AP Chemistry Big 5 Jeopardy Game

Thermochemistry

100

4.) A reaction has both positive ΔS° and ΔH° values. From this information, it can be concluded that the reaction A.) can be spontaneous at any temperature. B.) cannot be spontaneous at any temperature. C.) cannot be spontaneous at high temperatures. D.) can be spontaneous only at low temperatures. E.) can be spontaneous only at high temperatures.

The correct answer is (E). For this problem, it is quite essential to know the equation ΔG = ΔH – TΔS. With a positive value of ΔS, the – TΔS term favors spontaneity. But at low temperatures, its contribution will be small. Hence, the ΔH will predominate and the reaction will be non-spontaneous (ΔG = positive). However, at higher temperatures, the magnitude of the – TΔS term increases and can overwhelm the positive ΔH term. In this case, the – TΔS predominates and the reaction is spontaneous (ΔG = negative) at high temperatures.

200

2Al (s) + 3Cl2 (g) à 2AlCl3 (s) The reaction above is not spontaneous under standard conditions, but becomes spontaneous as the temperature decreases toward absolute zero. Which of the following is true at standard conditions? A.) ΔS and ΔH are both negative. B.) ΔS and ΔH are both positive. C.) ΔS is negative, and ΔH is both positive. D.) ΔS is positive, and ΔH is both negative. E.) ΔS and ΔH are both equal to zero.

The correct answer is (A). For this problem, it is quite essential to know the equation ΔG = ΔH – TΔS. If the reaction is spontaneous only when the temperature is very low, then ΔG is only negative when T is very small. This can only happen when ΔH is negative (which favors spontaneity) and ΔS is negative (which favors nonspontaneity). A very small value for T will eliminate the influence of ΔS.

300

The specific heat of lead metal, Pb, is 0.127 J/(g x °C). How many joules of heat would be required to raise the temperature of a 5.00 g sample from 25°C to 35°C? A.) 2.5 x 10-3 B.) 0.127 C.) 6.4 D.) 16.1 E.) 390

The correct answer is (C). The equation q = (m x c x ΔT) is needed to solve this problem. Substitute the given information to solve for q. q = (5.00 g)(0.127 J/(g x °C))( 35°C - 25°C) = 6.4 J

400

2H2 (g) + O2 (g) à 2H2O (g) Based on the information given in the table below, what is ΔH° for the above reaction? BOND H-H O=O O-H AVERAGE BOND ENERGY (kJ/mol) 500 500 500 A.) -2,000 kJ B.) -1,500 kJ C.) -500 kJ D.) +1,000 kJ E.) +2,000 kJ

The correct answer is (C). The bond energy is the energy that must be put into a bond to break it. First let’s figure out how much energy must be put into the reactants to break their bonds. To break 2 moles of H-H bonds, it takes (2 x 500) kJ = 1,000 kJ To break 1 mole of O=O bonds, it takes 500 kJ. So to break up the reactants, it takes +1,500 kJ. Energy is given off when a bond is formed; that’s the negative of the bond energy. Now let’s see how much energy is given off when 2 moles of H2O are formed. 2 moles of H2O molecules contains 4 moles of O-H bonds, so (4 x -500) kJ = -2,000 kJ are given off. So the value of ΔH° for the reaction is (-2,000 kJ, the energy given off) + (1,500 kJ, the energy put in) = -500 kJ.

500

C (s) + 2H2 (g) à CH4 (g) C (s) + O2 (g) à CO2 (g) H2 (g) + (1/2) O2 (g) à H2O (l) Based on the information given above, what is ΔH° for the following reaction? A.) x + y + z B.) x + y – z C.) z + y – 2x D.) 2z + y – x E.) 2z + y – 2x ΔH° = x ΔH° = y ΔH° = z

The correct answer is (D). This is an application of Hess’s Law. The equations given above the question give the heats of formation of all the reactants and products (remember, the heat of formation of O2, an element in its most stable form, is 0). ΔH° for a reaction = (ΔH° for the products) – (ΔH° for the reactants) First, the products: From 2H2O, we get 2z. From CO2, we get y. So the ΔH° for the products = 2z + y Now the reactants: From CH4, we get x. The heat of formation of O2 is defined to be 0. ΔH° for the reaction = (2z + y) – (x) = 2z + y – x.

