Question

The dissociation of calcium carbonate has an equilibrium constant of Kp= 1.20 at 800°C. CaCO3(s) ⇋ CaO(s) + CO2(g)

Answer 1

Explanation:

(a)   Formula that shows relation between $K_{c}$ and $K_{p}$ is as follows.

                 $K_c = K_p \times (RT)^{-\Delta n}$

Here, $\Delta n$ = 1

Putting the given values into the above formula as follows.

        $K_c = K_p \times (RT)^{-\Delta n}$

                  = $1.20 \times (RT)^{-1}$

                  = $\frac{1.20}{0.0820 \times 1073}$

                  = 0.01316

(b) As the given reaction equation is as follows.

               $CaCO_{3}(s) \rightleftharpoons CaO(s) + CO_{2}(g)$

As there is only one gas so ,

                $p[CO_{2}] = K_{p}$ = 1.20

Therefore, pressure of $CO_{2}$ in the container is 1.20.

(c)   Now, expression for $K_{c}$ for the given reaction equation is as follows.  

             $K_{c} = \frac{[CaO][CO_{2}]}{[CaCO_{3}]}$

                        = $\frac{x \times x}{(a - x)}$

                        = \frac{x^{2}}{(a - x)}[/tex]

where,    a = initial conc. of $CaCO_{3}$

                  = $\frac{22.5}{100} \times 9.56$

                  = 0.023 M

          0.0131 = $\frac{x^{2}}{0.023 - x}$

                  x = 0.017

Therefore, calculate the percentage of calcium carbonate remained as follows.

       % of $CaCO_{3}$ remained = $(\frac{0.017}{0.023}) \times 100$

                                  = 75.46%

Thus, the percentage of calcium carbonate remained is 75.46%.