
For $CaC{O_3}(s)\overset {} \leftrightarrows CaO(s) + C{O_2}(g),\,{K_c}\,$is equal to:
A. $\left[ {C{O_2}} \right]$
B. $\left[ {CaO} \right]$
C.$\left[ {CaC{O_3}} \right]$
D. None of these.
Answer
556.2k+ views
Hint:The equilibrium constant is a ratio of the concentration of the products to the concentration of the reactants. If the K value is less than one the reaction will move to the left and if the K value is greater than one the reaction will move to the right.
Complete step by step answer:
Equilibrium constant ${K_c}$, is the ratio of equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients.
For a reaction:$aA + bB\xrightarrow{{}}cC + dD$
Equilibrium constant: ${K_c}$=$\dfrac{{{{[C]}^c}{{[D]}^d}}}{{{{[A]}^a}{{[B]}^b}}}$
Calculation of ${K_c}$:
For the gas phase reaction- The concentrations of gases are corporated in its value. The concentrations of gases are expressed in the formula of equilibrium constant.
For pure solid and liquids: Concentrations of pure solids and liquids are fixed by their density and molar mass (both constants) and do not vary with the amount.
Thus , the concentration of solids and liquids are incorporated in the value of ${K_c}$, they are not part of variable expressions.
Solve the given question;
$CaC{O_3}(s)\xrightarrow{{}}CaO(s) + C{O_2}(g)$
Write all the values of stoichiometric coefficients ;
a=1, b=0, c=1, d=1;
we know that in the equation part concentration of only gaseous part will be present.
So the equation of ${K_c}$will be equal to ${K_c} = [C{O_2}]$
Because the concentration of only gaseous parts will be present in the equation , concentration of solids and liquids would be absent , as we have discussed earlier. Stoichiometric coefficient is 1 . So power is raised to power 1.
So, our final answer would be $\left[ {C{O_2}} \right]$ option (A).
Note:Characteristics of ${K_c}$:
Equilibrium can be approached from both direction
${K_c}$ is independent of initial concentration of reactants and products and also of temperature.
Complete step by step answer:
Equilibrium constant ${K_c}$, is the ratio of equilibrium concentrations of products over the equilibrium concentrations of reactants each raised to the power of their stoichiometric coefficients.
For a reaction:$aA + bB\xrightarrow{{}}cC + dD$
Equilibrium constant: ${K_c}$=$\dfrac{{{{[C]}^c}{{[D]}^d}}}{{{{[A]}^a}{{[B]}^b}}}$
Calculation of ${K_c}$:
For the gas phase reaction- The concentrations of gases are corporated in its value. The concentrations of gases are expressed in the formula of equilibrium constant.
For pure solid and liquids: Concentrations of pure solids and liquids are fixed by their density and molar mass (both constants) and do not vary with the amount.
Thus , the concentration of solids and liquids are incorporated in the value of ${K_c}$, they are not part of variable expressions.
Solve the given question;
$CaC{O_3}(s)\xrightarrow{{}}CaO(s) + C{O_2}(g)$
Write all the values of stoichiometric coefficients ;
a=1, b=0, c=1, d=1;
we know that in the equation part concentration of only gaseous part will be present.
So the equation of ${K_c}$will be equal to ${K_c} = [C{O_2}]$
Because the concentration of only gaseous parts will be present in the equation , concentration of solids and liquids would be absent , as we have discussed earlier. Stoichiometric coefficient is 1 . So power is raised to power 1.
So, our final answer would be $\left[ {C{O_2}} \right]$ option (A).
Note:Characteristics of ${K_c}$:
Equilibrium can be approached from both direction
${K_c}$ is independent of initial concentration of reactants and products and also of temperature.
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