Equilibrium

100

An equilibrium mixture of the reactants is placed in a sealed container at 150°C. The amount of the products may be increased by which of the following changes? raising the temperature of the container increasing the volume of the container adding 1 mol of C(s) to the container II only I and II I only II and III III only

B-The addition or removal of some solid, as long as some remains some present, will not change the equilibrium. An increase in volume will cause the equilibrium to shift towards the side with more moles of gas (right). Raising the temperature of an endothermic process will shift the equilibrium to the right. Any shift to the right will increase the amounts of the products.

200

An equal number of moles of each of the reactants are sealed in a container and allowed to come to the equilibrium shown above. At equilibrium which of the following must be true? [CO2] must equal [H2O] [O2] must be less than [C2H4] [CO2] must be greater than [C2H4] II and III I only III only II only I and II

E—Assuming 1 mol of each reactant is used, the equilibrium quantities would be: [C2H4] = 1 – x, [O2] = 1 – 3x, [CO2] = 2x, and [H2] = 2x. Unless the value of x is known, it is not possible to relate the actual concentrations of any reactant to any product.

300

The above materials were sealed in a flask and allowed to come to equilibrium at a certain temperature. A small quantity of O2(g) was added to the flask, and the mixture allowed to return to equilibrium at the same temperature. Which of the following has increased over its original equilibrium value? the quantity of NO2(g) present the quantity of NO(g) present the equilibrium constant, K the rate of the reaction the partial pressure of NO(g)

A—The addition of a product will cause the equilibrium to shift to the left. The amounts of all the reactants will increase, and the amounts of all the products will decrease (the O2 will not go below its earlier equilibrium value since excess was added). The value of K is constant, unless the temperature is changed. The rates of the forward and reverse reactions are equal at equilibrium.

400

A 1.00-L flask is filled with 0.30 mol of CH4 and 0.40 mol of CO2, and allowed to come to equilibrium. At equilibrium, there are 0.20 mol of CO in the flask. What is the value of Kc, the equilibrium constant, for the reaction? 1.2 0.027 0.30 0.060 3.0

B—Using the following table: The presence of 0.20 mol of CO (0.20 M) at equilibrium means that 2x = 0.20 and that x = 0.10. Using this value for x, the bottom line of the table becomes: The equilibrium expression is: K = [CO]2[H2]2/[CH4][CO2]. Entering the equilibrium values into the equilibrium expression gives: K = (0.20)2(0.20)2/(0.20)(0.30)

500

The equilibrium constant, K, for the above equilibrium is 7.2 × 10–2. This value implies which of the following? A solution with equimolar amounts of HC3H5O2(aq) and HCOO–(aq) is neutral. C3H5O2–(aq) is a stronger base than HCOO–(aq). HC3H5O2(aq) is a stronger acid than HCOOH(aq). HCOO–(aq) is a stronger base than C3H5O2–(aq). The value of the equilibrium does not depend on the temperature.

B—The low value of K means that the equilibrium lies to the left. The equilibrium always lies away from the stronger acid and the stronger base.

Kinetics

100

The slow rate of a particular chemical reaction might be attributed to which of the following? a low activation energy a high activation energy the presence of a catalyst the temperature is high the concentration of the reactants are high

B—Slow reactions have high activation energies. High activation energies are often attributed to strong bonds within the reactant molecules. All the other choices give faster rates.

200

The energy difference between the reactants and the transition state is the free energy the heat of reaction the activation energy the kinetic energy the reaction energy

C—This is the definition of the activation energy.

300

A reaction follows the rate law: Rate = k[A]2. Which of the following plots will give a straight line? 1/[A] versus 1/time [A]2 versus time 1/[A] versus time ln[A] versus time [A] versus time

C—The "2" exponent means this is a second-order rate law. Second-order rate laws give a straight-line plot for 1/[A] versus t.

