Chem 003 Final Exam Practice

What is the rate of reaction in terms of C in the following reaction: A + B → 2 C.

+0.5∆[C]/∆t

Which of the following represents the rate at which ClO₂⁻(aq) is appearing in the reaction below? 2 ClO₂(aq) + 2 OH⁻(aq) → ClO₂⁻(aq) + ClO₃⁻(aq) + H₂O(l)

+∆[ClO₃⁻]/∆t

For the reaction 2 A + 3 B → 4 D + 5 E, how is the rate of reaction expressed in terms of rate of disappearance of B?

-1/3∆[B]/∆t

Given the reaction A + 2 B → 3 C, the rate of appearance of C is also equal to which of the following?

-3/2∆[B]/∆t

What is the rate for the first order reaction A → Products when [A] = 0.201 M? (k = 0.360 1/min)

0.0724 M/min The units indicate that this is a first order reaction. first order- rate=k[A] rate=(0.201M)(0.3601min)=0.0724Mmin

A reaction has a rate law of Rate = (1.25 M⁻¹s⁻¹)[A][B]. What is the rate of the reaction if [A] = 0.301 M and [B] = 0.280 M?

0.105 M/s rate=1.251M⋅s(0.301M)(0.280M)=0.105Ms

What is the rate constant of a first-order reaction if the half-life is 2.50 min

0.277 1/min The half-life for a first-order reaction can be calculated using the formula t12=ln(2)k . 0.693/2.50=0.277min−

A reaction has a rate law of Rate = (1.25 M⁻²s⁻¹)[A][B]². What concentration of [B] would give the reaction a rate of 0.0891 M/s if the concentration of [A] is 0.250 M?

0.534 M rate=1.25 1/(M2⋅s)[A][B]2 0.0891 M/s= (1.251M2⋅s)(0.250M)(B)2 B2=0.28512 B=0.534M

The radioisotope ¹⁴C (half life = 5730 years) is used for carbon dating. What is the first order rate constant for ¹⁴C?

1.21 × 10⁻⁴ 1/years 5730=0.693/k k=1.21×10^−4

What is the absolute magnitude of the rate of change for [NH₃] if the rate of change for [H₂] is 2.00 M/s in the reaction 2 NH₃(g) → N₂(g) + 3 H₂(g)?

1.34 Rate=−12Δ[NH3]Δt=+13Δ[H2]Δt Substituting 2.00 Ms for Δ[H2]Δt and solving for the absolute magnitude ofΔ[NH3]Δt12Δ[NH3]Δt=13(2.00Ms)=0.6667MsΔ[NH3]Δt=2×0.6667Ms=1.33Ms

How is a second-order half-life (t1/2) calculated?

1/(k[A]₀) The half-life for a second-order reaction can be calculated using the formula t1/2=1/k[Ao].

The half-life of element X is 500 years. If there are initially 8 g of X, how much will remain after 1500 years?

1g

What is the half-life for a particular reaction if the rate law is rate = (1301 M⁻¹*min⁻¹)[A]² and the initial concentration of A is 0.250 M?

3.07 × 10⁻³ min t 1/2=11301(0.250)=3.07×10−3min

The reaction A + 2 B → C has the rate law rate = k[A][B]. By what factor does the rate of reaction increase when both [A] and [B] are doubled?

4 The rate law shows that the concentration of A and B do not have any exponents. Therefore, doubling a reactant will result in a doubling of the rate. In this case, both reactants are doubled, which would result in a rate of reaction increase of 4. 2×2=4.

What is the overall reaction order for the following rate law: Rate = k[A][B][C]²

4 The reaction order can be determined by finding the sum of the exponents from the rate law. In this case, it is 1+1+2=4.

For the reaction 3 A + 4 B → 2 C + 4 D, what is the magnitude of the rate of change for [B] when [C] is increasing at 2.0 M/s?

4.0 M/s

For the reaction 2 A + 4 B → 2 C + 4 D, what is the absolute magnitude of the rate of change for [B] when [C] is increasing at 2.0 M/s?

