Question

A sample of CaCO3(s) is introduced into a sealed container of volume 0.654 L and heated to 1000 K until equilibrium is reached. The Kp for the reaction CaCO3(s) ∆ CaO(s) + CO2(g) is 3.9 * 10-2 at this temperature. Calculate the mass of CaO(s) that is present at equilibriu

Answer 1

Answer:

$m_{CaO}=0.0174gCaO$

Explanation:

Hello,

In this case, the undergoing reaction is:

$CaCO_3(s)\rightleftharpoons CaO(s)+CO_2(g)$

Thus, since the Kp is given, it equals the partial pressure of carbon dioxide since the equilibrium expression only consider the gaseous species, hence:

$Kp=p_{CO_2}^{eq}=3.9x10^{-2}atm$

Therefore, by considering the given pressure and temperature, one computes the moles of carbon dioxide:

$n_{CO_2}=\frac{p_{CO_2}V}{RT} =\frac{3.9x10^{-2}atm*0.654L}{0.082\frac{atm*L}{mol*K} *1000K}=3.11x10^{-4}molCO_2$

In such a way, carbon dioxide and calcium oxide have a 1-to-1 molar relationship, thereby, the mass of calcium oxide turns out:

$m_{CaO}=3.11x10^{-4}molCO_2*\frac{1molCaO}{1molCO_2}*\frac{56.1gCaO}{1molCaO} \\m_{CaO}=0.0174gCaO$

Best regards.