PORTAGE LEARNING EXAM zm zm
Exam Solution zm
Chem 103 Final Portage 2026 A+ GRADE ASSURED CO zm zm zm zm zm zm zm zm
MPLETE SOLUTIONS AND VERIFIED ANSWERS (B84EC) zm zm zm zm zm
QUESTION 1 zm
ln [A] - ln [A]0 = -
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k t 0.693 = k t1/2 An ancient sample of paper was found to contain 19.8 % 14C cont
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ent as compared to a present-
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day sample. The t1/2 for 14C is 5720 yrs. Show the calculation of the decay constant (
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k) and the age of the paper.
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ANSWER
0.693 = k (5720) k = 0.693/5720 = 1.2115 x 10 -4 ln [A] - ln [A]0 = - k t ln 19.8 - ln 100 = -
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(1.2115 x 10-4) t t = -1.6195 / -(1.2115 x 10-4) = 13,368 years
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QUESTION 2 zm
Using the potential energy diagram below, state whether the reaction described by th
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e diagram is endothermic or exothermic and spontaneous or nonspontaneous, being s
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ure to explain your answer.
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ANSWER
Large Eact = nonspontaneous ∆H- = exothermic
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QUESTION 3 zm
Show the calculation of Kc for the following reaction if an initial reaction mixture of 0
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.900 mole of CO and 2.70 mole of H2 in a 9.00 liter container forms an equilibrium m
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ixture containing 0.346 mole of H2O and corresponding amounts of CO, H2, and CH4.
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CO (g) + 3 H2 (g) CH4 (g) + H2O (g)
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ANSWER
At equilibrium H2O = 0.346 mole (as stated) CH4 = 0.346 mole (1 mole of CH4 forms for every mol
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e of H2O that is formed) CO = 0.900 -
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0.346 mole (1 mole of CO reacts for every mole of H2O that is formed) H2 = 2.70 -
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, 3 x 0.346 mole (3 mole of H2 reacts for every mole of H2O that is formed) Change all amounts to
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moles/L before entering in Kc expression: H2O = 0.346 mole / 9.00 L = 0.0384 M CH4 = 0.346 mol
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e / 9.00 L = 0.0384 M CO = 0.554 mole / 9.00 L = 0.0616 M H2 = 1.662 mole / 9.00 L = 0.185 M
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Kc = [CH4] [H2O] = [0.0384] [0.0384] = 3.78 [CO] [H2]3 [0.0616] [0.185]3
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QUESTION 4 zm
Explain the terms substrate and active site in regard to an enzyme.
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ANSWER
The substrate is the substance whose reaction rate is increased by an enzyme. The active site is the
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mgroup of atoms on the surface of an enzyme where the substrate binds to undergo the reaction cat
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alyzed by the enzyme. zm zm zm
QUESTION 5 zm
The reaction below has the indicated equilibrium constant. Is the equilibrium mixture
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made up of predominately reactants, predominately products or significant amounts
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of both products and reactants. Be sure to explain your answer. 2 H2 (g) + S2 (g) 2 H
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2S (g) Kc = 9.39 x 10-5
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ANSWER
The very small Kc indicates that this equilibrium mixture will be composed of mostly reactants.
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QUESTION 6 zm
The equilibrium reaction below has the Kc = 3.93. If the volume of the system at equil
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ibrium is decreased from 6.00 liters to 2.00 liters, how and for what reason will the e
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quilibrium shift? Be sure to calculate the value of the reaction quotient, Q, and use thi
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s to confirm your answer. CO (g) + 3 H2 (g) CH4 (g) + H2O (g)
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ANSWER
When volume decreases from 6.00 to 2.00, the pressure triples and the concentration of all gases (C
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O, H2, CH4, and H2O) triples so: (at equilibrium) Qc =Kc = [CH4] [H2O] = 3.93[CO] [H2]3 (volume 1
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/3 = pressure tripled = conc tripled) Qc = [3 CH4] [3 H2O] = Kc [3 CO] [3 H2]3 9 The reaction mus
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t shift briefly in the direction that decreases the pressure by going toward the side with the lesser
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moles of gas (forward direction : 4 moles of gas yields 2 moles of gas) to come back to equilibrium.
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mThis is in agreement with Qc < Kc: the reaction will proceed to the right (in the direction of the pr
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oducts).
