Thermochemistry and Kinetics

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AP Chemistry › Thermochemistry and Kinetics

Questions 1 - 10

Chaning which of the following factors can alter the rate of a zero-order reaction?

Temperature

Increasing the concentration of reactants

Increasing the concentration of products

Decreasing the concentration of products

Explanation

A zero-order reaction has a rate of formation of product that is independent of changes in concentrations of any of the reactants; however, since the rate constant itself is dependent on temperture, changing the temperature can alter the rate.

$O_{_{3}} \rightarrow O_{_{2}} + O$ (slow)

$O + O_{_{3}} \rightarrow 2O_{_{2}}$ (fast)

The mechanism for decomposition of ozone is shown. What is the intermediate of the process?

$O_{2}$

$2O_{2}$

$O_{3}$

Explanation

Intermediate is created and destroyed, and therefore does not appear on the net equation, which is $2O_{3} \rightarrow 3O_{2}$ . Thus, the intermediate is . Note that when asked for an intermediate, the coefficient in front of it is not used, rather we are looking for the species that is a product of one reaction and a reactant in a subsequent step.

For the following cell reaction:

2 Al (s) + 3 Mn2+ (aq) -> 2 Al3+ (aq) + 3 Mn (s)

predict if the cell potential Ecell will be larger, or smaller than Eocell for the following conditions: \[Al3+\] = 1.0 M; \[Mn2+\] = 5.0 M.

Larger

Smaller

Identical

Can not be determined

Explanation

$E_{cell} = E^0_{cell} - \frac{0.0592}{n} : log : Q$

At these concentrations Q will become smaller, and thus log Q will become smaller. This will give rise to a larger cell potential.

1. $\small Zn \rightarrow Zn^{2+} + 2e^{-}$ $\small \small E^{o} = 0.762 V$

2. $\small Cu \rightarrow Cu^{2+} + 2e^{-}$ $\small \small E^{o}= -0.340 V$

3. $\small Zn^{2+} + Cu \rightarrow Zn + Cu^{2+}$

What is $\small \small E_{cell}^{o}$ ? Is reaction 3 spontaneous?

$\small -1.1 V$ , nonspontaneous

$\small -1.1 V$ , spontaneous

$\small 1.1 V$ , nonspontaneous

$\small 1.1 V$ , spontaneous

Explanation

$\small E^{o}_{cell} = E_{red}^{o} + E_{ox}^{o}$

$\small E_{cell}^{o} = -0.762 V - 0.340 V$

$\small E^{o}_{cell} = -1.1 V$

There are two concepts to consider in this problem. First, the question asks for the $\small E^{o}_{cell}$ . Reaction 3 has $\small Zn^{2+}$ being reduced so the potential for the half reaction becomes negative. $\small Cu$ half reaction appears the same in reaction 3 so the potential is the same. Second, negative voltages indicate non-spontaneity and positive voltages are spontaneous.

The following galvanic cell is created:

$2Au^{3+} + 3Cu \rightarrow 3Cu^{2+} + 2Au$ $E^{\circ} = 1.16V$

Which of the following takes place at the cathode?

Gold ions receive electrons

Copper loses electrons

Gold loses electrons

Copper gains electrons

Explanation

Reduction always takes place at the cathode. This means that a substance is receiving electrons.

$2Au^{3+} + 3Cu \rightarrow 3Cu^{2+} + 2Au$

In the reaction, the gold ions are receiving electrons in order to create gold atoms. This takes place at the cathode.

Copper ions and gold are products, and will not be oxidized or reduced. Copper is reduced at the anode during this reaction.

How many grams of Cr can be obtained by the electrolysis of a Cr(NO3)3 if 10 amps are passed through the cell for 6 hours?

38.8 g

19.4 g

103 g

12.5 g

56.3 g

Explanation

$Cr^{3+} + 3 e- \rightarrow Cr(s)$

$mass = \frac{10 A , x, 6 hr , x , \frac{3600 s}{1 hr}}{ \frac{96485 C}{mol e-} } , x , \frac{1 mol , Cr}{3 mol e-} , x , \frac{51.996 g , Cr}{1 mol , Cr}= 38.8 g$

Calculate the standard cell potential of the following reaction:

Cd(s) + MnO2 (s) + 4 H+ (aq) + -> Cd2+ (aq) + Mn2+ (aq) + 2 H2O (l)

Given:

MnO2 (s) + 4 H+ (aq) + 2e- -> Mn2+ (aq) + 2 H2O (l) Eo = 1.23 V

Cd2+ (aq) + 2 e- -> Cd (s) Eo = -0.40 V

1.63 V

0.83 V

-1.63 V

-0.83 V

0.0 V

Explanation

Eocell = Eo cathode - Eoanode

Eocell = 1.23 – (-0.40) = 1.63 V

Which of the following is a law of thermodynamics?

ΔH (system) + ΔH (surroundings) = ΔH (universe)

ΔS (system) + ΔS (surroundings) = ΔS (universe)

ΔE (surroundings) = ΔE (system)

The entropy of the universe is always decreasing

Explanation

The second law of thermodynamics states that the entropy of the system and the entropy of the surroundings is equal to the entropy of the universe. The rest of the answer choices are not one of the fundamental laws of themodynamics.

Which of the following is a classic example of a first-order reaction?

Radioactive decay

A collision between 2 reactant molecules

A change in temperature

None of the other answers

Explanation

First order reactions have rates that are directly proportional to only 1 reactant. In radioactive decay, the rate of decrease of a radioactive material is proportional only to the amount of the material.

Given the reaction A + B → C. What is the rate law for the following experiment?

\[A\] \[B\] Rate

0.05 0.05 0.0125

0.05 0.1 0.0250

0.1 0.05 0.0125

rate = k\[A\]\[B\]2

rate = k\[A\]2\[B\]

rate = k\[A\]\[B\]

rate = k\[A\]

rate = k\[B\]

Explanation

When the concentration of B doubles, the rate doubles. Making this reaction first order in regards to compound B. When the concentration of A doubles the rate is unaffected, making this reaction zero order in regards to compound A. This leaves a rate law of rate=k\[B\]

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