ K$ in a $0.654 \ L$ vessel,$CaCO_{3(s)}$ is taken. For the reaction $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$,the equilibrium constant $K_p$ is $3.9 \times 10^{-2} \ bar$. Find the weight of $CaO$ produced at equilibrium. $(Ca=40, C=12, O=16)$
(N/A) The equilibrium constant for the reaction is $K_p = P_{CO_2} = 3.9 \times 10^{-2} \ bar$.
Using the ideal gas law $PV = nRT$,we find the moles of $CO_2$ at equilibrium:
$n_{CO_2} = \frac{P_{CO_2} V}{RT} = \frac{(3.9 \times 10^{-2} \ bar) \times (0.654 \ L)}{(0.08314 \ L \ bar \ K^{-1} \ mol^{-1}) \times (1000 \ K)} \approx 3.067 \times 10^{-4} \ mol$.
According to the stoichiometry of the reaction $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$,$1 \ mol$ of $CO_2$ is produced for every $1 \ mol$ of $CaO$ formed.
Therefore,$n_{CaO} = n_{CO_2} = 3.067 \times 10^{-4} \ mol$.
The molar mass of $CaO = 40 + 16 = 56 \ g \ mol^{-1}$.
Weight of $CaO = n_{CaO} \times \text{Molar mass} = 3.067 \times 10^{-4} \ mol \times 56 \ g \ mol^{-1} \approx 0.01718 \ g$.
Using the ideal gas law $PV = nRT$,we find the moles of $CO_2$ at equilibrium:
$n_{CO_2} = \frac{P_{CO_2} V}{RT} = \frac{(3.9 \times 10^{-2} \ bar) \times (0.654 \ L)}{(0.08314 \ L \ bar \ K^{-1} \ mol^{-1}) \times (1000 \ K)} \approx 3.067 \times 10^{-4} \ mol$.
According to the stoichiometry of the reaction $CaCO_{3(s)} \rightleftharpoons CaO_{(s)} + CO_{2(g)}$,$1 \ mol$ of $CO_2$ is produced for every $1 \ mol$ of $CaO$ formed.
Therefore,$n_{CaO} = n_{CO_2} = 3.067 \times 10^{-4} \ mol$.
The molar mass of $CaO = 40 + 16 = 56 \ g \ mol^{-1}$.
Weight of $CaO = n_{CaO} \times \text{Molar mass} = 3.067 \times 10^{-4} \ mol \times 56 \ g \ mol^{-1} \approx 0.01718 \ g$.
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