# 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

Apr 5, 2017

The answer is B) ${894}^{\circ} \text{C}$

#### 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

${\text{CaCO"_ (3(s)) + color(red)(Delta) rightleftharpoons "CO"_ ((s)) + "CO}}_{2 \left(g\right)}$

the equilibrium constant ${K}_{p}$ is defined as

${K}_{p} = {P}_{{\text{CO}}_{2}}$

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 $\text{1 atm}$. This means that when the partial pressure of carbon dioxide above the solid will reach $\text{1 atm}$, the gas will start to escape to the atmosphere.

You can thus say that

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

Consequently, you will have

${K}_{p} = \text{1 atm}$

Now, you know that

$\log \left({K}_{p}\right) = 7.282 - \frac{8500}{T}$

But since for a partial pressure of $\text{1 atm}$ you have

$\log \left(1\right) = 0$

you can say that

$0 = 7.282 - \frac{8500}{T}$

Solve for $T$, the absolute temperature at which the evolved carbon dioxide can escape to the atmosphere

$\frac{8500}{T} = 7.282$

$T = \frac{8500}{7.282} = \text{1167.26 K}$

Finally, convert this to degrees Celsius by using

$t \left[\text{^@"C"] = T["K}\right] - 273.15$

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 ${894}^{\circ} \text{C}$.