can someone help me plz? At 900OC, Kp = 1.04 for the reaction
CaCO3 (s) ↔ CaO (s) + CO2 (g)
At low temperature, dry ice (solid CO2), calcium oxide, and calcium carbonate are
introduced into a 50.0-L reaction chamber. The temperature is raised to 900OC,
resulting in the dry ice converting gaseous CO2, For the following mixtures, will
the initial amount of calcium oxide increase, decrease, or remain the same as the
system moves toward equilibrium at 900OC?
a. 655 g CaCO3, 95.0 g CaO, PCO2 = 2.55 atm

Question

can someone help me plz? At 900OC, Kp = 1.04 for the reaction
CaCO3 (s) ↔ CaO (s) + CO2 (g)
At low temperature, dry ice (solid CO2), calcium oxide, and calcium carbonate are
introduced into a 50.0-L reaction chamber. The temperature is raised to 900OC,
resulting in the dry ice converting gaseous CO2, For the following mixtures, will
the initial amount of calcium oxide increase, decrease, or remain the same as the
system moves toward equilibrium at 900OC?
a. 655 g CaCO3, 95.0 g CaO, PCO2 = 2.55 atm

anonymous · Answer

at 900*C

anonymous · Answer

how would you relate the Kp to the concentration or partial pressures of the reactants/products for this reaction?

anonymous · Answer

Kp=mol/v ?

anonymous · Answer

i dont get it.

anonymous · Answer

okay, Kp is the equilibrium constant. do you know what it is?

anonymous · Answer

yes i know

anonymous · Answer

mol/V is a molar concentration. look up "equilibrium constant" in your book.

anonymous · Answer

yeah. i gotta find mol of CaCO3 CaO and CO2, right?

anonymous · Answer

it's possible to relate the equilibrium constant to the equilibrium concentrations/pressures of the things at equilibrium with each other. have you heard of the "law of mass action"?