The dissociation of calcium carbonate has an equilibrium constant of kp= 1.20 at 800°c. caco3(s) ⇋ cao(s) + co2(g)

a.) what is the kc for the reaction?
b.) if you place 20.0 g of caco3 in a 2.50 l container at 800°c, what is the pressure of co2 in the container?
c.) what percentage of the original 20.0 g sample of caco3 remains undecomposed at equilibrium?

Explanation:

(a)   Formula that shows relation between and is as follows.

Here, = 1

Putting the given values into the above formula as follows.

                  =

                  =

                  = 0.01316

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

As there is only one gas so ,

                = 1.20

Therefore, pressure of in the container is 1.20.

(c)   Now, expression for for the given reaction equation is as follows.  

                        =

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

where,    a = initial conc. of

                  =

                  = 0.023 M

          0.0131 =

                  x = 0.017

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

       % of remained =

                                  = 75.46%

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

consider the reaction below.

 no rx

at 500 k, the reaction is at equilibrium with the following concentrations.

[pci5]= 0.0095 m

[pci3] = 0.020

[ci2] = 0.020 m

what is the equilibrium constant for the given reaction?

0.042

0.42

2.4

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