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. equilibria. • The outcome is governed by EQUILIBRIUM AND EXTERNAL EFFECTS Henri Le Chatelier 1850-1936 Studied mining engineering. Interested in glass and ceramics. EQUILIBRIUM AND EXTERNAL EFFECTS • Temperature change ---> change ---> • Consider the fizz in a soft drink CO2(g) + H2O(liq) ¸ CO2(aq) + heat O(liq) CO aq) in K LE LE CHATELIER’S PRINCIPLE • “...if a system at equilibrium is disturbed, the system tends to shift its equilibrium position to counter the effect of the disturbance.” • Decrease T. What happens to equilibrium position? To value of K? • K = [CO2] / P (CO2) K increases as T goes down because [CO2] increases and P(CO 2) decreases. • Increase T. Now what? • Equilibrium shifts left and K decreases. Temperature Effects on Equilibrium N2O4 (colorless) + heat (colorless) ¸ 2 NO2 (brown) NO (brown) Ho = + 57.2 kJ 57.2 EQUILIBRIUM AND EXTERNAL EFFECTS EQUILIBRIUM AND EXTERNAL EFFECTS • Add catalyst ---> no change in ---> • A catalyst only affects the RATE of approach to equilibrium. K NH3 NH3 Production Production • N2(g) + 3 H2(g) ¸ 2 NH3(g) NH • K = 3.5 x 10 8 at 298 K at Kc [NO2 ]2 [N2O4 ] Kc (273 K) = 0.00077 (273 Kc (298 K) = 0.0059 (298 Catalytic exhaust system Page 1 EQUILIBRIUM AND EXTERNAL EFFECTS EQUILIBRIUM AND EXTERNAL EFFECTS • Concentration changes ---> no change ---> in K — only the position of equilibrium changes. Le Chatelier’s Principle Le Le Chatelier’s Principle Le Adding a “reactant” to a chemical system. Removing a “reactant” from a chemical system. Le Chatelier’s Principle Le Le Chatelier’s Principle Le butane ButaneIsobutane Equilibrium K= [isobutane] [butane] 2.5 isobutane Adding a “product” to a chemical system. Removing a “product” from a chemical system. Page 2 Butane ¸ Isobutane Isobutane Assume you are at equilibrium with [iso] = 1.25 [iso] M and [butane] = 0.50 M. Now add 1.50 M butane. When the system comes to equilibrium again, what are [iso] and [iso] [butane]? K = 2.5 2.5 butane Butane ¸ Isobutane Isobutane Assume you are at equilibrium with [iso] = 1.25 M and [iso] [butane] = 0.50 M. Now add 1.50 M butane. When the system comes to equilibrium again, what are [iso] [iso] and [butane]? K = 2.5 Butane ¸ Isobutane Isobutane You are at equilibrium with [iso] = 1.25 M and [iso] [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. isobutane. Set up concentration table [butane] [isobutane] isobutane] Initial 0.50 + 1.50 1.25 Change -x +x Equilibrium 1.25 + x 2.00 - x Solution Calculate Q immediately after adding more butane and compare with K. Q= isobutane [isobutane] [butane] 1.25 = 0.63 0.50 + 1.50 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 Isobutane You are at equilibrium with [iso] = 1.25 M and [iso] [butane] = 0.50 M. Now add 1.50 M butane. Solution Nitrogen Dioxide Equilibrium Nitrogen Dioxide Equilibrium N2O4(g) ¸ 2 NO 2(g) N2O4(g) ¸ 2 NO 2(g) ¸ Nitrogen Dioxide Equilibrium Nitrogen Dioxide Equilibrium N2O4(g) ¸ 2 NO 2(g) N2O4(g) ¸ 2 NO 2(g) Kc = [NO2 ]2 = 0.0059 at 298 K [N2O4 ] K = 2.50 = [isobutane] [butane] 1 .25 + x 2.00 - x Kc = x = 1.07 M At the new equilibrium position, [butane] = 0.93 M and [ isobutane] = 2.32 M. isobutane] [NO2 ]2 = 0.0059 at 298 K [N2O4 ] Equilibrium has shifted toward isobutane. Equilibrium isobutane. Increase P in the system by reducing the volume. 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 LEFT and decreases and P of N 2O4 increases. increases. Page 3 ...
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This note was uploaded on 01/11/2011 for the course ENGINEERIN MAE 107 taught by Professor Pozikrizdis during the Fall '08 term at San Diego.

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