AP Chem Review - 2017 Practice Exam

KEY CONCEPT ANSWER at the 1/2 Eq pt. pH = pKa take the pH value and find the Ka from it by doing 10^-pH

# 1 E KEY CONCEPT: HOW ARE pH AND Ka RELATED? Explain how the student can estimate the value of Ka for HC2H3O2(aq) using the titration curve.

The reaction is a redox reaction because the oxidation numbers of some atoms changed during the reaction (both oxygen and chlorine undergo changes in oxidation number). **DO NOT NEED TO BE SPECIFIC IN WHAT IS BEING OXIDIZED/REDUCED

# 2 A H2O2(aq) + OCl−(aq) → H2O(l) + Cl−(aq) + O2(g) A student investigates the reaction between H2O2(aq) and NaOCl(aq), which is represented by the net-ionic equation shown above. Is the reaction represented above a redox reaction? Justify your answer.

The temperature increases because the reaction is exothermic (ΔH° < 0 ). The exothermic system releases energy when its bonds break thus giving energy to its surroundings. HOWEVER- This increases the thermal energy thus increasing the the temperature of the system as well because all the particles are moving faster (I know it sounds odd)

# 2 C H2O2(aq) + OCl−(aq) → H2O(l) + Cl−(aq) + O2(g) A student investigates the reaction between H2O2(aq) and NaOCl(aq), which is represented by the net-ionic equation shown above. To better understand the reaction, the student looks up thermodynamic data for the reaction. For the reaction represented above, Δ�298° is −197 KJ/������ represented above, the value of and the value of Δ�298° is 144 J/(K·������) Does the temperature inside the flask increase, decrease, or remain the same as the reaction proceeds? Justify your answer.

KEY CONCEPT ANSWERED: DELTA G = -RT lN K -This the in the equation sheet - The value of R is the first one listed in ideal gas law constant **We have to e both sides because we are solving for K - Plug in the given delta G and T values

# 2 D KEY CONCEPT: DELTA HAS MANY EQUATIONS TO FIND IT; WHICH CAN USE USE HERE THAT RELATES IT TO K? A student investigates the reaction between H2O2(aq) and NaOCl(aq), which is represented by the net-ionic equation shown above. To better understand the reaction, the student looks up thermodynamic data for the reaction. For the reaction represented above, Δ�298° is −197 KJ/������ represented above, the value of and the value of Δ�298° is 144 J/(K·������) Calculate the value of the equilibrium constant, K, for the reaction at 298 K

FIRST- find the moles with PV = nRT (after converting to the correct units) THEN - use the given M of 0.800 and the moles you just found to find the mL

# 2 E Calculate the volume of 0.800 M H2O2(aq) that the student should add to excess NaOCl(aq) to produce 40.0 mL of O2(g) at 0.988 atm and 298 K.

KEY CONCEPT ANSWER: percent yield = (actual yield / theoretical yield) * 100 (ii) 36.5 mL/40.0 mL x 100 = 91.3% (iii) No, the gas also contains water vapor and air that was originally in the flask. Look at the buret and flask in the picture. We cannot only have O2 because there is also water in the buret and water and air in the flask.

# 2 F (ii) and (iii) KEY CONCEPT: HOW DO YOU CALCULATE % YIELD? The student decides to produce 40.0mL of O2(g) at a pressure of 0.988 atm and a temperature of 298 K using the reaction represented above. The student uses the equipment shown below. The student sets up a 250 mL Erlenmeyer flask fitted with a one-hole stopper. The flask is connected to a 50 mL gas-collection tube that initially is completely filled with water. The student added the amount of H2O2(aq) calculated in part (e) to excess NaOCl(aq). However, instead of producing 40.0 mL of O2(g), the volume indicated in the diagram below was produced. 36.5 mL was actually produced (ii)Assuming that all the gas in the tube is O2(g), calculate the percent yield of O2(g). (iii) Is the assumption that all the gas in the tube is O2(g) correct? Explain.