QUESTION 7 zm
The equilibrium reaction below has the Kc = 0.254 at 25oC. If the temperature of the
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system at equilibrium is decreased to 0oC, how and for what reason will the equilibri
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Exam Solution zm
Chem 103 Final Portage 2026 A+ GRADE ASSURED CO zm zm zm zm zm zm zm zm
MPLETE SOLUTIONS AND VERIFIED ANSWERS (B84EC) zm zm zm zm zm
QUESTION 1 zm
ln [A] - ln [A]0 = -
zm zm zm zm zm zm
k t 0.693 = k t1/2 An ancient sample of paper was found to contain 19.8 % 14C cont
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ent as compared to a present-
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day sample. The t1/2 for 14C is 5720 yrs. Show the calculation of the decay constant (
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k) and the age of the paper.
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ANSWER
0.693 = k (5720) k = 0.693/5720 = 1.2115 x 10 -4 ln [A] - ln [A]0 = - k t ln 19.8 - ln 100 = -
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(1.2115 x 10-4) t t = -1.6195 / -(1.2115 x 10-4) = 13,368 years
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QUESTION 2 zm
Using the potential energy diagram below, state whether the reaction described by th
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e diagram is endothermic or exothermic and spontaneous or nonspontaneous, being s
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ure to explain your answer.
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ANSWER
Large Eact = nonspontaneous ∆H- = exothermic
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QUESTION 3 zm
Show the calculation of Kc for the following reaction if an initial reaction mixture of 0
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.900 mole of CO and 2.70 mole of H2 in a 9.00 liter container forms an equilibrium m
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ixture containing 0.346 mole of H2O and corresponding amounts of CO, H2, and CH4.
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CO (g) + 3 H2 (g) CH4 (g) + H2O (g)
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ANSWER
At equilibrium H2O = 0.346 mole (as stated) CH4 = 0.346 mole (1 mole of CH4 forms for every mol
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e of H2O that is formed) CO = 0.900 -
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0.346 mole (1 mole of CO reacts for every mole of H2O that is formed) H2 = 2.70 -
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, 3 x 0.346 mole (3 mole of H2 reacts for every mole of H2O that is formed) Change all amounts to
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moles/L before entering in Kc expression: H2O = 0.346 mole / 9.00 L = 0.0384 M CH4 = 0.346 mol
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e / 9.00 L = 0.0384 M CO = 0.554 mole / 9.00 L = 0.0616 M H2 = 1.662 mole / 9.00 L = 0.185 M
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Kc = [CH4] [H2O] = [0.0384] [0.0384] = 3.78 [CO] [H2]3 [0.0616] [0.185]3
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QUESTION 4 zm
Explain the terms substrate and active site in regard to an enzyme.
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ANSWER
The substrate is the substance whose reaction rate is increased by an enzyme. The active site is the
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mgroup of atoms on the surface of an enzyme where the substrate binds to undergo the reaction cat
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alyzed by the enzyme. zm zm zm
QUESTION 5 zm
The reaction below has the indicated equilibrium constant. Is the equilibrium mixture
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made up of predominately reactants, predominately products or significant amounts
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of both products and reactants. Be sure to explain your answer. 2 H2 (g) + S2 (g) 2 H
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2S (g) Kc = 9.39 x 10-5
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ANSWER
The very small Kc indicates that this equilibrium mixture will be composed of mostly reactants.
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QUESTION 6 zm
The equilibrium reaction below has the Kc = 3.93. If the volume of the system at equil
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ibrium is decreased from 6.00 liters to 2.00 liters, how and for what reason will the e
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quilibrium shift? Be sure to calculate the value of the reaction quotient, Q, and use thi
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s to confirm your answer. CO (g) + 3 H2 (g) CH4 (g) + H2O (g)
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ANSWER
When volume decreases from 6.00 to 2.00, the pressure triples and the concentration of all gases (C
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O, H2, CH4, and H2O) triples so: (at equilibrium) Qc =Kc = [CH4] [H2O] = 3.93[CO] [H2]3 (volume 1
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/3 = pressure tripled = conc tripled) Qc = [3 CH4] [3 H2O] = Kc [3 CO] [3 H2]3 9 The reaction mus
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t shift briefly in the direction that decreases the pressure by going toward the side with the lesser
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moles of gas (forward direction : 4 moles of gas yields 2 moles of gas) to come back to equilibrium.
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mThis is in agreement with Qc < Kc: the reaction will proceed to the right (in the direction of the pr
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oducts).
QUESTION 7 zm
The equilibrium reaction below has the Kc = 0.254 at 25oC. If the temperature of the
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system at equilibrium is decreased to 0oC, how and for what reason will the equilibri
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