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Quantitative Changes in Equilibrium System

Quantitative Changes in Equilibrium System. Q, the Reaction Quotient. We have written the formula for the equilibrium constant: K = [C] c [D] d [A] a [B] b

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Quantitative Changes in Equilibrium System

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  1. Quantitative Changes in Equilibrium System

  2. Q, the Reaction Quotient • We have written the formula for the equilibrium constant: K = [C]c[D]d [A]a[B]b • Given concentrations we can determine if a system is at equilibrium, by using the equilibrium constant expression, and comparing the value to the equilibrium constant. • The calculated value we compare to the eqm constant is called the Reaction Quotient, and is given the symbol Q.

  3. Q, the Reaction Quotient An eqm system is formed by the decomposition of carbon dioxide gas into carbon monoxide and oxygen gases. At 300oC, the K is 1.2x 10-5. When the conc of the gases were measured it was determined carbon dioxide was at 2.5 M, carbon monoxide was at 0.24 M, and oxygen was at 0.045 M. Is the system at equilibrium? 2CO2(g) 2CO(g) + O2(g) K = [CO]2[O2] = 1.2 x 10-5 [CO2]2 We can only use K at eqm, or we are using it incorrectly.

  4. Q, the Reaction Quotient 2CO2(g) 2CO(g) + O2(g) K = [CO]2[O2] = 1.2 x 10-5 [CO2]2 [CO2] =2.5 M [CO] = 0.24 M, [O2]= 0.045 M Q = [CO]2[O2] = (0.24)2(0.045) [CO2]2 (2.5)2 =4.1472 x 10-4 Q > K Therefore the system is not at equilibrium.

  5. Q, the Reaction Quotient Q = products reactants • If Q = K,  system is at eqm. • If Q > K, too many products. To reach eqm the system shifts to the left, or the reverse reaction is favoured, or more reactants are made. • If Q < K,too many reactants. To reach eqm the system shifts to the right, or the forward reaction is favoured, or more products are made.

  6. Calculations involving Equilibrium • We have already done a number of calculations with equilibrium systems. • We have used an ICE table to organize our data. • There are other types of calculations we need to be able to do.

  7. Determining equilibrium concentrations using K A possible source of hydrogen gas is from a reversible rxn of methane gas with steam. Determine the eqm conc of hydrogen if the equm concentrations are as follows:methane is 0.400 M, carbon monoxide is 0.300 M, and water is 0.800 M. The eqm constant is 5.67 at 1500oC. CH4(g) + H20(g) CO(g)+ 3H2(g)

  8. Determining equilibrium concentrations using K CH4(g) + H20(g) CO(g)+ 3H2(g) K= [CO][H2]3 =5.67 [CH4][H20] [CH4(g)]= 0.400 [CO(g)]= 0.300 [H20(g) ]=0.800 5.67= [0.300][H2 ]3 [0.400][0.800] [H2 ]3 =6.048 [H2 ] = 1.821953383  the concentration of hydrogen at equilibrium is 1.82M

  9. Determining equilibrium concentrations using K The eqm constant for the formation rxn of hydrogen iodide gas is 50 at 448oC. What amount of hydrogen gas is present at eqm when 1.0 moles of hydrogen is mixed with 1.0 moles of iodine gas in a 0.50 L flask and allowed to react at 448oC? H2(g) + I2(g) 2HI(g)

  10. Determining equilibrium concentrations using K n=1.0 V=0.5 C=n/V =1.0/0.5 =2.0 2.0 2.0 0 - x/2 - x/2 + x 2.0 - x/2 2.0 - x/2 x K = [HI]2 = 50 [H2][I2] K = [x]2= 50 [2.0 - x/2][2.0 - x/2]

  11. Determining equilibrium concentrations using K K = [x]2= 50 [2.0 - x/2][2.0 - x/2] [HI] = x = 3.1 [I2]=[H2] = 2.0-x/2 (x)2= (2.0 - x/2)2 = 0.440962419 = 0.44 Since the [H2] is 0.44 M, and we have a 0.5 L container, we have 0.22 moles of H2. (x) =7.071067812 (2.0 - x/2) x = 7.071067812(2.0 - x/2) x = 14.14213562 – 3.535533906x Therefore there are 0.22 moles of hydrogen. 4.535533906x = 14.14213562 x = 3.118075162

  12. Determining equilibrium concentrations using K The same system, at 1100oK, has a eqm constant of 25.0. If 2.00 moles of hydrogen and 3.00 moles of iodine are placed in a 1.00 L vessel, determine the eqm conc of each gas.

  13. Determining equilibrium concentrations using K The atmosphere contains large amounts of oxygen and nitrogen. The two gases do not react at ordinary temperatures. At very high temperatures, such as a lightening strike, or in a car’s engine, they do react to produce nitrogen monoxide. A chemist studying this reaction puts 0.085 moles of nitrogen and 0.038 moles of oxygen in a 1.0 L flask. At 200oC, the eqm constant is 4.2 x 10-8. Determine the nitrogen monoxide conc at eqm.

  14. Very small K values When K is very small ignore the impact of x, if the initial conc is at least 100 times larger than the K. Initial conc K > 100 You can only ignore an x that is added or subtracted, any x being multiplied or divided is significant.

  15. Homework Do Pg 465 #1,2 Do Pg 466 #3,4 Do Pg 472 #5,6 Do Pg 476 #7,8 Do Pg 480 #9,10 Prereading Pg 482-486 Each page has a different type – Make sure you try each!!

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