We have a total of 4 bonds breaking in 2 O3 molecules so we should put a 4 infront of its variable when we are solving for it O=O=O vs O=O

# 3 C Ozone decomposes according to the reaction represented below 2 O3(g) → 3 O2(g) ΔH° = − 285 kJ/molrxn The bond enthalpy of the oxygen-oxygen bond in O2 is 498 kJ/mol. Based on the enthalpy of the reaction represented above, what is the average bond enthalpy, in kJ/mol, of an oxygen-oxygen bond in O3?

ii) T1/2 = 0.693/K 0.01 = 0.693/K 69.3 1/s ASK!! - where are we getting this info iii) The data are consistent with rate = k 2 [O 3 ][HSO 3 - ] because [HSO 3 - ] remains essentially constant during the experiment. ( [ HSO 3 - ] can be contained in the rate constant k 1 .) ASK!! - How do we know to do this? IV) Repeat the experiment with a different initial concentration of HSO 3 - . (If the change in the rate of the reaction is directly proportional to the change in [ HSO 3 - ], then the reaction is first order with respect to HSO 3 -)

# 3 D (ii) (iii) (iV) The data are consistent with the following rate law: rate = k1[O3] The half like is 0.01 sec (ii)Determine the value of the rate constant, k1, for the rate law. Include units with your answer. (iii) Considering the relative concentrations of the reactants, briefly explain why the data in the graph are also consistent with the following rate law: rate = k2[O3][HSO3−] (IV) Briefly describe an experiment that could provide evidence to support the rate law given in part (d)(iii)

For hydrogen bonding to happen, there needs to be an H inTRAmolecularly bonded to a N.O.F. atom to produce a very polar covalent bond. Even though the picture doesn't show it, another glycerol molecule next to the glycerol they drew could then form hydrogen bonds. Each substance has intermolecular London dispersion forces and dipole-dipole forces. Glycerol also has hydrogen bonding. The presence of hydrogen bonding in glycerol results in the total intermolecular forces in glycerol being stronger than the total intermolecular forces in trichloropropane.

# 4 A The structural formulas of glycerol and trichloropropane are given above. Both compounds are liquids at 25°C. For each compound, identify all types of intermolecular forces present in the liquid. Explain why glycerol has the higher boiling point in terms of the relative strengths of the intermolecular forces.

D The amount of O3(g) will decrease, because Q > Kc. Use the given concentrations to calculate Q (plug the values into the K expression) Since Q > K the reaction has too many products so the system will shift to consume excess produces and form more reactants.

MCQ # 41 3 O2(g) ⇄ 2 O3(g) Kc = 1.8 × 10−56 at 570 K For the system represented above, [O2] and [O3] initially are 0.150 mol/L and 2.5 mol/L respectively. Which of the following best predicts what will occur as the system approaches equilibrium at 570 K? Responses A The amount of O3(g) will increase, because Q < Kc. B The amount of O3(g) will decrease, because Q < Kc. C The amount of O3(g) will increase, because Q > Kc. D The amount of O3(g) will decrease, because Q > Kc.

B 24.7°C We are increasing the volume by multiplying by 10 (100 to 1000), so we have to multiply the temperature by 1/10

MCQ # 40 When 5.0 g of NH4ClO4(s) is added to 100. mL of water in a calorimeter, the temperature of the solution formed decreases by 3.0°C. If 5.0 g of NH4ClO4(s) is added to 1000. mL of water in a calorimeter initially at 25.0°C, the final temperature of the solution will be approximately A 22.0°C B 24.7°C C 25.3°C D 28.0°C

D H+(aq) + NO2−(aq) → HNO2(aq) This is a buffer solution meaning that it keeps the pH the same when a strong base (HCl is added to it) - this is because the H+ from the HCl is reacting with the conjugate base of NO2-

MCQ #49 A student prepares a solution by combining 100 mL of 0.30 M HNO2(aq) and 100 mL of 0.30 M KNO2(aq). Which of the following equations represents the reaction that best helps to explain why adding a few drops of 1.0 M HCl(aq) does not significantly change the pH of the solution? Responses A K+(aq) + Cl−(aq) → KCl(s) B HNO2(aq) → H+(aq) + NO2−(aq) C H+(aq) + OH−(aq) → H2O(l) D H+(aq) + NO2−(aq) → HNO2(aq)

