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

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 = $K_{c} \times (8.314 \times 1103)^{1}$

                     $K_{c} = 5.452 \times 10^{-3}$

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

       $K_{c}$ = $[CO_{2}]$

          $5.452 \times 10^{-3}$ = x

Hence, same amount of CaO is produced.

Moles of CaO = $5.452 \times 10^{-3}$

    Mass of CaO = $5.452 \times 10^{-3} \times 57 g/mol$

                          = 0.310 g

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