AP Chemistry Midterm Review

Gases

100

Nitrogen gas was collected over water at 25ºC. If the vapor pressure of water at 25ºC is 23 mmHg, and the total pressure in the container is measured at 781 mmHg, what is the partial pressure of the nitrogen gas? A. 23 mmHg B. 46 mmHg C. 551 mmHg D. 735 mmHg E. 758 mmHg

E. The pressure for any gas collected over water is caused by the gas and the water vapor pressure. A simple subtraction allows the calculation of the gas itself. 781 − 23 = 758.

200

A gas sample is confined in a 5-liter container. Which of the following will occur if the temperature of the container is increased? I. The kinetic energy of the gas will increase. II. The pressure of the gas will increase III. The density of the gas will increase. A. I only B. II only C. I and II only D. I and III only E. I, II, and III

C. When you increase the temperature of a sealed rigid container, the mass nor the volume will change, thus the density will not change. Kinetic energy will increase, which in turn increased the pressure.

300

Chlorine gas and fluorine gas will combine to form one gaseous product. One L of Cl2 reacts with 3 L of F2 to produce 2 L of product. assuming constant temperature and pressure condition, what is the formula of the product? A. ClF B. Cl2F2 C. ClF2 D. Cl2F E. ClF3

E. Since the reaction is at constant temp and pressure, you can write an equation using the volumes as moles. Knowing the stoichiometry of the reactants and products will tell you what the formula must be. Cl2 + 3F2 → 2 Cl?F? thus ClF3

400

A mixture of gases contains 1.5 moles of oxygen, 3.0 moles of nitrogen, and 0.5 mole of water vapor. If the total pressure is 700 mmHg, what is the partial pressure of the nitrogen gas? A. 70 mmHg B. 210 mmHg C. 280 mmHg D. 350 mmHg E. 420 mmHg

E. The mole fraction (symbolized X) of a gas causes that same fraction of the total pressure. XP(total) = P, solve for the mole fraction of nitrogen: (3/5)x700 It is easiest to divide 700 by 5 to get 140, then multiply by 3 to get 420

500

Consider the combustion of 6.0 g of ethane, C2H6. What volume of carbon dioxide will be formed at STP? A. 0.20 L B. 0.40 L C. 2.2 L D. 9.0 L E. 22.4 L

D. This is a stoichiometry problem that requires a balanced equation. C2H6 + (7/2)O2 → 2CO2 + 3H2O Convert the ethane to moles, 6g x (1mol/ 30g), but keep it as a fraction,(1/5). Thus (1/5)C2H6× 2(CO2)/C2H6× 22.4 = While this math may not look so smooth, consider the 22.4 to be ~20, to get an answer of ~8 and survey the answers to see that a value of 9 L compared to the rest of the choices, is close enough to select and move on.

Thermodynamics

100

Lead(II) oxide is reduced with carbon to produce metallic lead, with the thermochemical equation: PbO + C -> PB + CO Delta H = +106.8 kj/mol This is a __________ reaction, which ______heat. A. formation;absorbs B. An exothermic, absorbs C. An exothermic, releases D. an endothermic, absorbs E. an endothermic, releases

D. an endothermic, absorbs The sign for delta H is positive, which signifies that energy is being gained by the system from the surroundings, which means it absorbs energy. This is called an endothermic reaction.

200

Given the equation B2H6(g) + 302(g) -> B2O3(s) + 3H2O(g) delta H = -1941kj What is the enthalpy change for the reaction of boron oxide with water vapor to produce diborane gas and oxygen gas? A. -3822 kj B. -1941 kj C. 970.5 kj D. 1941 kj E. 3882 kj

D. 1941 kj. The enthalpy change for the reverse reaction is the same magnitude, but opposite in sign.

300

For a reaction, delta S is positive and delta H is positive. In what temperature range is the reaction spontaneous? A. The reaction is always spontaneous. B. The reaction is never spontaneous. C. The reaction is spontaneous at T>deltaH/deltaS D. The reaction is spontaneous at T < deltaH/deltaS

C. The reaction is spontaneous at T>deltaH/deltaS. Knowing DeltaG = deltaH - TdeltaS; if deltaG>0 it is spontaneous. If delta H is positive and nonspontaneous, and delta S is positive but spontaneous, the reaction will be spontaneous at temperatures larger than deltaH/deltaS.

400

When .400 g of CH4 is burned in excess oxygen in a bomb calorimeter that has a heat capacity of 3245 J/C, a temperature of 6.795 C is observed. What is the value of q? A. 220 kJ/mol B. -882 kJ C. 477 kJ D.-22.05 kJ E. 8.820 kJ/g

D. q=-(heat capacity)(delta T)= -(3245 J/C)(6.795 C)= -22.05 kJ

500

Water boils at 100 C with a molar heat of vaporization of +43.9 kJ. At 100 C what is the entropy change when water condenses? H2O(g)---->H2O(l) A. Problem cannot be solved; delta G must also be shown. B. Problem cannot be solved; this is not a chemical reaction. C. -439 J/K D. 0.0439 J/K E. -118 J/K

E. During a phase change delta G=0 and T(delta S)= delta H. So, delta S = delta H/T = 43900 J/373 K = 118 J/K

Stoichiometry

100

Balance the following equation: ___NH3 + ___O2 → ___NO2 + ___H2O The balanced equation shows that 1.00 mole of NH3 requires ___ mole(s) of O2 a. 0.57 b. 1.25 c. 1.33 d. 1.75 e. 3.5