A forces among Br2 molecules are stronger than those among Cl2 molecules

MCQ 3 At 298 K and 1 atm, Br2 is a liquid with a high vapor pressure, and Cl2 is a gas. Those observations provide evidence that under the given conditions, the Responses A forces among Br2 molecules are stronger than those among Cl2 molecules B forces among Cl2 molecules are stronger than the Cl−Cl bond C Br−Br bond is stronger than the Cl−Cl bond D Cl−Cl bond is stronger than the Br−Br bond

1) convert grams to moles 2)Use the stoichometeric ratio to relate Pb to e- 3) multiply this by farraday's constant on the equation sheet 4)use I=q/t to find the sec

# 5 C Overall reaction: Pb(s) + PbO2(s) + 2 H+(aq) + 2 HSO4−(aq) → 2 PbSO4(s) + 2 H2O(l) Cathode half-cell reaction: PbO2(s) + 3 H+(aq) + HSO4−(aq) + 2e- → PbSO4(s) + 2 H2O(l) The equations above represent reactions associated with the operation of a lead storage battery. The first is the overall reaction that occurs as the battery produces an electrical current, and the second is the half-reaction that occurs at the cathode. Calculate the time, in seconds, needed to regenerate 100. g of Pb(s) in the battery by applying a current of 5.00 amp

KEY CONCEPT ANSWER: WHEN YOU INCREASE THE PRODUCTS, IT CONSUMES THE EXCESS PRODUCTS AND FORMS REACTANTS. SINCE THE REACTANT IS A SOLID, THIS IS LESS SOLUBLE AgCl(s) <-> Ag+(aq) + Cl-(aq) An increased [Cl- ] will decrease the solubility of AgCl(s) since the Ksp is a product of the [Ag+] and [Cl-]. (This is an example of the common ion effect.)

# 6 B(ii) KEY CONCEPT: WHAT HAPPENS TO AN EQ SYSTEM WHEN WE INCREASE THE PRODUCTS. HOW DOES THIS AFFECT THE SOLUBILITY Answer the following questions about the solubility of AgCl(s). The value of Ksp for AgCl(s) is 1.8 × 10−10 The concentration of Cl−(aq) in seawater is 0.54 M. Calculate the molar solubility of AgCl(s) in seawater. 1.8 x 10-10 = [Ag+] x (0.54) ⇒ [Ag+] = 3.3 x 10-10 M Explain why AgCl(s) is less soluble in seawater than in distilled water.

KEY CONCEPT ANSWER: FIND THE HIGHEST ENERGY LEVEL (COEFFICIENT) AND THE ELECTRONS THAT CORRESPOND TO THEM Write out the electron configuration and list the highest energy level 7s^2 7p^4

# 7 A KEY CONCEPT: HOW DO YOU FIND THE AMOUNT OF VALENCE ELECTRONS FROM THE ELECTRON CONFIGURATION A new element with atomic number 116 was discovered in 2000. In 2012 it was named livermorium, Lv. Although Lv is radioactive and short-lived, its chemical properties and reactivity should follow periodic trends. Write the electron configuration for the valence electrons of Lv in the ground state.

1 point is earned for dispensing from the buret. - Use the buret to deliver 5.75 mL of 2.000 M HC2H3O2 to the 100 mL volumetric flask. 1 point is earned for diluting the solution to the calibration mark of the volumetric flask. - Then add distilled water from the wash bottle to the flask (adding the last few drops with an eyedropper) until the volume of liquid in the flask is at the calibration mark EXPLANATION - - The 2M is already in the Buret- move it from the Buret to a volumetric flask (better for finding the correct volume) -then DILUTE solution by adding distilled water until reach the 100ml line on the flask -Use the eyedropper to add the last couple of drops to make sure the volume is accurate TAKE AWAYS: -make sure you add the solution into the best device for measuring volume (usally volumetric flask) -- Make sure the flask is the mL required for experiment after its diluted -THEN you dilute it

#1 A (ii) The student is first asked to prepare 100.0 mL of 0.115 M HC2H3O2(aq) using a 2.000 M standard solution. Calculate the volume, in mL, of 2.000 M HC2H3O2(aq) the student needs to prepare 100.0 mL of 0.115 M HC2H3O2(aq). The answer for this is 5.75 mL ***Describe the procedure the student should use to prepare 100.mL of 0.115 M HC2H3O2(aq) using appropriate equipment selected from the list below. Assume that the student uses appropriate safety equipment. 100 mL beaker 100 mL graduated cylinder 100 mL volumetric flask Eye dropper 500 mL wash bottle filled with distilled water 2.000 M HC2H3O2(aq) in a 50 mL buret

