equilibrium

equilibrium - Pictu re 6 Pictu re 8 A student ran the...

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Unformatted text preview: Pictu re 6 Pictu re 8 A student ran the following reaction in the laboratory at 284 K: 2 NO (g) + Br 2 (g) 2 NOBr (g) When she introduced 0.142 moles of NO (g) and 0.125 moles of Br 2 (g) into a 1.00 Liter container, she found the equilibrium concentration of NOBr (g) to be 0.112 M. Calculate the equilibrium constant, K c , she obtained for this reaction. K c = Feedback: Step 1 Set up the problem using the "ICE" Method. . 2 NO (g) + Br 2 (g) 2 NOBr (g) Initial 0.142 M 0.125 M Change- 2 x- x + 2 x Equilibrium ( 0.142- 2 x) ( 0.125- x) 2 x = 0.112 M Step 2 From the equilibrium concentration of NOBr (g) calculate x: x = [NOBr] /2 = (0.112 / 2) = 5.60E-2 M Step 3 Use x to calculate the other equilibrium concentrations: [ NO ] = ( 0.142- 2 x) = ( 0.142- 2 ( 5.60E-2 )) = 3.00E-2 M [ Br 2 ] = ( 0.125- x) = ( 0.125- ( 5.60E-2 )) = 6.90E-2 M Step 4 Calcuate K c using the equilibrium concentrations: K c = [ NOBr ] 2 = ( 0.112 ) 2 = 202 Pictu re 85 Pictu re 86 Pictu re 92 [ NO ] 2 [ Br 2 ] ( 3.00E-2 ) 2 ( 6.90E-2 ) LeChatelier's Principle: A change in any of the factors that determine the equilibrium conditions of a system will cause the system to change in such a manner as to reduce or counteract the effect of the change. Factors that determine the equilibrium conditions of a system are: 1. Concentration 2. Temperature (K changes) 3. Volume (Pressure) for gaseous systems. Concentration When a system is at equilibrium, Q = K. Changing the concentration of reactants or products that are part of the equilibrium constant expression cause the reactant quotient, Q, to change. Q will no longer be equal to K and the system will seek to restore itself to a new equilibrium that offsets the change. For example: Adding reactant to an equilibrium system. 2 NO 2 (g) N 2 O 4 (g) ...... K c = [N 2 O 4 ] / [NO 2 ] 2 Adding NO 2 (g) will cause the reaction to run in the forward direction to consume the added reactant . Since NO 2 is in the denominator, Q is less than K, and more product needs to be produced to reestablish equilibrium. Therefore the concentration of N 2 O 4 (g) will increase. 2 H 2 O(l) 2 H 2 (g) + O 2 (g) ...... K c = [H 2 ] 2 [O 2 ] Adding H 2 O(l) will have no effect on the equilibrium because the reactant is a liquid. Notice that the concentration of H 2 O(l) is not a part of the equilibrium constant expression. Since the concentration of product gases do not change when H 2 O(l) is added, Q is still equal to K and there is no driving force for the system to change. As long as some liquid is present to establish there is no driving force for the system to change....
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This note was uploaded on 05/07/2010 for the course CHEM 116 taught by Professor Stevenson during the Spring '08 term at Purdue.

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equilibrium - Pictu re 6 Pictu re 8 A student ran the...

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