Question

When solid calcium carbonate is heated, it decomposes according to the reaction: CaCO3(s) ⇄ CaO(s) + CO2(g) Kp = 0.50 A sample of CaCO3 is placed in a 1.0 L evacuated flask and heated to 830 Celsius. What is the mass of CaO produced when equilibrium is established?

Answer 1

To find the mass of CaO produced when equilibrium is established, use the ideal gas equation to relate pressure and moles, substitute the given values to find the moles, and then use stoichiometry and the molar mass to calculate the mass of CaO.

Step-by-step explanation:

To find the mass of CaO produced when equilibrium is established, we need to use stoichiometry and the given equilibrium constant, Kp. From the balanced equation, we can see that 1 mole of CaCO3 decomposes to produce 1 mole of CaO. So the moles of CaO produced will be equal to the moles of CaCO3 decomposed at equilibrium.

Using the ideal gas equation, we can relate the pressure (P) and number of moles (n) at equilibrium:

P = n / V

Since the flask is evacuated, the pressure at the start is zero. At equilibrium, the pressure is equal to Kp:

Kp = n / V

Rearranging the equation, we get:

n = Kp * V

Substituting the given values, we have:

n = 0.50 * 1.0

n = 0.50 moles

Since 1 mole of CaO has a molar mass of approximately 56 g/mol, the mass of CaO produced will be:

Mass = molar mass * moles

Mass = 56 g/mol * 0.50 moles

Mass = 28 g

Answer 2

Step-by-step explanation:

Relation between
`K_(p)` and
`K_(c)` is as follows.

`K_(p) = K_(c) [RT]^(\Delta n)`

Given, temperature =
`830^(o)C` = (830 + 273) K = 1103 K

R = 8.314 J/mol K

`\Delta n` = 1 - 0 = 1

Now, putting the given values into the above formula as follows.

`K_(p) = K_(c) [RT]^(\Delta n)`

0.5 =
`K_(c) * (8.314 * 1103)^(1)`

`K_(c) = 5.452 * 10^(-3)`

ICE table for the given reaction will be as follows.

`CaCO_(3)(s) \rightleftharpoons CaO(s) + CO_(2)(g)`

Initial: c - -

Equilibrium: (c - x) x x

`K_(c)` =
`[CO_(2)]`

`5.452 * 10^(-3)` = x

Hence, same amount of CaO is produced.

Moles of CaO =
`5.452 * 10^(-3)`

Mass of CaO =
`5.452 * 10^(-3) * 57 g/mol`

= 0.310 g

Thus, we can conclude that 0.310 g of CaO produced when equilibrium is established.