FIRST - Find the moles of the 10ml 2M HC2... to get 0.0200 mol THEN - use the 1:1 mole ratio to say this is the moles for NaOH - Find the ml of NaOH added at the Eq piont (this is when they completely react with each other) and use it with the moles you found above to find the M of NaOH needed to IN GENERAL - - Find moles of acid -find mL of the added bases - use the moles of acid (which is really the moles of the base because of the 1:1 ratio) and divide by the ml of the base at Eq to find the M of the base

#1 D In a separate experimental procedure, the student titrates 10.0 mL of the 2.000 M HC2H3O2(aq) with an NaOH(aq) solution of unknown concentration. The student monitors the pH during the titration. The following titration curve was created using the experimental data presented in the table. NaOH = 14 ml in the solution at Eq point Calculate the molar concentration of the NaOH(aq) solution.

KEY CONCEPT ANSWER: Reaction goes to completion and we are left with only products Disagree. The very large value of K implies that the reaction goes essentially to completion, so essentially all of the H2O2 reacts to form O2 (meaning the expected amount would be produced)

#2 G KEY CONCEPT: WHAT DOES IT MEAN IF WE HAVE A LARGE K VALUE? To account for the percent yield being less than 100 percent, the student claims that the reaction reached equilibrium before the expected amount of O2(g) was produced. Considering your answer to part (d) above (SEE K VALUE BELOW), do you agree or disagree with the student's claim? Justify your answer. K = 3 * 10^34

*** READ CAREFULLY do not forget to add the resonance structures

#3 A The O3 molecule has a central oxygen atom bonded to two outer oxygen atoms that are not bonded to one another. In the box below, draw the Lewis electron-dot diagram of the O3 molecule. Include all valid resonance structures.

Because the kJ are given as kJ/mol, and you can convert moles to grams. See image 1) find q with q=mcAT 2)your answer is in kJ/mol --convert the kJ/mol to moles with the Delta H as the conversion factor --convert moles to grams

#4 B Glycerol (molar mass 92.09 g/mol) has been suggested for use as an alternative fuel. The enthalpy of combustion, ΔH°comb , of glycerol is −1654 kJ/mol. What mass of glycerol would need to be combusted to heat 500.0 g of water from 20.0°C to 100.0°C ? (The specific heat capacity of water is 4.184 J/(g⋅°C). Assume that all the heat released by the combustion reaction is absorbed by the water.)

KEY CONCEPT ANSWER: YOU COULD SEE WHEN IT IS BONDED TO ELSEWHERE IN THE EQUATION BUT... SOMETIMES THIS ELEMENT (like Pb in this example) CHANGES OXIDATION NUMBERS ITSELF WITHOUT US KNOWING ABOUT IT. BECAUSE OF THIS< WE SHOULD KNOW THE CHARGE FOR THE POLYATOMIC IONS (for this equation we have to look at SO4 which has a 2- charge- we can use this charge to figure out an oxidation number of +6 for S without even knowing anything about Pb) +6 The Oxidation number for S does not change on either side of the equation.

#5 A KEY CONCEPT: FINDING OXIDATION NUMBERS - WHEN THERE IS AN ELEMENT INCLUDED THAT WE DO NOT KNOW THE CHARGE OF HOW DO WE FIGURE OUT THE OXIDATION NUMBERS OF THE ELEMENTS BONDED TO IT (SEE PbSO4)? Overall reaction: Pb(s) + PbO2(s) + 2 H+(aq) + 2 HSO4−(aq) → 2 PbSO4(s) + 2 H2O(l) Cathode half-cell reaction: PbO2(s) + 3 H+(aq) + HSO4−(aq) + 2e- → PbSO4(s) + 2 H2O(l) The equations above represent reactions associated with the operation of a lead storage battery. The first is the overall reaction that occurs as the battery produces an electrical current, and the second is the half-reaction that occurs at the cathode. Determine the oxidation number of sulfur in the overall reaction.