D It might be easiest to balance the equation with mostly whole numbers: 2 NH3 + ⁷⁄₂O2 → 2NO2 + 3H2O. The question asks about the amount of oxygen reacting with ONE mole of ammonia, thus cut the ⁷⁄₂ (3.5) of oxygen in half to 1.75

200

Calculate the mass of hydrogen formed when 27 g of aluminum reacts with excess hydrochloric acid according to the balanced equation below. 2Al + 6HCl → 2 AlCl3 + 3 H2 a. 1.5 g b. 2.0 g c. 3.0 g d. 6.0 g e. 12 g

C In multiple choice questions without a calculator, you must look for the “easy math” − You will be most successful at this if you put all the numbers in the dimensional analysis on the page and look for common factors you can cancel out. 27g Al x (1 mol/27g) x (3 H2/2 Al) x (2g/1 mol)=3g H2

300

The reaction of 7.8 g benzene, C6H6, with excess HNO3 resulted in 0.90 g of H2O. What is the percentage yield? Molar Mass (g/mol): C6H6=78 HNO3=63 C6H5NO2=123 H2O=18 C6H6 + HNO3 → C6H5NO2 + H2O a. 100% b. 90% c. 50% d. 12% e. 2%

C 7.8g C6H6 x (1 mol/78g)=0.1g C6H6 0.1g C6H6 x (1 H2O/1 C6H6) x (18g/1mol)=1.8g H2O (0.9g H2O/1.8g H2O) x 100=50% H2O

400

Write a balanced equation for the combustion of propane, C3H8. When balanced, the equation indicates that ___ moles of O2 are required for each mole of C3H8. a. 1.5 b. 3 c. 3.5 d. 5 e. 8

D Balance: C3H8 + 5O2 → 3CO2 + 4H2O

500

N2(g) + 2O2(g) → N2O4(g) The above reaction takes place in a closed flask. The initial amount of N2(g) is 8 mole, and that of O2(g) is 12 mole. There is no N2O4(g) initially present. The experiment is carried out at constant temperature. What is the total amount of mole of all substances in the container when the amount of N2O4(g) reaches 6 mole? a. 0 mole b. 2 mole c. 6 mole d. 8 mole e. 20 mole

D When the reaction reaches a quantity of 6 mole of product (remember the problem states that there was no product, N2O4 to start with), use the coefficients in the balanced equation to determine that 12 mole of O2 and 6 mole of N2 must have reacted to produce the 6 mole of product. 6 mol N2O4 x (2 mol O2/1 mol N2O4)=12 mol O2 so none left since reaction started with 12 mol O2 6 mol N2O4 x (1 mol N2/1 mol N2O4)=6 mol N2 so 2 mol N2 left Thus 2 mole N2 left over, no O2 left over, and 6 mole of N2O4 produced, will mean a total of 8 mole of substances left in the flask.

Atomic Structure

100

Which of the following is the ground-state electron configuration of an oxide ion? A. 1s2 2s2 2p4 B. 1s2 2s2 2p5 C. 1s2 2s2 2p6 D. 1s2 2s2 2p6 3s1 E. 1s2 2s2 2p6 3s2

C. 1s2 2s2 2p6. The atomic number of oxygen is 8, and the oxide ion is O2-, which has 2 more electrons than the oxygen atom, giving it a total of 10 electron in the ion. Add up the superscripts and that is the correct answer.

200

Which atom has the largest covalent radius? A. Argon B. Arsenic C. Phosphorous D. Selenium E. Sulfur

B. Arsenic. Atomic radius increases for those elements in the Periodic table that are down and to the left . Arsenic will has the largest size.

300

All of the following statements concerning alkali metals are true except A. They are strong oxidizing agents B. They form ions with a +1 oxidation state C. As the atomic numbers of the alkali metals increase, their electonegativity decreases. D. As the atomic numbers of the alkali metals increase, the first ionization energy decreases.

A. They are strong oxidizing agents. The alkali metals are not strong oxidizing agents, they are strong REDUCING agents. They give up their single valence electrons easily, so they are easily oxidized. If they are easily oxidized, they are strong reducing agents.

400

What are the electron pair geometry and the molecular geometry of SF4? A. Trigonal bipyramidal; seesaw B. Trigonal bipyramidal; t shaped C. Trigonal bipyramidal; trigonal bipyramidal D. Octahedral; square planar

A. Trigonal bipyramidal; seesaw SF4 has 4 bonding pairs and 1 nonbonding pair for a total of 5 electron pairs. This gives a trigonal bipyramidal electron structure. According to VSEPR theory, a molecule with 4 atoms bonded to the central atom in a trigonal bipyramidal geometry has a seesaw molecular geometry.

500

Why does phosphorous have a lower electron affinity than silicon? This is explained by an electron being added to phosphorous to: A. a filled orbital B. a new subshell C. an empty orbital D. a half filled orbital E. A new valence shell

D. A half filled orbital. Electron affinity represents the amount of energy liberated as an electron is added to the valence orbital with the lowest energy. In phosphorous, an electron is being added to an orbital that is already occupied by an electron, which is less favorable than adding an electron to an empty orbital.