400

The table below gives the initial concentrations and rate for three experiments. The reaction is CO(g) + Cl2(g) → COCl2(g). What is the rate law for this reaction? Rate = k[CO] Rate = k[CO]2[Cl2] Rate = k[Cl2] Rate = k[CO][Cl2]2 Rate = k[CO][Cl2]

E—Beginning with the generic rate law: Rate = k[CO]m[Cl2]n, it is necessary to determine the values of m and n (the orders). Comparing Experiments 2 and 3, the rate doubles when the concentration of CO is doubled. This direct change means the reaction is first order with respect to CO. Comparing Experiments 1 and 3, the rate doubles when the concentration of Cl2 is doubled. Again, this direct change means the reaction is first order. This gives: Rate = k[CO]1[Cl2]1 = k[CO][Cl2].

500

The mechanism below has been proposed for the reaction of CHCl3 with Cl2. Step l: Cl2 2 Cl(g) fast Step 2: Cl(g) + CHCl3(g) → CCl3(g) + HCl(g) slow Step 3: CCl3(g) + Cl(g) → CCl4(g) fast Which of the following rate laws is consistent with this mechanism? Rate = k[Cl2] Rate = k[CHCl3][Cl2] Rate = k[CHCl3] Rate = k[CHCl3]/[Cl2] Rate = k[CHCl3][Cl2]1/2

E—The rate law depends on the slow step of the mechanism. The reactants in the slow step are Cl and CHCl3 (one of each). The rate law is first order with respect to each of these. The Cl is half of the original reactant molecule Cl2. This replaces the [Cl] in the rate law with [Cl2]1/2. Do not make the mistake of using the overall reaction to predict the rate law.

Electrochemistry

100

All of the following may serve as reducing agents, EXCEPT: Mg Cs Fe2+ MnO4– Br–

D—For a substance to serve as a reducing agent, it must be capable of being oxidized. The manganese, in the MnO4–, is already in its highest oxidation state, so it could not be oxidized. All other answers contain a substance that may be oxidized.

200

H+ + NO3– + e– → NO + H2O What is the coefficient for water arising when the above half-reaction is balanced? 3 4 2 1 6

C—The balanced chemical equation is: 4 H+ + NO3– + 3 e– → NO + 2 H2O

300

A sample of silver is to be purified by electrorefining. This will separate the silver from an impurity of gold. The impure silver is made into an electrode. Which of the following is the best way to set up the electrolytic cell? an impure silver cathode and an inert anode an impure silver cathode and a pure gold anode a pure silver cathode with an impure silver anode a pure gold cathode with an impure silver anode an impure silver cathode with a pure silver anode

C—The impure silver must be oxidized so it will go into solution. Oxidation occurs at the anode. Reduction is required to convert the silver ions to pure silver. Reduction occurs at the cathode. The cathode must be pure silver, otherwise it could be contaminated with the cathode material.

400

2 Fe3+ + Zn → Zn2+ + 2 Fe2+ The reaction shown above was used in an electrolytic cell. The voltage measured for the cell was not equal to the calculated E ° for the cell. This discrepancy could be caused by which of the following? The anion in the anode compartment was chloride, instead of nitrate as in the cathode compartment. One or more of the ion concentrations was not 1 M. Both of the solutions were at 25°C instead of 0°C. The solution in the salt bridge was Na2SO4 instead of KNO3. The anode and cathode were different sizes.

B—If the voltage was not equal to E °, then the cell was not standard. Standard cells have 1 M concentrations, and operate at 25°C with a partial pressure of each gas equal to 1 atm. No gases are involved in this reaction, so the cell must be operating at a different temperature or a different concentration (or both).

500

2 M(s) + 3 Zn2+(aq) → 2 M3+(aq) + 3 Zn2+(aq) E° = 0.90 V Zn2+(aq) + 2e– → Zn(s) E° = –0.76 V Using the above information, determine the standard reduction potential for the following reaction: M3+(aq) + 3e–→ M(s) 0.90 V +1.66 V 0.00 V –0.62 V –1.66V

E—The half-reactions giving the overall reaction must be: Thus, –0.76 + ? = 0.90, giving ? = 1.66 V. The half-reaction under consideration is the reverse of the one used in this combination, so the sign of the calculated voltage must be reversed. Do not make the mistake of multiplying the voltages when the half-reactions were multiplied to equalize the electrons.

Acids & Bases

100

Which of the following salts when dissolved in water forms a basic solution? (a) NaCl (b) (NH4)2SO4 (c) CuSO4 (d) K2CO3 (e) NH4NO3

4. D; CO3 is conj. base of a weak acid; accepts H+ from water; liberates OH- .