Pb(s) + HSO4- (aq) → PbSO4(s) + H+(aq) + 2e- A battery is made up of an anode (negative end that releases electrons) and cathode (positive end that is reduced). The chemical reactions happening at the anode and cathode add up to the overall battery reaction. In this problem, you were given the overall reaction and the cathode reaction. Subtract the cathode reaction from the overall reaction to get the anode reaction.

#5 B Overall reaction: Pb(s) + PbO2(s) + 2 H+(aq) + 2 HSO4−(aq) → 2 PbSO4(s) + 2 H2O(l) Cathode half-cell reaction: PbO2(s) + 3 H+(aq) + HSO4−(aq) + 2e- → PbSO4(s) + 2 H2O(l) The equations above represent reactions associated with the operation of a lead storage battery. The first is the overall reaction that occurs as the battery produces an electrical current, and the second is the half-reaction that occurs at the cathode. Write the equation for the half-reaction that occurs at the anode as the battery operates. After the battery has operated for some time, it can be recharged by applying a current to reverse the overall reaction.

SEE ANSWER ON THE PICTURE KEY CONCEPTS ANSWER: ( [H3O+] / initial concentration of acid ) * 100 i) - We have to first use the pH to find the H30+ by doing 10^-pH ---This ends up being 0.00120M - Then we plug this value and the initial value given to use for the HC2... which is 0.115 into a RICE table (because this is Eq equation) --- In the Eq section of the RICE table the M of C2...- is the same as the M of H3O+ we found ---We must subtract this value by 0.115 for the Eq concentration by 0.00120 ---Plug these values into the Ka expression to get 1.3 * 10^-5 ii) Plug known values into the equation above to get 1.0%

KEY CONCEPTS: HOW DO YOU FIND % DISSOCIATION (IONIZATION)? # 1 B (i) and (ii) HC2H3O2(aq) + H2O(l) ⇄ H3O+(aq) + C2H3O2−(aq) i. Using a pH probe, the student determines that the pH of 0.115 M HC2H3O2(aq) is 2.92. ii. Using the pH value, calculate the value of Ka for HC2H3O2(aq) Calculate the percent dissociation of ethanoic acid in 0.115 M HC2H3O2(aq)

B 300 kJ/mol

MCQ # 37 Two molecules of the amino acid glycine join through the formation of a peptide bond, as shown above. The thermodynamic data for the reaction are listed in the following table. Based on the bond energies listed in the table above, which of the following is closest to the bond energy of the C−N bond? A 200 kJ/mol B 300 kJ/mol C 400 kJ/mol D 500 kJ/mol

D The sample contains NaCl(s) and LiCl(s) Li has a MM of 7g/mol while Na has a MM of 22.98g/mol and since we have more Cl compared to what else is in the sample (75% vs. 61%) we need to pick the compound that has more Cl in it compared to what it is bonded with 7g/mol of Li vs 35.45g/mol of Cl

MCQ # 16 A sample of a solid labeled as NaCl may be impure. A student analyzes the sample and determines that it contains 75 percent chlorine by mass. Pure NaCl(s) contains 61 percent chlorine by mass. Which of the following statements is consistent with the data? Responses A The sample contains only NaCl(s). B The sample contains NaCl(s) and NaI(s). C The sample contains NaCl(s) and KCl(s). D The sample contains NaCl(s) and LiCl(s).

A SO3 1) find moles of each (treat the % as grams because it says "percent by mass") 40g/1 = 1mol/32.07g = 1.247mol S 60g/1 = 1mol/15.999g= 3.752mol O 2) divide the smaller by the larger 3,752/1.247 = 3 (this means we have 3 O for every 1 S) 3) make the empirical formula: SO3

MCQ # 17 if a pure sample of an oxide of sulfur contains 40. percent sulfur and 60. percent oxygen by mass, then the empirical formula of the oxide is Responses A SO3 B SO4 C S2O6 D S2O8

D Compound 2, because it forms hydrogen bonds, whereas compound 1 does not H-N is present in compound 2 and not compound 1