4.0 M/s The following solution may contain one or more values that are different from the problem provided to you, however, the steps to solve the problem are the same. .The rate of a reaction is equal to the rate of change of a reactant or product, divided by its coefficient in the balanced equation. Since reactants are disappearing they have negative values and products have positive values. For the reaction, 2A+4B⟶2C+4D the rate in terms of the change in B and C is given by Rate=−1/4Δ[B]/Δt=+1/2Δ[C]/Δt Substituting 2.0 Ms forΔ[C]/Δtand solving for the absolute magnitude ofΔ[B]/Δt 1/4Δ[B]/Δt=1/2(2.0Ms)=1.0Ms Δ[B]/Δt=4×1.0Ms=4.0Ms

The following reaction 2CO(g) + Br₂(g) → 2COBr(g) has the rate law, rate = k[CO] [Br₂]³. The overall order of this reaction is

4th rate=k[CO][Br2]3. The exponent for CO is 1 and the exponent for Br is 3. Adding those together gives an overall reaction order of 4.

What is the half-life for a particular reaction if the rate law is rate = (1301 min⁻¹)[A]?

5.33 × 10⁻⁴ min The half-life for a first-order reaction can be calculated using the formula t12=ln(2)k . 0.6931301=x x=5.33×10−4min

The reaction A + B → 2 C has the rate law rate = k[A][B]³. By what factor does the rate of reaction increase when [A] remains constant but [B] is doubled?

8 The rate law rate shows that the concentration of B is cubed, indicating that if it was doubled, the rate would be 8 times as large. 2×2×2=8.

Consider the radioisotope ²³⁵U (half life = 7.04 × 10⁸ years). What is the first order rate constant for ²³⁵U?

9.84 × 10⁻¹⁰ 1/years The half-life for a first-order reaction can be calculated using the formula t 1/2=ln(2)/k . Solve for the rate constant. 7.04×108=0.693/k k=9.84×10^−10

Consider the following equilibrium reaction: A (g) + B (s) ⇌ C (g) If Kp = 6 × 10⁻³, which species will have the highest partial pressure at equilibrium?

A The equilibrium constant is calculated by dividing the partial pressures of the products by the partial pressure of the reactants. Coefficients become exponents. Compounds which are solids and liquids have constant composition and therefore do not appear in the equilibrium expression. Kp=PCPA If the Kp is less than 1, this indicates that the partial pressure of A is larger than the partial pressure of C.

Which species has the greatest rate of disappearance in the reaction below? CH₄ + 2 O₂ → 2 CO₂ + H₂O

B)O₂ Reactants disappear during a reaction. For the equation given CH4+2O2⟶CO2+2H2O the reactants are CH4 andO2. From the coefficients in the equation you can see that twice as manyO2molecules react asCH4soO2has a greater rate of disappearance thanCH4and thus the fastest rate of disappearance of all the compounds in the reaction.

Which of the following affects the collision rate of molecules in the gas phase?

D) All of these affect collision rate Feedback: The collision rate is affected by speed, mass, and molecular diameter.

If Kp is the equilibrium constant for the reaction 2 A (g) + B (g) ⇌ 4 C (g). What is equilibrium constant (in terms of Kp) for the reaction 2 C (g) ⇌ A (g) + ½ B (g)?

Equilibrium constants are the products divided by the reactants. Coefficients become exponents. Because of this, multiplying the coefficients by 1/2 results in an exponent of 1/2. Because the equation is reversed, the Kp needs to be inverted. (Kp)^-1/2

If Kp is the equilibrium constant for the reaction 2 A (g) + B (g) ⇌ 3 C (g). What is the equilibrium constant (in terms of Kp) for the reaction 6 A (g) + 3 B (g) ⇌ 9 C (g)?

Equilibrium constants are the products divided by the reactants. Coefficients become exponents. Because of this, multiplying the coefficients by 3 results in an exponent of 3. (Kp)^3

What order of reaction has a half life equation of t½ = 0.693/k?

First

What is the order of the reaction if A decomposes to B and C with a rate constant of 8.43 × 10⁻⁴ s⁻¹ at a certain temperature?

First The order of a reaction can be determined from the units of the rate constant. The rate constant of this reaction has the units s−1, so it is a first-order reaction.

Which of the following statements are always true for a reaction at equilibrium? I. The rate of the forward and reverse reactions are equal. II. The concentrations of the reactants and the products remain constant. III. The amount of reactants is equal to the amount of products.

I and II

Which of the following statements is correct concerning the reaction 2 A + B → 2 C + 2 D?

It is not possible to determine the overall order nor the rate law from the given information.

What are the units for the rate constant of a reaction with the rate law, Rate = k[A][B]?

