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6 Diagnostic Tests 225 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #1 : Net Ionic Equations

What is the net ionic reaction when a solution of Cd(NO₃)₂ is reacted with (NH₄)₂S?

Possible Answers:

Cd(NO₃)₂ (aq) + (NH₄)₂S (aq) -> CdS (s) + 2 NH₄NO₃(aq)

Cd²⁺(aq) + S^2- -> CdS (s)

NH4⁺ (aq) + NO₃^- (aq) -> NH₄NO₃ (s)

Cd(NO₃)₂ (aq) + (NH₄)₂S (aq) -> CdS (aq) + 2 NH₄NO₃(aq)

Cd(NO₃)₂ (aq) + (NH₄)₂S (aq) -> CdS (s) + 2 NH₄NO₃ (aq)

Correct answer:

Cd²⁺(aq) + S^2- -> CdS (s)

Explanation:

Only Cd and S ions are interacting according to the solubility rules to form a solid precipitant.

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Example Question #3 : Chemical Reactions

What is the net ionic reaction when a solution of rubidium sulfate is mixed with barium hydroxide?

Possible Answers:

Ba²⁺ (aq) + SO₄^2- (aq) -> BaSO4 (s)

Rb₂SO₄ (aq) + Ba(OH)₂ (aq) -> 2 RbOH (s) + BaSO₄ (aq)

Rb⁺ (aq) + OH^- (aq) -> RbOH (s)

No Reaction

Rb₂SO₄ (aq) + Ba(OH)₂ (aq) -> 2 RbOH (aq) + BaSO₄ (s)

Correct answer:

Ba²⁺ (aq) + SO₄^2- (aq) -> BaSO4 (s)

Explanation:

Only the Ba and sulfate ions will react according to the solubility rules.

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Example Question #103 : Thermochemistry And Kinetics

The rate constant for a second-order reaction is 0.15 M^-1s^-1. If the initial concentration of the reactant is 0.30 M, how long does it take for the concentration to decrease to 0.15 M?

Possible Answers:

22.2 seconds

44.4 second

11.1 seconds

253 second

88.8 seconds

Correct answer:

22.2 seconds

Explanation:

$\frac{1}{[A]_t} = kt + \frac{ 1 }{[A]_o}$ $\displaystyle \frac{1}{[A]_t} = kt + \frac{ 1 }{[A]_o}$

$t =\frac{ \frac{1}{[A]_t} -\frac{1}{[A]_o}}{k} =\frac{\frac{1}{0.15} - \frac{1}{0.30}}{0.15}= 22.2 seconds$ $\displaystyle t =\frac{ \frac{1}{[A]_t} -\frac{1}{[A]_o}}{k} =\frac{\frac{1}{0.15} - \frac{1}{0.30}}{0.15}= 22.2 seconds$

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Example Question #1 : Integrated Rate Laws

Based on the figure above, what is the order of reaction?

Possible Answers:

Second Order

Third Order

First order

Squared Order

Zero Order

Correct answer:

First order

Explanation:

For a first order reaction, the ln [A]_t is linear with t.

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Example Question #2 : Integrated Rate Laws

A compound decomposes by a first-order process. If 25.0% of the compound decomposes in 60 minutes, the half-life of the compound is?

Possible Answers:

65 minutes

145 minutes

120 minutes

180 minutes

198 minutes

Correct answer:

145 minutes

Explanation:

$\displaystyle ln[A]_t=-kt + ln[A]_o$

$k = - \frac{ln[A]_t - ln[A]_o}{t}= - \frac{ln[75]-ln[100]}{60}= 4.79 x10^{-3}$ $\displaystyle k = - \frac{ln[A]_t - ln[A]_o}{t}= - \frac{ln[75]-ln[100]}{60}= 4.79 x10^{-3}$

$t_{\frac{1}{2}}=\frac{ ln2 }{k} = \frac{ln2}{4.79x10^{-3}}= 145 minutes$ $\displaystyle t_{\frac{1}{2}}=\frac{ ln2 }{k} = \frac{ln2}{4.79x10^{-3}}= 145 minutes$

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Example Question #3 : Integrated Rate Laws

Cyclopentane is unstable and decomposes by a first order reaction. The rate constant for this reaction is 9.5 s^-1. What is the half life of the reaction?

Possible Answers:

0.0729 second

0.0314 seconds

0.0582 seconds

0.132 seconds

0.0614 seconds

Correct answer:

0.0729 second

Explanation:

$t_{\frac{1}{2}}=\frac{ln2 }{k} = \frac{ln2}{9.5} = 0.0729 seconds$ $\displaystyle t_{\frac{1}{2}}=\frac{ln2 }{k} = \frac{ln2}{9.5} = 0.0729 seconds$

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Example Question #4 : Integrated Rate Laws

The half life of a first order reaction is 1.5 hours. What is the rate constant of this reaction?

