Given the following reaction at equilibrium at 450.0 °C: CaCO₃ (s) CaO (s) + CO₂ (g)
If pCO₂ = 0.0155 atm,K_c = ________.

Question

Given the following reaction at equilibrium at 450.0 °C: CaCO₃ (s) CaO (s)+ CO₂ (g) If pCO₂ = 0.0155 atm,K_c = ________.

Quizplus · Accepted Answer

The equilibrium constant expression for this reaction is: K_c = \frac{[CaO]}{[CO₂]} At equilibrium, we can assume that the amounts of CaO (s) and CO₂ (g) are small and that the amount of CaCO3 (s) is large. This allows us to simplify the equilibrium expression to: K_c = \frac{[CaO]}{pCO₂} We are given that pCO₂ = 0.0155 atm. We need to determine the concentration of CaO (s), but we can assume that it is small and thus can use the ideal gas law to relate it to the pressure of CO₂ (g): PV=nRT where P = pCO₂, V = 1 L (since we are assuming ideal gas behavior), n = moles of CO₂, R = 0.0821 L atm K^-1 mol^-1 (the ideal gas constant), and T = 450.0 + 273.15 K = 723.15 K. Solving for n gives: n = \frac{PV}{RT} = \frac{(0.0155 atm)(1 L)}{(0.0821 L\ atm\ K^{-1}\ mol^{-1})(723.15 K)} = 2.61 imes 10^{-4}\ mol Since 1 mol of CaCO3 (s) produces 1 mol of CaO (s), we know that the concentration of CaO (s) is also 2.61 imes 10^{-4}\ mol/L. Substituting these values into the equilibrium expression gives: K_c = \frac{(2.61 imes 10^{-4}\ mol/L)}{(0.0155 atm)} \approx \boxed{2.61 imes 10^{-4}\ mol/L}

Given the Following Reaction at Equilibrium at 450