Cacoals) o cao(s) co,(g) when heated, calcium carbonate decomposes according to
ID: 984596 • Letter: C
Question
Cacoals) o cao(s) co,(g) when heated, calcium carbonate decomposes according to the equation above. In a study of the decomposition of calcium carbonate, a student added a 50.0 g sample of powdered Cacols) to a L rigid container. The student sealed the container, pumped out all the gases, then heated the container In an oven at 1100 K. As the container was heated, the total pressure of the co.g) in the container was measured over time. That data are plotted in the graph below 0.75 line (min) The student repeated the experiment, but this time the student used a 100.0 g sample of powdered Caco,(s). In this experiment, the final pressure in the container was 1.04 atm, which was the same final pressure as in the first experiment. a. Calculate the number of moles of colg) present in the container after 20 minutes of heating. b. The student claimed that the final pressure in the container in each experiment became constant because all of the Cacols) had decomposed. Based on the data in the experiments, do you agree with this claim? Explain. c. After 20 minutes some co,(g) was injected into the container, initially raising the pressure to 1.5 atm. Would the final pressure inside the container be less than, greater than, or equal to 1.04 atm? Explain your reasoning. d. there sufficient data obtained in the experiments to determine the value of the Are equilibrium constant, Kp, for the decomposition of Caco,(s) at 1100 K? Justify your answer.Explanation / Answer
a) Assume CO2 be ideal gas.
Then, Pco2V = nRT => n = PV/RT = 1.04*1/(0.0821*1100) = 0.0115
number of moles of CO2 (g) = 0.0115
b) No.
If all of CaCO3 (s) were decomposed in each experiment then pressure due to CO2 would have been different. But, this is not the case.
c) Final pressure will become 1.04 atm. There will be bacward shifting to attain the equilibrium pressure that is 1.04 atm.
d) Yes. In the above graph, pressure become constant after sometime for the equilibrium is established. And at that equilibrium, pressure = 1.04 atm.
KP = Pco2 = 1.04 atm