M⁻¹ s⁻¹ The rate of a reaction is expressed in the units M−1s−1. For the rate law, Rate = k[A][B], rearrange the equation, and add units to see what the units of k must be. k=Rate M s−1[A] M [B] M=1M s=M−1s−1

The units for the rate constant for a zero order reaction are

M･s⁻¹

What is the rate at which Br⁻(aq) disappears in the reaction below if the rate of disappearance of BrO₃⁻(aq) is 0.020 M/s? BrO₃⁻ + 5 Br⁻ + 6 H⁺ → 3 Br₂ + 3H₂O

Rate=−15Δ[Br−]Δt=−Δ[BrO3−]Δt Substituting -0.020 Ms for Δ[BrO−3]Δt and solving for Δ[Br−]Δt−15Δ[Br−]Δt=−(−0.020)Ms)=0.020 MsΔ[Br−]Δt=−5×0.020Ms=−0.10Ms Therefore the rate of disappearance is 0.10Ms.

Which species has the greatest rate of appearance in the reaction below? 2 H₂S + O₂ → 2 S + 2 H₂O

S and H₂O have identical rates Products appear during a reaction. For the equation given 2H2S+O2⟶2S+2H2O the products are S andH2O. From the coefficients in the equation you can see that 2 molecules ofSare produced for every 2 molecules ofH2O. Therefore, S andH2Ohave identical rates.

Consider the following diagrams which show the progress for the reaction A (blue) ⇌ B (red). The equilibrium constant (K) for this reaction is 0.8. At which point does the reaction reach equilibrium?

The equilibrium constant is the products divided by the reactants. An equilibrium constant or K value of 0.8 is consistent where the ratio of product to reactant is 4:5.

Which of the following is true for a reaction at equilibrium?

The rate of the forward reaction is equal to the rate of the reverse reaction

Which of the following statements best describes what occurs at equilibrium?

The rate of the reaction in the forward direction is equal to the rate of the reaction in the reverse direction.

Considering the rate law, rate = k[A]²[B], which of the following statements is correct?

The reaction is first order in B and third order overall. rate=k[A]2[B] the reaction is second-order in A because the exponent of [A] is two. It is first-order in B because the exponent of [B] is one, making it third-order overall.

For the reaction below, Q = 600. What must happen for the reaction to reach equilibrium? 2 A (g) + B (s) ⇌ 2 C (s) + D (g) Kp = 8210

The reaction needs to shift in the forward direction. Kp>Q and this indicates that the reaction will shift in the forward direction.

Which of the following will decrease the rate of a reaction?

Which of the following will decrease the rate of a reaction?

Consider the following rate law: Rate = k[B] Which of the following equations could be used to find the concentration of B at some time (t) in the reaction?

[B] = [B]₀ × e^(-kt) This reaction is a first order reaction. The integrated rate law for this first order reaction is: [B]=[B0]×e−kt

What can be said about the concentrations of reactants relative to the concentrations of products at equilibrium?

[Reactants]/[Products] = Constant

The following reaction 2 NO(g) + O₂(g) → 2 NO₂(g) was found to be first order in each of the two reactants and second order overall. The rate law is therefore

rate = k[NO][O₂] The order with respect to each reactant are given as 1, therefore the exponents of each reactant in the rate law are 1. The rate law is rate=k[NO][O2].

What is the rate for the second order reaction A → Products when [A] = 0.201 M? (k = 0.761 M⁻¹s⁻¹)

second order- rate = k[A]2 0.0307

What are the units for the rate constant of a first-order reaction?

s⁻¹

All of the following are factors that affect the rate of a reaction except _____

the magnitude of the equilibrium constant.

When a system is at dynamic equilibrium,

the rates of the forward and reverse reactions are equal.

The reaction rate of a reaction at 60 °C will be greater than at 30°C because _____

there is a greater proportion of reactants with the necessary kinetic energy to react. The molecules are moving faster at higher temperatures. This increases the likelihood that they have enough kinetic energy to react.

What is the rate for the zero order reaction A → Products when [A] = 0.200 M? (k = 3.57 M/min)

zero order rate=k 3.57

What is the half-life for a particular reaction if the rate law is rate = (1301 M*min⁻¹)[A]⁰ and the initial concentration of A is 0.250 M?

zero order rate law t1/2 = [A]o / 2 9.61 × 10⁻⁵ min t12=0.250(1301)(2)=9.61×10−5min