College

Thank you for visiting The dissociation of calcium carbonate has an equilibrium constant of tex K p 1 16 tex at 1073 K text CaCO 3 s leftrightarrow text. This page is designed to guide you through key points and clear explanations related to the topic at hand. We aim to make your learning experience smooth, insightful, and informative. Dive in and discover the answers you're looking for!

The dissociation of calcium carbonate has an equilibrium constant of [tex]K_p = 1.16[/tex] at 1073 K.

\[ \text{CaCO}_3(s) \leftrightarrow \text{CaO}(s) + \text{CO}_2(g) \]

If you place 35.9 g of [tex]\text{CaCO}_3[/tex] in a 9.56-L container at 1073 K, what is the pressure of [tex]\text{CO}_2[/tex] in the container?

Answer :

Answer:

[tex]Pressure of CO_2 in the container=1.6 atm[/tex]

Explanation:

First balance the chemical equation:

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

two components are solid so these two will not exert any kind of pressure in the container so at equilibrium only CO2 will apply pressure on the container

Therefore only partial pressure of CO2 will be taken for the calculation of equilibrium pressure constant i.e. Kp

[tex]K_p=[CO_2][/tex]

[tex][CO_2]=p[/tex]

[tex]K_p=p[/tex]

[tex]p=K_p = 1.16atm[/tex]

[tex]Pressure of CO_2 in the container=1.6 atm[/tex]

Thank you for reading the article The dissociation of calcium carbonate has an equilibrium constant of tex K p 1 16 tex at 1073 K text CaCO 3 s leftrightarrow text. We hope the information provided is useful and helps you understand this topic better. Feel free to explore more helpful content on our website!

Rewritten by : Jeany