At 900 ∘c,kc=0. 0108 for the reaction caco3(s)←→cao(s)+co2(g) a mixture of caco3, cao, and co2 is placed in a 10. 0-l vessel at 900 ∘c. For the following mixtures, will the amount of caco3 increase, decrease, or remain the same as the system approaches equilibrium?.

A chemical reaction is said to have attained equilibrium when the rate of forward reaction is equal to the rate of reverse reaction. We are told that the Kc of the reaction is 0. 0108.

In the first case:

CaCO3 - 15.0 g/100g/mol/10 L = 0.015 M

CaO - 15.0 g/56 g/mol/10 L = 0.027 M

CO2 - 4.25 g/44 g/mol / 10 L = 0.0096 M

Q = [0.027] [0.0096]/0.015

Q = 0.017

Since Q > K, the concentration of CaCO3 increases

In the second case;

CaCO3 - 2.5 g/100g/mol/10 L =0.0025 M

CaO - 25.0 g/56 g/mol/10 L = 0.045 M

CO2 - 5.66 g/44 g/mol / 10 L = 0.013 M

Q = [0.045 ] [ 0.013]/[0.0025 ]

Q = 0.23

Q > K the concentration of CaCO3 increases

In the third case;

CaCO3 - 30.5 g/100g/mol/10 L =0.031 M

CaO - 25.5 g/56 g/mol/10 L =0.046 M

CO2 - 6.48 g//44 g/mol / 10 L = 0.015 M

Q = [0.046] [ 0.015]/[0.031]

Q = 0.022

Q> K hence the concentration of CaCO3 increases

Missing parts: At 900oC, Kc = 0.0108 for the reaction: CaCO3(s) <===> CaO(s) + CO2(g) A mixture of CaCO3, CaO, and CO2 is placed in a 10.0 Liter vessel at 900oC. For the following mixtures, will the amount of CaCO3 increase, decrease, or remain the same as the system approaches equilibrium? CaCO3 CaO CO2 At Equilibrium, CaCO3 will ?? 15.0 g 15.0 g 4.25 g Answer 2.5 g 25.0 g 5.66 g Answer 30.5 g 25.5 g 6.48 g Answer

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