Final answer:
To find the equilibrium constant (Kp) for CO₂ decomposition, we calculate the partial pressures using the initial pressure and extent of decomposition. Applying the equilibrium expression for the reaction, Kp can be determined for the decomposition of CO₂ at 2500 K.
Step-by-step explanation:
The question asks us to calculate the equilibrium constant (Kp) for the decomposition of carbon dioxide (CO₂) into carbon monoxide (CO) and oxygen (O₂) at a specific temperature. Given that the initial pressure of CO₂ is 1.00 atm and 2.10% of the molecules decompose at 2500 K, we can determine the partial pressures at equilibrium and hence the equilibrium constant.
At the start:
CO₂(g) initially = 1.00 atm
After 2.10% decomposition:
CO₂(g) at equilibrium = 1.00 atm - (2.10% of 1.00 atm) = 1.00 atm - 0.021 atm = 0.979 atm
From the stoichiometry of the decomposition reaction (2 CO₂ → 2 CO + O₂), every mole of CO₂ that decomposes will produce one mole of CO and half a mole of O₂.
CO(g) at equilibrium = 2 * 0.021 atm = 0.042 atm (as 2.10% decomposes to CO)
O₂(g) at equilibrium = 0.021 atm / 2 = 0.0105 atm (as 2.10% decomposes to O₂)
The equilibrium constant, Kp, is calculated using partial pressures of the gases at equilibrium. The Kp expression is Kp = (P_CO)^2 * (P_O2) / (P_CO₂)^2. Substituting the calculated pressures we get: Kp = (0.042)^2 * (0.0105) / (0.979)^2. Calculating this out, we get the value for the equilibrium constant Kp at 2500 K.