For the reaction

CaCO₃ (s) <=> CaO (s) + CO₂ (g), Kp = 0.5 atm at 25⁰C . If 20 gm of CaCO₃ is placed in a closed container at 25⁰C, calculate the amount of CO₂ produced at equilibrium.

(a) 20 gm (b) 44gm (c) Data insufficient (d) None of these

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Sir,

I proceeded as shown below , am I right or wrong Please give detailed solution

Kp = Kc (RT)ⁿ Here n = 1

So,

Kc = Kp/RT = 0.5/(0.0821* 298) = 0.0204 -------------------------------- (1)

Now,

CaCO₃ (s) <=> CaO (s) + CO₂ (g)

20/100 0 0

At equilibrium

0.2 - x x x

So, Kc = x -------------(2) as CaO and CaCO₃ are solid

From (1) and (2)

x = 0.0204 moles = 0.0204 * 44 gm = 0.9 gm

Please reply soon

Let's break down the problem step by step to determine the amount of CO₂ produced at equilibrium for the reaction you provided. The reaction is:

CaCO₃ (s) <=> CaO (s) + CO₂ (g)

Understanding the Equilibrium Constant

The equilibrium constant \( K_p \) is given as 0.5 atm at 25 °C. Since we are dealing with solids and gases, we only need to consider the gaseous component for \( K_p \). The expression for \( K_p \) in this case is:

Kp = P_CO2

Since \( K_p \) is equal to the partial pressure of CO₂ at equilibrium, we can directly relate it to the amount of CO₂ produced.

Calculating Moles of CaCO₃

First, we need to find out how many moles of CaCO₃ we have. The molar mass of CaCO₃ is approximately 100 g/mol. Therefore, for 20 g of CaCO₃:

Number of moles of CaCO₃ = mass / molar mass

Number of moles = 20 g / 100 g/mol = 0.2 moles

Setting Up the Equilibrium Expression

At the start of the reaction, we have 0.2 moles of CaCO₃ and no CO₂. As the reaction proceeds to equilibrium, let \( x \) be the amount of CO₂ produced. Since CaCO₃ decomposes to form CaO and CO₂, we can express the changes in moles as follows:

• Initial: CaCO₃ = 0.2 moles, CO₂ = 0 moles
• Change: CaCO₃ = -x, CO₂ = +x
• Equilibrium: CaCO₃ = 0.2 - x, CO₂ = x

Using Kp to Find x

Since \( K_p = P_{CO2} \), we can relate the amount of CO₂ produced to the equilibrium constant:

Kp = x

Given that \( K_p = 0.5 \) atm, we can set \( x = 0.5 \) atm. Now, we need to convert this pressure into moles using the ideal gas law:

P = nRT

Where:

• P = pressure (0.5 atm)
• R = ideal gas constant (0.0821 L·atm/(K·mol))
• T = temperature (298 K)

Rearranging the ideal gas law gives us:

n = P / (RT)

Substituting the values:

n = 0.5 atm / (0.0821 L·atm/(K·mol) * 298 K) ≈ 0.0204 moles

Calculating the Mass of CO₂

Now that we have the number of moles of CO₂ produced at equilibrium, we can find the mass:

Mass of CO₂ = moles × molar mass

The molar mass of CO₂ is approximately 44 g/mol:

Mass = 0.0204 moles × 44 g/mol ≈ 0.8976 g

Final Answer

Rounding this to a reasonable number of significant figures, we find that the mass of CO₂ produced at equilibrium is approximately 0.9 g. Therefore, the correct answer is:

(d) None of these

Your approach was mostly correct, but it seems there was a misunderstanding in how to relate the equilibrium constant to the amount of CO₂ produced. I hope this detailed explanation clarifies the steps for you!