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Part 1: Perfect Squares Method 1.  Students will: 1) Determine the equilibrium concentrations of a chemical equilibrium reaction given the initial concentrations.

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Presentation on theme: "Part 1: Perfect Squares Method 1.  Students will: 1) Determine the equilibrium concentrations of a chemical equilibrium reaction given the initial concentrations."— Presentation transcript:

1 Part 1: Perfect Squares Method 1

2  Students will: 1) Determine the equilibrium concentrations of a chemical equilibrium reaction given the initial concentrations 2

3  Students will: 1) Apply a problem solving methodology 2) Know if they need to determine the reaction quotient (Q) to solve the question 3) Apply appropriate algebraic skills to solve the problem 3

4 1)Write the equation and state the K value 2)Determine reaction quotient, Q (if required) 3)Set up an ICE table a) enter initial concentrations b) determine changes in concentration 4)Write K equation 5)Solve for K by entering initial concentrations 6)Use “perfect squares method” to solve for x 7)Find equilibrium concentrations 8)Check answer by plugging calculated equilibrium concentrations into K equation (values should match) 4

5  Carbon monoxide reacts with water vapour to produce carbon dioxide and hydrogen. At 900℃, K is 4.200. calculate the concentrations of all entities at equilibrium if 4.000 mol of each entity are initially placed in a 1.000-L closed container. 5

6  CO (g) + H 2 O (g) ⇔ CO 2(g) + H 2(g) K = 4.200 6

7 [CO (g) ]=[H 2 O (g) ]=[CO 2(g) ]=[H 2(g) ] = 4.000mol/L Q = [CO 2(g) ][H 2(g) ] = (4.000)(4.000) = 1.000 [CO (g) ][H 2 O (g) ] (4.000)(4.000) Q < K ∴ the reaction must move forward to reach equilibrium. 7

8 a) enter initial concentrations b) determine changes in concentration  Since this reaction must proceed forward to reach equilibrium, the concentrations of CO (g) and H 2 O (g) must decrease 8 mol/L CO (g) +H 2 O (g) ⇔CO 2(g) +H 2(g) I4.000 C- x + x E4.000 - x 4.000 + x

9 K = [CO 2(g) ][H 2(g) ] = 4.200 [CO (g) ][H 2 O (g) ] 9

10 K = [CO 2(g) ][H 2(g) ] = 4.200 [CO (g) ][H 2 O (g) ] (4.000+x)(4.000+x) = 4.200 (4.000-x)(4.000-x) 10

11 11

12 @ equilibrium: [CO (g) ]= 4.000 – x = 4.000 – 1.377 = 2.623 mol/L [CO (g) ]=[H 2 O (g) ]= 2.623mol/L CO 2(g) ]= 4.000 + x = 4.000 + 1.377 = 5.377 mol/L CO 2(g) ]=[H 2(g) ] = 5.367mol/L 12

13  plug calculated equilibrium concentrations into K equation (values should match) K = [CO 2(g) ][H 2(g) ] = 4.200 [CO (g) ][H 2 O (g) ] K = [5.377][5.377] [2.623][2.623] K = 4.200 13


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