Knowing that [reactant]Ó = [reactant]initial - [reactant]onsumedAnd thatFe"(aq) + SCN'(aq) Ó Fe(SCN)+*(aq)Complete the table below and determine an average value for the equilibriumconstant K.Exp#1exp#2exp#3[Fe*1° (M)1.65 x 101.65 x 1031.65 x 103[SCN]° (M)3.30 x 106.60 x 109.90 x 104[Fe(SCN)*jÓ (M)9.00 x 1051.78 x 102.67 x 10[Fe*jóISCNJÓKAverage K =Fe"(aq) +SCN (aq) ó Fe(SCN)**(aq)I1.65 x 10-9.00 x 10°3.30 x 10-9.00 x 105C0.00+9.00 x 109.00 x 10$

Answered: Image /qna-images/answer/95e1afc2-440d-49f2-bc48-36b77e6cce22.jpg

consumed Knowing that [reactant]Ó = [reactant]nitial- [reactant] And that Fe"(aq) + SCN(aq) Ó Fe(SCN)"(aq) Complete the table below and determine an average value for the equilibrium constant K. Еxp#1 exp#2 exp#3 [Fe*1° (M) 1.65 x 103 1.65 x 10 1.65 x 10 [SCN] (M) 3.30 x 104 6.60 х 104 9.90 x 104 [Fe(SCN)*JÓ (M) 9.00 x 105 1.78 x 10 2.67 x 104 [Fe"jÓ ISCNJÓ ........ ..... Average K = ...... Fe" (aq) + SCN (aq) ó Fe(SCN)**(aq) I 1.65 x 10 -9.00 x 10 3.30 x 104 C 0.00 -9.00 x 10 +9.00 x 10 9.00 x 10

consumed Knowing that [reactant]Ó = [reactant]nitial- [reactant] And that Fe"(aq) + SCN(aq) Ó Fe(SCN)"(aq) Complete the table below and determine an average value for the equilibrium constant K. Еxp#1 exp#2 exp#3 [Fe1° (M) 1.65 x 103 1.65 x 10 1.65 x 10 [SCN] (M) 3.30 x 104 6.60 х 104 9.90 x 104 [Fe(SCN)JÓ (M) 9.00 x 105 1.78 x 10 2.67 x 104 [Fe"jÓ ISCNJÓ ........ ..... Average K = ...... Fe" (aq) + SCN (aq) ó Fe(SCN)**(aq) I 1.65 x 10 -9.00 x 10 3.30 x 104 C 0.00 -9.00 x 10 +9.00 x 10 9.00 x 10

Chemistry: Principles and Practice

3rd Edition

ISBN:9780534420123

Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Publisher:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer

Chapter14: Chemical Equilibrium

Section: Chapter Questions

Problem 14.40QE

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Similar questions

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A common type of reaction we will study is that having a very small K value (K 1). Solving for equilibrium concentrations in an equilibrium problem usually requires many mathematical operations to be perfomed. However, the math involved when solving equilibrium problems for reactions having small K values (K 1) is simplified. What assumption is made when solving the equilibrium concentrations for reactions with small K values? Whenever assumptions are made, they must be checked for validity. In general, the 5% rule is used to check the validity of assuming x (or 2 x, 3x, and so on) is very small compared to some number. When x (or 2 x. 3x. and so on) is less than 5% of the number the assumption was made against, then the assumption is said to be valid. If the 5% rule fails, what do you do to solve for the equilibrium concentrations?
The diagram represents an equilibrium mixture for the reaction N2(g) + O2(g) ⇌ 2 NO(g) Estimate the equilibrium constant.
Again the experiment in Exercise 12.33 was redesigned. This time, 0.15 mol each of N, and O2 was injected into a 5.0-L container at 2500 K, at which the equilibrium constant is 3.6 X 10-?. What was the composition of the reaction mixture at equilibrium? l'he following reaction establishes equilibrium at 2000 K: N2(g) + O2(g) *2 2 NO K = 4.1 X IO-4 If the reaction began with 0.100 mol L-1 of N2 and 0.100 mol L-’ ofO2, what were the equilibrium concentrations of all species?
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Because carbonic acid undergoes a second ionization, the student in Exercise 12.39 is concerned that the hydrogen ion concentration she calculated is not correct. She looks up the equilibrium constant for the reaction HCO,-(aq) «=* H+(aq) + COf'(aq) Upon finding that the equilibrium constant for this reaction is 4.8 X 10“H, she decides that her answer in Exercise 12.39 is correct. Explain her reasoning. A student is simulating the carbonic acid—hydrogen carbonate equilibrium in a lake: H,CO,(aq) 5=6 H+(aq) + HCO,'(aq) K = 4.4 X 10'7She starts with 0.1000 A1 carbonic acid. W hat are the concentrations of all species at equilibrium?
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