Dean of Students Office | Iowa State University · Web viewIndicate whether the reactions are...

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Final Exam Review #3 Supplemental Instruction Iowa State University Leader: Allison Course: Chem 163 Instruct or: Dr. Appy Date: 12/14/14 1. Define and state the ΔErxn for each: a. Exothermic: energy-releasing, ΔErxn < 0 b. Endothermic: Energy-absorbing, ΔErxn > 0 2. Use the chart to answer the following questions: Reaction Energy of Reactants (kJ/mol) Energy of Products (kJ/mol) Energy of Transition State (kJ/mol) 1 50 60 70 2 20 10 90 3 10 20 90 a. Indicate whether the reactions are endothermic or exothermic. Reaction 1: Endothermic Reaction 2: Exothermic Reaction 3: Endothermic b. Determine the value of ΔErxn. Reaction 1: 60 kJ/mol – 50 kJ/mol = 10 kJ/mol Reaction 2: 10 kJ/mol – 20 kJ/mol = -10 kJ/mol Reaction 3: 20 kJ/mol – 10 kJ/mol = 10 kJ/mol c. Determine the value of Ea. Reaction 1: 70 kJ/mol – 50 kJ/mol = 20 kJ/mol Reaction 2: 90 kJ/mol – 20 kJ/mol = 70 kJ/mol Reaction 3: 90 kJ/mol – 10 kJ/mol = 80 kJ/mol

Transcript of Dean of Students Office | Iowa State University · Web viewIndicate whether the reactions are...

Page 1: Dean of Students Office | Iowa State University · Web viewIndicate whether the reactions are endothermic or exothermic. Reaction 1: Endothermic Reaction 2: Exothermic Reaction 3:

Final Exam Review #3Supplemental InstructionIowa State University

Leader: AllisonCourse: Chem 163Instructor: Dr. AppyDate: 12/14/14

1. Define and state the Erxn for each:Δa. Exothermic: energy-releasing, Erxn < 0Δ

b. Endothermic: Energy-absorbing, Erxn > 0Δ

2. Use the chart to answer the following questions:

ReactionEnergy of Reactants

(kJ/mol)Energy of Products

(kJ/mol)Energy of Transition

State (kJ/mol)

1 50 60 70

2 20 10 90

3 10 20 90

a. Indicate whether the reactions are endothermic or exothermic.Reaction 1: EndothermicReaction 2: ExothermicReaction 3: Endothermic

b. Determine the value of Erxn.ΔReaction 1: 60 kJ/mol – 50 kJ/mol = 10 kJ/molReaction 2: 10 kJ/mol – 20 kJ/mol = -10 kJ/molReaction 3: 20 kJ/mol – 10 kJ/mol = 10 kJ/mol

c. Determine the value of Ea.Reaction 1: 70 kJ/mol – 50 kJ/mol = 20 kJ/molReaction 2: 90 kJ/mol – 20 kJ/mol = 70 kJ/molReaction 3: 90 kJ/mol – 10 kJ/mol = 80 kJ/mol

3. In the reaction of X + Y --> Z, the energy of the reactants is 60 kJ/mol, and the energy of the products is 90 kJ/mol. The energy of the transition state is 130 kJ/mol.

a. Is the reaction endothermic or exothermic?Endothermic

b. What is the Erxn?90 kJ/mol – 60 kJ/mol = 30 kJ/mol

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c. What is the Ea?130 kJ/mol – 60 kJ/mol = 70 kJ/mol

4. Catalysts speed up reactions by:a. Providing an alternative mechanism for a reactionb. Lowering the Ea for the reactionc. Remaining active throughout the chemical reactiond. All of the above are correct.

5. The slow step in the mechanism for the production of NO2F from NO2 and F22

is 2NO2 + F2 --> NO2 F + F. Write the predicted rate law for the reaction.The predicted rate law for the reaction contains the reactants with their respective coefficients as exponents. The predicted rate law is therefore the reaction rate = k[NO2]^2[F2]

6. Write the equilibrium constant expression for the reaction

H2 (g) + I2 (g) <--> 2 HI (g)

Keq = [HI]^2 [H2]x[I2]

7. Write the Keq for the following reaction and solve for the Keq with the following concentration at equilibrium

a. CH4 + 2O2 -> CO2 +2 H2O 0.234 0.20 0.378 0.209

Keq= [0.378] x [0.234] x [0.20] = 0.35

8. Using Le Chatelier’s principle:

a. Concentrationi. If we increase the concentration of the reactants, which way

will the equilibrium switch? To the right/products

ii. If we increase the concentration of the products, which way will the equilibrium switch? To the left/reactants

iii. Will the Keq change? No

b. Temperaturei. Which way will the equilibrium switch if we increase the

temperature? To the left/reactants

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ii. If we decrease the temperature? To the right/reactants

iii. Will the Keq change? YES

c. Pressurei. If we increase the pressure which way will the equilibrium

switch? Shift to the side with less moles

ii. If we decrease the pressure? Shift to the side with more moles

iii. Will the Keq change? No

9. Describe the difference between electrolytes and nonelectrolytes:10. Answer the following questions:

a. Is the compound H2SO4 molecular or ionic?Molecular

b. How did you decide on your answer to part (a)?The formula has no metals in it, just nonmetals.

c. Based on your answer to part (a), would you call H2SO4 an electrolyte or a nonelectrolyte?

Most molecular compounds are nonelectrolytes

d. An aqueous solution of H2SO4 causes a light bulb to light. What does this tell you about H2SO4?

If the bulb lights up, it means that it is an electrolyte.

11. What is the difference between a strong electrolyte and a weak electrolyte?Strong: completely dissociates into ions upon dissolving in waterWeak: one that only partially dissociates into ions upon dissolving in water

12. What is the difference between a strong acid and a weak acid?Strong: water-soluble compound that dissociates extensively in water to give large numbers of H30+ ionsWeak: water-soluble compound that dissociates only partially in water and produces only a few H3O+ ions.

13. Brønsted-Lowry definition of an acid:Anything that donates a proton

14. Brønsted-Lowry definition of a base:Anything that accepts a proton

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15. In the reaction NH3 + PH3 --> NH2- + PH4+ which reactant is the Brønsted-Lowry acid and which is the base?

Because it accepts a proton, PH3 is the base. Because it donates the proton, NH3 is the acid

16. Write the formula of the salt formed from the combination of H3PO4 and LiOH.

Anion: PO4 3-Cation: Li+Salt: Li3PO4

17. Calculate the [OH-] in a solution in which [H3O+] = 2.3 x 10^-4

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For Hydrogen:

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Useful Equations and Constants:

1 mi = 5280 ft1 in = 2.54 cm1 mL = 1 cm3

d = m/VK = ºC + 273.15 ºF = 32 + (9/5)ºC ºC = (5/9)(ºF – 32)q = mCsΔTavg. atomic mass = m1f1 + m2f2 + …..

E = hc/λc = 3.00 x 108 m/sh = 6.626 x 10-34 J*s1 eV = 1.602 x 10-19 J1 m = 109 nmF = kq1q2/r2

1 mole = 6.022 x 1023 things% yield = (actual yield)/(theoretical yield) x 100%

PV = nRTPiVi/(niTi) = PfVf/(nfTf)MM = mRT/(PV)MiVi = MfVf

R = 0.0821 L*atm/(mol*K)STP = 0 °C and 1 atmK = °C + 273.15ΔTf,b = Kf,b x moles of solute / (Kg solvent)P = F/A

Kw = [H3O+][[OH-] = 10-14

pH = -log[H3O+]