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AQA A1 Equilibrium

Equilibrium Kc

QuestionAnswer
What happens to the equilibrium when the concentration of Ethanol is increased in the equilibrium C2H5OH + CH3COOH <=> CH3COOC2H5 +H2O? to minimise change, ie to decrease the concentration of Ethanol, the system will move to the side with less moles of Ethanol, the equilibrium shifts to the right
What happens to the equilibrium when the temperature is increased for an equilibrium where the forward reaction is endothermic? 1) to minimise the change, ie. to absorb heat/lower temperature, 2) the system will favour the endothermic reaction, here forward, 3) so the equilibrium will shift to the right (are you sticking with the model answer?)
What happens to Kc when the temperature is increased for an equilibrium where the forward reaction is endothermic? the equilibrium will shift to the right; CONCENTRATION (not simply “more”) products increases; Kc will increase
What happens to Kc when the concentration of Ethanol is increased in the equilibrium C2H5OH + CH3COOH <=> CH3COOC2H5 +H2O? equilibrium shifts to the right BUT Kc does NOT change (increased [C2H5OH] AND [products)
What happens to the equilibrium CH4 + 2H2O <=> CO2 + 4H2 (all chemicals are in gaseous state) when pressure is increased? 1) to minimise the change, ie. to decrease the pressure, 2) the system will move toward the side with less moles of gas, 3) so the equilibrium will shift to the left (are you sticking with the model answer?)
What happens to Kc for the equilibrium CH4 + 2H2O <=> CO2 + 4H2 (all chemicals are in gaseous state) when pressure is increased? the equilibrium will shift to the left BUT Kc does NOT change (only depends of temperature)
What happens to Kc for the equilibrium CO + 2H2 <=> CH3OH delta-H=-91kJ.mol^-1 when the temperature is increased? equilibrium will shift to the left (as backward is endo); CONCENTRATION products decreases so Kc decreases
What is the effect of a catalyst on the equilibrium C2H5OH + CH3COOH <=> CH3COOC2H5 +H2O? No change; a catalyst will increase the rate of BOTH the forward and backward reaction so equilibrium is reached faster but equilibrium concentrations remain the same with or without catalyst
CHALLENGE: Explain, in terms of Kc, why is yield increased for C2H5OH + CH3COOH <=> CH3COOC2H5 +H2O when water is removed the term on top decreases; the equilibrium moves to restore Kc by increasing the top and decreasing the bottom; so equilibrium shifts to the right.
What is the equilibrium constant for the equilibrium C2H5OH + CH3COOH <=> CH3COOC2H5 +H2O? Kc=([ CH3COOC2H5]x[H2O]) / ([ C2H5OH]x[CH3COOH]) typical mistakes are using round brackets instead of square ones; or ignoring H2O (you only do that for Ka)
What is the unit of Kc for the equilibrium CH4 + 2H2O <=> CO2 + 4H2 (all chemicals are in gaseous state)? Kc=( mole.dm^-3 x [mole.dm^-3]^4) / ( mole.dm^-3 x [ mole.dm^-3]^2) = mole^2.dm^-6
What is the value of Kc for C2H5OH + CH3COOH <=> CH3COOC2H5 +H2O when you start with 0.10moles of Ethanol and CH3COOH in 20cm3 and only 0.033moles of CH3COOH are left? 4.1, no unit; method: ICE; remember to work out CONCENTRATIONS by dividing moles at equilib by 0.020dm3!
CHALLENGE: What amount of CH3COOC2H5 would be collected if Kc for C2H5OH + CH3COOH <=> CH3COOC2H5 +H2O is 4.0 and you start with 1.0 mole of Ethanol and CH3COOH? 0.67; at equilibrium x moles of Ester and H2O will be formed and there will be(1-x) moles of eahch reactant left; assume a volume V, the volumes cancel out; solve the equation Kc=x^2/(1-x)^2
Define a dynamic equilibrium When rate of a forward and rate of a backward reaction are the same, the concentrations stay the same, in a closed system
Created by: UrsulineChem