MCQ # 20 Based on the structures shown above, which of the following statements identifies the compound with the higher boiling point and provides the best explanation for the higher boiling point? Responses A Compound 1, because it has stronger dipole-dipole forces than compound 2 B Compound 1, because it forms hydrogen bonds, whereas compound 2 does not C Compound 2, because it is less polarizable and has weaker London dispersion forces than compound 1 D Compound 2, because it forms hydrogen bonds, whereas compound 1 does not

C NH3 Hydrogen Bonding!! (it has the most similar IMF to that of water (which also has Hydrogen bonding) which means water is more likely to be attracted to NH3 instead of being attracted to itself more)

MCQ # 21 On the basis of molecular structure and bond polarity, which of the following compounds is most likely to have the greatest solubility in water? Responses A CH4 B CCl4 C NH3 D PH3

B 2.0 atm ***Equilibrium problem - CANNOT USE STOCHIOMETRY UNLESS YOU HAVE A REALLY LARGE K VALUE K = [N2O4]^2/[NO2] 3= x^2/ 1.33 x=2

MCQ # 22 A mixture of NO2(g) and N2O4(g) is placed in a glass tube and allowed to reach equilibrium at 70°C, as represented above. If ��2�4 is 1.33 atm when the system is at equilibrium at 70°C, what is ���2? Responses A 0.44 atm B 2.0 atm C 2.3 atm D 4.0 atm

ANSWER TO KEY CONCEPT: The things (other atoms) in between the layers of an atom in a interstitial alloy make it harder for the layers to slide past each other (thus making it harder to bend) D The additional carbon atoms within the alloy make it more difficult for the iron atoms to slide past one another. They are asking why is this harder to bend (when layers can shift past each other without breaking) - because there are things (carbons) in between these layers, it is harder for the layers to slide past each other

MCQ # 27 KEY CONCEPT: HOW ARE ALLOYS DIFFERENT FROM REGULAR PURE METALS The table above provides some information about two types of steel, both of which are alloys of iron and carbon. Which of the following best helps to explain why high-carbon steel is more rigid than low-carbon steel? Responses A Elemental carbon is harder than elemental iron. B The additional carbon atoms within the alloy make the high-carbon steel less dense. C The additional carbon atoms within the alloy increase the thermal conductivity of the high-carbon steel. D The additional carbon atoms within the alloy make it more difficult for the iron atoms to slide past one another.

A KEY CONCEPTS ANSWER: DELTA H IS ENTHALPY WHICH IS THE ENERGY GAINED/RELEASED IN A REACTION + WHEN BREAKING BONDS and - WHEN FORMING BONDS DELTA S IS ENTROPY WHICH IS THE MEASURE OF DISPERSION OF MATTER/ENERGY + WHEN ATTRACTIONS BREAK and - WHEN ATTRACTIONS ARE FORMING (just like enthalpy)

MCQ # 34 KEY CONCEPTS WHAT IS DELTA H VS DELTA S? HOW DO WE KNOW IF THEY ARE -/+ The process of dissolution of NaCl(s) in H2O(l) is represented in the diagram above. Which of the following summarizes the signs of ∆H° and ∆S° for each part of the dissolution process?

C We will have what is in the container originally and then (with the proper stoichiometric ratios) - at the 1/2 equ point, the reaction is 1/2 way done meaning we will have what was originally in the container and what is on the products side of the reaction (water and H+ are not in the drawing as mentioned in the reaction)

MCQ # 35 A particle view of a sample of H2O2(aq) is shown above. The H2O2(aq) is titrated with KMnO4(aq), as represented by the equation below. 2 MnO4−(aq) + 5 H2O2(aq) + 6 H+(aq) → 2 Mn2+(aq) + 5 O2(g) + 8 H2O(l) Which of the following particle views best represents the mixture when the titration is halfway to the equivalence point? (H2O molecules and H+ ions are not shown.)

