# In the preparation of CaO from CaCO3 using the equilibrium, CaCO3(s) <------> CaO(s) + CO2(g) Log Kp = 7.282- 8500/T For complete decomposition of CaCO3, the temperature in Celsius is ? A) 1167 B) 894 C) 8500 D) 850

##### 1 Answer

#### Answer:

The answer is **B)**

#### Explanation:

This problem can only be solved if an important *assumption* is made -- the decomposition reaction **must** take place in an open container, i.e. a container exposed to the air.

Moreover, we must assume that the atmospheric pressure **must** be expressed in *atmospheres* in order for the equation given to you to work.

So, for the thermal decomposition of calcium carbonate

#"CaCO"_ (3(s)) + color(red)(Delta) rightleftharpoons "CO"_ ((s)) + "CO"_ (2(g))#

the equilibrium constant

#K_p = P_ ("CO"_ 2)#

Notice that the forward reaction is **endothermic**. This tells you that as temperature **increases**, the partial pressure of carbon dioxide above the solid will **increase**. This happens because when you increase the temperature, the equilibrium **shifts to the right**, i.e. the forward reaction is favored.

Now, here's the tricky part. You can also force the equilibrium to shift to the right by **removing** some of the carbon dioxide gas produced by the forward reaction.

This can be done by allowing the carbon dioxide to **escape to the atmosphere**, hence the need to have an open container.

In order for the carbon dioxide to escape to the atmosphere, the **partial pressure** of carbon dioxide above the solid must be **equal** to the atmospheric pressure.

As you know, the atmospheric pressure at sea level is equal to

You can thus say that

#P_ ("CO"_ 2) = "1 atm"#

Consequently, you will have

#K_p = "1 atm"#

Now, you know that

#log(K_p) = 7.282 - 8500/T#

But since for a partial pressure of

#log(1) = 0 #

you can say that

#0 = 7.282 - 8500/T#

Solve for **absolute temperature** at which the evolved carbon dioxide can escape to the atmosphere

#8500/T = 7.282#

#T = 8500/7.282 = "1167.26 K"#

Finally, convert this to degrees Celsius by using

#t[""^@"C"] = T["K"] - 273.15#

You will end up with

#t[""^@"C"] = "1167.26 K" - 273.15 = color(dakgreen)(ul(color(black)(894^@"C")))#

You can thus say that calcium carbonate will decompose completely if heated at a temperature of