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? Part A 15.0 gCaCO3, 15.0 gCaO, and 4.25gCO2 15.0 , 15.0 , and 4.25 increases decreases remains the same SubmitRequest Answer Part B 2.50gCaCO3, 25.0 gCaO, and 5.66 gCO2 2.50, 25.0 , and 5.66 increases decreases remains the same SubmitRequest Answer Part C

To know how the system will proceed to reach equilibrium, we need to calculate the reaction quotient (Q) and compare it with the equilibrium constant (Kc).

Part A 15.0 g CaCO₃, 15.0 g CaO, and 4.25g CO₂

CaCO₃ and CaO are solids and they do not take part in Q nor Kc, so we will just calculate the concentration of CO₂.

[tex][CO_{2}]=\frac{moles}{volume} =\frac{mass}{molarmass.volume} =\frac{4.25g}{44.0g/mol.10.0L} =9.66\times 10^{-3} M[/tex]

Then,

Q = [CO₂] = 9.66 x 10⁻³

Q < Kc (0.0108), so the reaction will proceed to the right to achieve equilibrium, thus decreasing the amount of CaCO₃.

Part B 2.50 g CaCO₃, 25.0 g CaO, and 5.66g CO₂

[tex][CO_{2}]=\frac{moles}{volume} =\frac{mass}{molarmass.volume} =\frac{5.66g}{44.0g/mol.10.0L} =0.0129 M[/tex]

Then,

Q = [CO₂] = 0.0129

Q > Kc, so the reaction will proceed to the left to achieve equilibrium, thus increasing the amount of CaCO₃.

ardni313 ardni313 · Answer 2 · 2020-01-11T22:48:16+01:00

a. CaCO3 decreases

b. CaCO3 increases

Further explanation

The equilibrium constant is based on the concentration (Kc) in a reaction

pA + qB -----> mC + nD

[tex] \large {\boxed {\bold {Kc ~ = ~ \frac {[C] ^ m [D] ^ n} {[A] ^ p [B] ^ q}}}} [/tex]

To find out the reaction reaches equilibrium and the direction of the reaction, the value of the reaction quotient, Qc

Qc = Kc, the reaction equilibrium ---> reactant ⇌ product

Qc <Kc, reaction from left to right (product) until it reaches equilibrium (Qc = Kc) ---> reactant → product
Qc> Kc, the reaction from right to left (reactant) until it reaches equilibrium (Qc = Kc). If Qc> Kc, reactant ← product

equilibrium constants of heterogeneous reactions (more than 1 phase), liquid (L = liquid) and solid (S = solid) phases are ignored and not counted, only gas (G = gas) and solution (Aq = aqueous) phases so that the price of Kc for the reaction of CaCO₃ (s) → - → CaO (s) + CO₂ (g) is only based on concentration [CO₂]

Kc = [CO₂] = 0.0108

Similarly, the determination of Qc is also based on [CO₂]

We determine the mole and concentration of CO₂

molar mass of CO₂ = 44, and a volume of 10 L

a) 15.0 g CaCO₃, 15.0 g CaO, and 4.25g CO₂

4.25 g CO₂ (ignore masses of CaCO₃ and CaO)

mole CO₂= mass: molar mass

mole of CO₂ = 4.25: 44 = 0.0965

concentration = 0.0965: 10 = 0.00965 M

Because Qc = [CO₂] = 0.00965 and Qc <Kc, the reaction is more to the right (product), so CaCO₃ decreases

b) 2.50g CaCO₃, 25.0 g CaO, and 5.66 g CO₂

5.66 g CO₂

mole CO₂ = mass: molar mass

mole of CO₂ = 5.66: 44 = 0.1286

concentration = 0.1286: 10 = 0.01286 M

Because Qc = [CO₂] = 0.01286 and Qc> Kc, the reaction is more towards the left (reactant), so CaCO₃ increases

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