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Factors that Affect Equilibrium Can an equilibrium constant be altered ? What would happen if we changed –The concentration of a reactant or product?concentration.

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Presentation on theme: "Factors that Affect Equilibrium Can an equilibrium constant be altered ? What would happen if we changed –The concentration of a reactant or product?concentration."— Presentation transcript:

1 Factors that Affect Equilibrium Can an equilibrium constant be altered ? What would happen if we changed –The concentration of a reactant or product?concentration –The pressure of a reactant or productpressure –The temperature of a system?emperature What would happen if we used a catalyst ?catalyst The position of the equilibrium of a system changes to minimise the effect of any imposed change on the system

2 Concentration Changing the concentration of a reactant or product will shift the position of an equilibrium to minimise the change CH 3 COOH (aq) + CH 3 CH 2 OH (aq) CH 3 COOCH 2 CH 3 (ag) + H 2 O (l) K c = [CH 3 COOCH 2 CH 3 ] [H 2 O] [CH 3 COOH ] [CH 3 CH 2 OH] K c is the ratio of concentrations so… Changing the concentration of a reactant or product will not affect the value of Kc Increasing concentration of acid shifts equilibrium to the right, producing more ester & water Both the numerator & denominator increase

3 Changing Pressure Changing the pressure of a reactant or product will shift the position of an equilibrium to minimise the change N 2(g) + 3H 2 (g) 2NH 3 (g) K p = (pNH 3 ) 2 (pN 2 )(H 2 ) 3 Changing the pressure will not affect the value of Kp Increasing total pressure of the system shifts equilibrium to increasing the yield of ammonia If total pressure has increased then the partial pressures of all the gases have also increased

4 Changing Temperature Changing the temperature of a system will shift the position of an equilibrium and will change the value of Kc or Kp. N 2(g) + 3H 2 (g) 2NH 3 (g) K p = (pNH 3 ) 2 (pN 2 )(H 2 ) 3 ∆H = -92kJmol -1 exothermic Increasing temperature........... Favours backward reaction – trying to remove heat from system - equilibrium shifts to the left Decrease in yield of NH 3 This term is reduced K p decreased

5 Changing Temperature Changing the temperature of a system will shift the position of an equilibrium and will change the value of Kc or Kp. N 2(g) + 3H 2 (g) 2NH 3 (g) K p = (pNH 3 ) 2 (pN 2 )(H 2 ) 3 ∆H = -92kJmol -1 exothermic Decreasing temperature........... Favours exothermic forward reaction – trying give heat to system - equilibrium shifts to the right Increase in yield of NH 3 This term increases K p increased

6 Changing Temperature Predict the change in Kp if the temperature is increased in the following reaction CH 4(g) + H 2 O (g) CO (g) + 3H 2 (g) K p = (pCO)(pH 2 ) 3 (pCH 4 )(pH 2 O) 3 ∆H = +206kJmol -1 endothermic Increasing temperature........... Favours forward reaction – trying remove heat from system - equilibrium shifts to the right Increase in yield of CO & H 2 This term increases K p increased

7 Combustion engine problems N 2(g) + O 2 (g) 2NO (g) Kc = 4 x 10 -31 at 20 o C (298K) but at combustion temperature in a car engine Kc = 8 x 10 -9 at 800 o C (1100K) Is this reaction likely to happen at room temperature? Explain your reasoning The temperature of the spark which ignites the fuel/air mixture is about 2500K. At this temperature, how does the value of Kc change? How will this affect the yield of NO? Is the reaction endothermic or exothermic? Explain your answer using the change in the value of Kc

8 Combustion engine problems N 2(g) + O 2 (g) 2NO (g) Kc = 4 x 10 -31 at 20 o C (298K) but at combustion temperature in a car engine Kc = 8 x 10 -9 at 800 o C (1100K) Is this reaction likely to happen at room temperature? Explain your reasoning The temperature of the spark which ignites the fuel/air mixture is about 2500K. At this temperature, how does the value of Kc change? How will this affect the yield of NO? Is the reaction endothermic or exothermic? Explain your answer using the change in the value of Kc Kc very small at room temperature therefore reaction unlikely. Kc increases at higher temperatures Yield of NO increases K c = [NO] 2 [N 2 ] [O 2 ]

9 Combustion engine problems N 2(g) + O 2 (g) 2NO (g) Kc = 4 x 10 -31 at 20 o C (298K) but at combustion temperature in a car engine Kc = 8 x 10 -9 at 800 o C (1100K) Is the reaction endothermic or exothermic? Explain your answer using the change in the value of Kc K c = [NO] 2 [N 2 ] [O 2 ] Endothermic – increased temperature, increased yield of NO. Increasing temperature favours an endothermic reaction – the system tries to remove the imposed heat

10 Using a Catalyst A catalyst speeds up the forward and backward reaction equally The position of the equilibrium is not affected However the system reaches equilibrium faster when a catalyst is used Catalysts do not alter the value of Kc or Kp

11 Changing concentration has no effect on K c Changing pressure has no effect on K p Catalysts have no effect on K c Changing temperature changes the position of the equilibrium and the value of K c or K p Factors that Affect Equilibrium


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