Ch16_LeChat_2

Ch16_LeChat_2 - EQUILIBRIUM AND EXTERNAL EFFECTS •...

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Unformatted text preview: EQUILIBRIUM AND EXTERNAL EFFECTS • Temperature, catalysts, and changes in concentration affect equilibria . LE LE CHATELIER’S PRINCIPLE • The outcome is governed by The • “...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance.” Temperature Effects on Equilibrium N2O4 (colorless) + heat ¸ 2 NO2 (brown) EQUILIBRIUM AND EXTERNAL EFFECTS Henri Le Chatelier 1850-1936 Studied mining engineering. Interested in glass and ceramics. • Temperature change ---> change Temperature ---> • Consider the fizz in a soft drink CO2(g) + H2O(liq) ¸ CO2(aq) + heat • Decrease T. What happens to equilibrium position? To value of K? • K = [CO2] / P (CO2) K increases as T goes down because increases [CO2] increases and P(CO 2) decreases. • Increase T. Now what? • Equilibrium shifts left and K decreases. • Add catalyst ---> no change in Add ---> • A catalyst only affects the RATE of approach to equilibrium. K NH3 NH3 Production Production [NO2 ]2 [N2O4 ] Kc (273 K) = 0.00077 Kc (298 K) = 0.0059 in K EQUILIBRIUM AND EXTERNAL EFFECTS EQUILIBRIUM AND EXTERNAL EFFECTS Ho = + 57.2 kJ Kc EQUILIBRIUM AND EXTERNAL EFFECTS Catalytic exhaust system Page 1 • N2(g) + 3 H2(g) ¸ 2 NH3(g) (g) • K = 3.5 x 10 8 at 298 K EQUILIBRIUM AND EXTERNAL EFFECTS EQUILIBRIUM AND EXTERNAL EFFECTS ---> no change ---> in K — only the position of equilibrium changes. • Concentration changes Concentration Le Chatelier’s Principle Le Chatelier’s Principle Adding a “reactant” to a chemical system. Le Chatelier’s Principle Le Chatelier’s Principle Le Chatelier’s Principle Le Chatelier’s Principle Removing a “reactant” from a chemical system. ButaneIsobutane Equilibrium Le Chatelier’s Principle Le Chatelier’s Principle but a n e K= isobut a n e Adding a “product” to a chemical system. Removing a “product” from a chemical system. Page 2 [isobutane] [butane] 2.5 Butane ¸ Isobutane Butane Isobutane Butane ¸ Isobutane Butane Isobutane Assume you are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. When the system comes to equilibrium again, what are [ iso ] and [butane]? K = 2.5 [butane]? 2.5 b ut a n e Assume you are at equilibrium with [iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. When the system comes to equilibrium again, what are [iso] and [butane]? K = 2.5 and 2.5 Solution Calculate Q immediately after adding more butane and compare with K. Q= [isobutane] [butane] 1.25 = 0.63 0.50 + 1.50 isobut a n e Q is LESS THAN K. Therefore, the Q is LESS THAN K. Therefore, the reaction will shift to the ____________. reaction will shift to the ____________. Butane ¸ Isobutane Butane Isobutane Nitrogen Dioxide Equilibrium Nitrogen Dioxide Equilibrium N2O4(g) ¸¸ 2 NO 2(g) N2O4(g) 2 NO 2(g) You are at equilibrium with [ iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. Solution K = 2 .50 = [isobutane] [butane] x = 1.07 M At the new equilibrium position, At [butane] = 0.93 M and [ isobutane] = 2.32 M. 2.32 Equilibrium has shifted toward isobutane. Equilibrium Kc = Nitrogen Dioxide Equilibrium Nitrogen Dioxide Equilibrium N2O4(g) ¸¸ 2 NO 2(g) N2O4(g) 2 NO 2(g) Kc = ¸ 1 .25 + x 2.00 - x Butane ¸ Isobutane Butane Isobutane You are at equilibrium with [ iso] = 1.25 M and [butane] = 0.50 M. Now add 1.50 M butane. Solution Q is less than K, so equilibrium shifts right — away from butane and toward isobutane. Set up concentration table [butane] [isobutane ] [butane] Initial 0.50 + 1.50 1.25 Change Change -x +x Equilibrium 1.25 + x 2.00 - x [NO2 ]2 = 0.0059 at 298 K [N2O4 ] Increase P in the system by reducing the volume. Page 3 [NO2 ]2 = 0.0059 at 298 K [N2O4 ] Increase P in the system by reducing the volume. In gaseous system the equilibrium will shift to the side with fewer molecules (in order to reduce the P). Therefore, reaction shifts LEFT and P of NO2 Therefore, LEFT and decreases and P of N 2O4 increases. ...
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