Question

calcium carbonate decomposes to form calcium oxide and carbon dioxide, like this:

caco₃(s)→cao(s)+co₂(g)

at a certain temperature, a chemist finds that a 9.3 l reaction vessel containing a mixture of calcium carbonate, calcium oxide, and carbon dioxide at equilibrium has the following composition:

calculate the value of the equilibrium constant k_c for this reaction. round your answer to 2 significant digits.

k_c = □

Explanation:

Step1: Calculate the molarity of \(CO_2\)

The molar mass of \(CO_2\) is \(M = 44.01\space g/mol\).
The number of moles of \(CO_2\), \(n=\frac{m}{M}=\frac{14.1\space g}{44.01\space g/mol}\approx0.320\space mol\).
The molarity \(C=\frac{n}{V}\), where \(V = 9.3\space L\). So \(C=\frac{0.320\space mol}{9.3\space L}\approx0.0344\space mol/L\).

Step2: Write the equilibrium constant expression

For the reaction \(CaCO_3(s)
ightleftharpoons CaO(s)+CO_2(g)\), the equilibrium constant \(K_c = [CO_2]\) (because the concentrations of pure solids \(CaCO_3\) and \(CaO\) are omitted from the expression).

Answer:

\(K_c=0.034\) (rounded to 2 significant digits)