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CaCO3(s)⇄CaO(s)+CO2(g) When heated strongly, solid calcium carbonate decomposes to produce solid calcium oxide and carbon dioxide gas, as represented by the equation above. A 2.0mol sample of CaCO3(s) is placed in a rigid 100.L reaction vessel from which all the air has been evacuated. The vessel is heated to 898°C at which time the pressure of CO2(g) in the vessel is constant at 1.00atm , while some CaCO3(s) remains in the vessel. (a) Calculate the number of moles of CO2(g) present in the vessel at equilibrium. (b) Write the expression for Kp, the equilibrium constant for the reaction, and determine its value at 898C.

Respuesta :

A) The number of moles of COâ‚‚(g) present in the vessel at equilibrium is; 1.04 moles

B)  The expression for Kp, the equilibrium constant for the reaction, and its value at 898°C is; Kp = (P[CO₂] * P[CaO])/(P[CaCO₃]) and Kp = 0.0104 mol/L

How to Calculate Equilibrium Constant

We are given the reaction equation as;

CaCO₃ (s) ⇄ CaO (s) + CO₂(g)

We are given;

Number of moles of CaCO₃ used = 2 mols

Volume of vessel; V = 100 L

Temperature; T = 898°C = 1171 K

Pressure of COâ‚‚ in the vessel; P = 1 atm

A) Formula for number of moles of COâ‚‚ in the vessel at equilibrium is;

n = PV/RT

where;

P is Pressure

V is Volume

R is gas constant = 0.08206 L·atm/mol·k

n = (1 * 100)/(0.08206 * 1171)

n = 1.04 moles of COâ‚‚

B) Formula to find the equilibrium constant kp is;

Kp = (P[CO₂] * P[CaO])/(P[CaCO₃])

Where;

P[COâ‚‚] is the partial pressure of COâ‚‚

P[CaO] is the partial pressure of CaO

P[CaCO₃] is the partial pressure of CaCO₃

Since concentration of solids is taken as 1, then it means that;

Kp = P[COâ‚‚]

Kp = 1.04/100

Kp = 0.0104 mol/L

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