Possible Answers:

0.75

0.46

0.52

0.21

1.5

Correct answer:

0.46

Explanation:

$t_{\frac{1}{2}} = \frac{ln2 }{k}$ $\displaystyle t_{\frac{1}{2}} = \frac{ln2 }{k}$

$k =\frac{ln2}{t_{\frac{1}{2}}} = \frac{ln2}{1.5} = 0.46$ $\displaystyle k =\frac{ln2}{t_{\frac{1}{2}}} = \frac{ln2}{1.5} = 0.46$

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Example Question #1 : Reaction Mechanisms

Which of the following is true?

Possible Answers:

If we know that a reaction is an elementary reaction, then we know its rate law.

Since intermediate compounds can be formed, the chemical equations for the elementary reaction in a multistep mechanism do not always have to add to give the chemical equation of the overall process.

In a reaction mechanism an intermediate is identical to an activated complex

All of the above

The rate-determining step of a reaction is the rate of the slowest elementary step of its mechanism

Correct answer:

All of the above

Explanation:

All of the above describe elementary reactions and how they give an overall mechanism.

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Example Question #2 : Reaction Mechanisms

A possible mechanism for the overall reaction Br₂ (g) + 2 NO (g) -> 2 NOBr(g) is

The rate law for the formation of NOBr based on this mechanism is rate = .

Possible Answers:

$(k_2k_1/k_{-1}) [NO]^2[Br_2]$ $\displaystyle (k_2k_1/k_{-1}) [NO]^2[Br_2]$

$k_1[Br_2]^{1/2}$ $\displaystyle k_1[Br_2]^{1/2}$

$k_1[NO]^{1/2}$ $\displaystyle k_1[NO]^{1/2}$

$(k_1/k_{-1})^2[NO]^2$ $\displaystyle (k_1/k_{-1})^2[NO]^2$

$(k_2k_1/k_{-1}) [NO][Br_2]^2$ $\displaystyle (k_2k_1/k_{-1}) [NO][Br_2]^2$

Correct answer:

$(k_2k_1/k_{-1}) [NO]^2[Br_2]$ $\displaystyle (k_2k_1/k_{-1}) [NO]^2[Br_2]$

Explanation:

Based on the slowest step the rate law would be: Rate = k2 [NOBr2] [NO], but one cannot have a rate law in terms of an intermediate (NOBr2).

Because the first reaction is at equilibrium the rate in the forward direction is equal to that in the reverse, thus:

$k1 [NO][Br_2] = k_{-1}[NOBr_2]$ $\displaystyle k1 [NO][Br_2] = k_{-1}[NOBr_2]$ and: $[NOBr_2 ] = \frac{k_1 [NO][Br_2 ]}{k_{-1} }$ $\displaystyle [NOBr_2 ] = \frac{k_1 [NO][Br_2 ]}{k_{-1} }$

Substitution yields: $Rate = \frac{k_1 k_2 [NO]^2 [Br_2 ]}{k_{-1} }$ $\displaystyle Rate = \frac{k_1 k_2 [NO]^2 [Br_2 ]}{k_{-1} }$

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Example Question #2 : Reaction Mechanisms

For the reaction NO₂ (g) + CO (g) -> NO (g) + CO₂ (g), the reaction was experimentally determined to be Rate = k[NO2]². If the reaction has the following mechanism, what is the rate limiting step, and why?

Step 1: 2 NO₂ -> NO₃ + NO (slow)

Step 2: NO₃ + CO -> NO2 + CO2 (fast)

Possible Answers:

Step 2 is limiting because the fast step determines how quickly the reaction can occur.

Step 2 is limiting because the NO₃ intermediate has to be formed before the reaction can occur.

Step 1 is limiting because the reaction can not go faster than its slowest step.

Step 1 is limiting because the NO₂ is a reactant.

Not enough information

Correct answer:

Step 1 is limiting because the reaction can not go faster than its slowest step.

Explanation:

The reaction can never go faster than its slowest step.

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AP Chemistry : AP Chemistry

Study concepts, example questions & explanations for AP Chemistry

All AP Chemistry Resources

Example Questions

Example Question #1 : Net Ionic Equations

Example Question #3 : Chemical Reactions

Example Question #103 : Thermochemistry And Kinetics

Example Question #1 : Integrated Rate Laws

Example Question #2 : Integrated Rate Laws

Example Question #3 : Integrated Rate Laws

Example Question #4 : Integrated Rate Laws

Example Question #1 : Reaction Mechanisms

Example Question #2 : Reaction Mechanisms

Example Question #2 : Reaction Mechanisms

All AP Chemistry Resources

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