Answered

Get expert insights and reliable answers to your questions on IDNLearn.com. Get accurate and detailed answers to your questions from our knowledgeable and dedicated community members.

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 9.0L reaction vessel containing a mixture of calcium carbonate, calcium oxide, and carbon dioxide at equilibrium has the following composition:
Compound Amount
CaCO3 25.3 g
CaO 14.9 g
CO2 33.7 g
Calculate the value of the equilibrium constant Kc for this reaction. Round your answer to 2 significant digits.

Sagot :

Answer: The value of the equilibrium constant Kc for this reaction is 0.088

Explanation:

[tex]Molarity=\frac{x}{M\times V_s}[/tex]

where,

x = given mass

M = molar mass

[tex]V_s[/tex] = volume of solution in L

Equilibrium concentration of [tex]CaCO_3[/tex] = [tex]\frac{25.3}{100\times 9.0}=0.028M[/tex]

Equilibrium concentration of [tex]CaO[/tex] = [tex]\frac{14.9}{56\times 9.0}=0.029M[/tex]

Equilibrium concentration of [tex]CO_2[/tex] = [tex]\frac{33.7}{44\times 9.0}=0.085M[/tex]

The given balanced equilibrium reaction is,

[tex]CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)[/tex]

The expression for equilibrium constant for this reaction will be,

[tex]K_c=\frac{[CaO]\times [CO_2]}{[CaCO_3]}[/tex]  

Now put all the given values in this expression, we get :

[tex]K_c=\frac{0.029\times 0.085}{0.028}=0.088[/tex]

We greatly appreciate every question and answer you provide. Keep engaging and finding the best solutions. This community is the perfect place to learn and grow together. Find precise solutions at IDNLearn.com. Thank you for trusting us with your queries, and we hope to see you again.