D The value of ΔH° for the overall reaction is large and negative Add the Delta H values from the steps to find the overall delta H For this reaction, Delta H is possative meaning the forward reaction is endothermic (when we are increaing the temperature in this reaction we are moving in the endothermic direction thus producing more products - more products = larger K value)

MCQ # 42 H2(g) + Cl2(g) ⇄ 2 HCl(g) Kp = 2 × 1030 at 298 K HCl(g) can be synthesized from H2(g) and Cl2(g) as represented above. A student studying the kinetics of the reaction proposes the following mechanism Step 1: Cl2(g) → 2 Cl(g) (slow) ∆H° = 242 kJ/molrxn Step 2: H2(g) + Cl(g) → HCl(g) + H(g) (fast) ∆H° = 4 kJ/molrxn Step 3: H(g) + Cl(g) → HCl(g) (fast) ∆H° = -432 kJ/molrxn Which of the following statements identifies the greatest single reason that the value of Kp for the overall reaction at 298 K has such a large magnitude? A The activation energy for step 1 of the mechanism is large and positive. B The activation energy for step 2 of the mechanism is small and positive. C The value of ΔS° for the overall reaction is small and positive. D The value of ΔH° for the overall reaction is large and negative.

C CBr4 Periodic properties usually means look for elements in the same column. C is surrounded by 4 Cl. However, Br and Cl have similar periodic properties, because they both have 7 valence electrons. So CBr4 is best answer.

MCQ # 44 The compound CCl4 is nonflammable and was once commonly used in fire extinguishers. On the basis of the periodic properties, which of the following compounds can most likely be used as a fire-resistant chemical? Responses A BCl3 B CH4 C CBr4 D PbCl2

D Since K is less than one, this means the reaction favors the reverse direction which means the con acid and can base dissociate more than the acid and base in the reactants (thus making them stronger)

MCQ # 46 ClO2-(aq) + HCOOH(aq) ⇄ HClO2 (aq) + HCOO (aq) Keq < 1 What are the relative strengths of the acids and bases in the reaction represented by the equation above?

D The reaction has a large activation energy at 25°C. We know it has a large activation energy because it needed extra energy (which is in the form of the spark) to overcome the activation energy and form products.

MCQ # 48 2 H2(g) + O2(g) → 2 H2O(g) When H2(g) and O2(g) are mixed together in a rigid reaction vessel at 25°C, no reaction occurs. When the mixture is sparked, however, the gases react vigorously according to the equation above, releasing heat. Which of the following statements correctly explains why the spark is needed for the reaction to occur when the gases are originally at 25°C? Responses A The reaction is not thermodynamically favorable at 25°C. B ΔH° for the reaction has a large positive value at 25°C. C ΔS° for the reaction has a large negative value at 25°C. D The reaction has a large activation energy at 25°C.

D Draw the Lewis dot structures separately and then combine them

MCQ #30 NH3 reacts with BF3 to form a single species. Which of the following structural diagrams is the most likely representation of the product of the reaction?

Combine the e from the given equations (yes, the number of e should be the same so you could multiply they equations to get the same number of e BUT this does NOT affect the E value) E = E of the Cathode - E of Anode - then use an equation on the sheet to relate E to delta G

MCQ #31 Based on the information in the table above, which of the following shows the cell potential and the Gibbs free energy change for the overall reaction that occurs in a standard galvanic cell?

KEY CONCEPTS ANSWERED: WHEN BOTH ENTHALPY AND ENTROPY ARE UNFAVORABLE (-S and +H), FREE ENERGY(DELTA G) WILL BE THERMODYNAMICALY UNFAVORABLE (-G) THIS IS ACCORDING TO THE FOLLOWING EQUATION: Delta G = Delta H - T*Delta S T = temperature D It is not favored at any temperature.

MCQ #36 KEY CONCEPTS: HOW CAN WE USE ENTHALPY AND ENTROPY TO DETERMINE IF A REACTION IF THERMODYNAMICALLY FAVORED? Two molecules of the amino acid glycine join through the formation of a peptide bond, as shown above. The thermodynamic data for the reaction are listed in the following table. Under which of the following temperature conditions is the reaction thermodynamically favored? A It is only favored at high temperatures. B It is only favored at low temperatures. C It is favored at all temperatures. D It is not favored at any temperature.

