The given question is incomplete, the complete question is:
When limestone (solid CaCO3) is heated, it decomposes into lime (solid CaO) and carbon dioxide gas. This is an extremely useful industrial process of great antiquity, because powdered lime mixed with water is the basis for mortar and concrete — the lime absorbs CO2 from the air and turns back into hard, durable limestone.
Suppose some calcium carbonate is sealed into a limekiln of volume 850 L and heated to 740.0 C. When the amount of CaCO3 has stopped changing, it is found that 2.05 kg have disappeared.
Calculate the pressure equilibrium constant Kp this experiment suggests for the equilibrium between CaCO3 and CaO at 740.0 C. Round your answer to significant digits.
Answer:
The correct answer is 2.00.
Explanation:
Based on the given information, the pressure equilibrium constant or Kp is equivalent to pressure of carbon dioxide or PCO₂. The reaction taking place in the given case is:
CaCO₃ (s) ⇄ CaO (s) + CO2 (g)
The molecular weight of CO2 is 44 g/mol and the molecular weight of CaCO₃ is 100 g/mol. The amount of CaCO₃ that got disappeared in the given case is 2.05 Kg or 2050 grams. Therefore, the amount of CO₂ is,
2050 g CaCO₃ * (44 g CO2/100 g CaCO₃) = 902 g CO2
The pressure of CO2, which is equivalent to Kp can be determined by using the formula,
P = nRT/V
Here V is the volume, which is 850 L, T is the temperature, which is 740 degree C or 1013 K, R is the rate constant, which is 0.0822 L.atm/K.mole
n is the no. of moles, which is equal to weight/molecular wt,
n = 902 g/44 g/mol = 20.5 mole
Now putting the values we get,
P = 20.5 mole * 0.0821 L.atm/K.mole * 1013 K / 850 L
P = 2.00 atm or Kp = 2.00