B 0 < Keq < 1 As we determined in #36, this reaction is thermodynamiclly unfavorable, meaning it does not completely dissociate (meaning lower Keq)

MCQ #38 Two molecules of the amino acid glycine join through the formation of a peptide bond, as shown above. The thermodynamic data for the reaction are listed in the following table. Based on the thermodynamic data, which of the following is true at 298 K? Responses A Keq = 0 B 0 < Keq < 1 C Keq = 1 D Keq > 1

D Z < X < Y Compare K values (larger = can dissociate more (and therefore produce more Ag) smaller = cannot dissociate as much)

MCQ #39 Three saturated solutions (X, Y, and Z) are prepared at 25°C. Based on the information in the table above, which of the following lists the solutions in order of increasing [Ag+] ? A X < Z < Y B Y < X < Z C Z < Y < X D Z < X < Y

A −93 kJ Find the enthalpy like normal (add up the enthalpies from the provided reactions) then since this overall equation produces 2HCl, we need to divide the overall enthalpy we found by 2 to find the enthalpy per mole of 1 HCl

MCQ #43 H2(g) + Cl2(g) ⇄ 2 HCl(g) Kp = 2 × 1030 at 298 K HCl(g) can be synthesized from H2(g) and Cl2(g) as represented above. A student studying the kinetics of the reaction proposes the following mechanism Step 1: Cl2(g) → 2 Cl(g) (slow) ∆H° = 242 kJ/molrxn Step 2: H2(g) + Cl(g) → HCl(g) + H(g) (fast) ∆H° = 4 kJ/molrxn Step 3: H(g) + Cl(g) → HCl(g) (fast) ∆H° = -432 kJ/molrxn What is the value of the enthalpy change per mole of HCl(g) produced? A −93 kJ B −121 kJ C −186 kJ D −242 kJ

C Ag(NH3)2+(aq), because Keq3 = 1.6 x 107 When you add 2 reactions together to get a new reaction (Adding reactions 1 and 2 to get reaction 3), you multiply their K values together. K1 x K2 = K3

MCQ #47 Equal volumes of 0.1 M AgNO3(aq) and 2.0 M NH3(aq) are mixed and the reactions represented above occur. Which Ag species will have the highest concentration in the equilibrium system shown below, and why? Ag+(aq) + 2 NH3(aq) ⇌ Ag(NH3)2+(aq)���3=? A Ag+(aq), because Keq3 = 4 B Ag+(aq), because Keq1 < Keq2 C Ag(NH3)2+(aq), because Keq3 = 1.6 x 107 D Ag(NH3)2+(aq), because Keq1 < Keq2

BE CAREFUL - SOMETIMES THE UNITS IN THE QUESTION ARE NOT THE SAME AS THE UNITS IN THE ANSWER! C 4 × 10^−6 mol The reason this answer is in ^-6 is because the answer is in mols whereas the values in the graph are in M Take the 4 x 10^-5 M and multiply by the .01 L (given as 100 mL)

MCQ 5 Fe3+(aq) + KSCN(s) → FeSCN2+(aq) + K+(aq) To determine the moles of Fe3+(aq) in a 100. mL sample of an unknown solution, excess KSCN(s) is added to convert all the Fe3+(aq) into the dark red species FeSCN2+(aq), as represented by the equation above. The absorbance of FeSCN2+(aq) at different concentrations is shown in the graph below. If the absorbance of the mixture is 0.20 at 453 nm, how many moles of Fe3+(aq) were present in the 100. mL sample? (Assume that any volume change due to adding the KSCN(s) is negligible.) Responses A 4 × 10−4 mol B 3 × 10−4 mol C 4 × 10−6 mol D 3 × 10−6 mol

B. Al(g) + Mg+(g) → Al+(g) + Mg(g) Ms. Brendzel said if you see one like this on the exam, then you should just guess and move on because this type of problem is long and complicated compared to other points you could get from other problems if you skip it.

MCQ 6 The first ionization energy of an element is the energy required to remove an electron from a gaseous atom of the element (i.e., X(g) → X+(g) + e−). The values of the first ionization energies for the third-row elements are shown in the graph above. On the basis of the information given, which of the following reactions is exothermic? Responses A Cl(g) + Mg+(g) → Cl+(g) + Mg(g) B Al(g) + Mg+(g) → Al+(g) + Mg(g) C P(g) + Mg+(g) → P+(g) + Mg(g) D S(g) + Mg+(g) → S+(g) + Mg(g)