Chem 2
What is the rate for the second order reaction A → Products when [A] = 0.201 M? (k = 0.761 M⁻¹s⁻¹)
.The units indicate that this is a second order reaction. rate=k[A]2 rate=(0.7611M⋅s)(0.201M)=0.0307Ms
For the reaction A + 2 B → 3 C + D, what is the magnitude of the rate of change for [B] when [C] is increasing at 4.6 M/s?
.The units indicate that this is a zero order reaction. For every 3C there are 2B. 4.6×23=3.1
If a plot of 1/[A] versus time produces a straight line with a positive slope for the reaction A → B + C, what is the order of the reaction?
A graph of 1/[A] versus time yields a straight line with a positive slope for second order reactions.
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 You can tell the order of a reaction 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.
How long will it take for the concentration of A to decrease from 0.500 M to 0.200 M in the first-order reaction A → B? (k = 0.800 s⁻¹)
Given data :- initial concentration = Co = 0.500 M final concentration = Ct = 0.200 M rate constant = k = 0.800 s-1 Solution :- The reaction is as follows :- A --------> B For a first order reaction, the concentration is related to time as per the following formula. ln (Ct / Co) = - k x t ln (0.200 / 0.500) = - 0.800 x t t = 1.145 s ≈ 1.15 s The time taken for concentration to become 0.200 M from 0.500 M is 1.145 (≈ 1.15) seconds.
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?
Plug the units into the given rate for this second order reaction. rate=1.251M⋅s(0.301M)(0.280M)=0.105Ms
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?
Plug the units into the given rate law. rate=1.251M2⋅s[A][B]2 0.0891Ms=(1.251M2⋅s)(0.250M)(B)2B2=0.28512B=0.534M
Which of the following affects the collision rate of molecules in the gas phase?
The collision rate is affected by speed, mass, and molecular diameter.
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
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 reactionBrO₃⁻+5Br⁻+6H⁺⟶3Br₂+3H₂O the rate in terms of the change in [Br⁻] and [BrO₃⁻] is given byRate=−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.
Copper roofs on houses form patina (copper carbonates) over the course of years due to the reaction with oxygen, carbon dioxide and water in the air. Which of the following statements is incorrect for this process?
The reaction has a low activation energy. The rate of reaction is affected by the concentration of the reactants, the presence of a catalyst, the surface area of the reactants, and the temperature of the reaction. Removing or decreasing each of these would result in a decreased reaction rate. Adding or increasing each of these would result in an increased reaction rate. In this case, it takes years for patina to form, so we know that the reaction rate is low. The reaction has a high activation energy because it takes a long time. Because temperature affects the reaction rate, the formation of patina would occur faster in the summer months than the winter months because more molecules would have the kinetic energy required to react. The formation of patina would occur slower at higher altitudes than lower altitudes because there is less reactant and thus a lower reaction rate. Finally, the addition of a catalyst would increase the reaction rate.
What is the overall reaction order for the following rate law: Rate = k[A][B][C]²
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.
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?
To solve this question, we just need to put the new number into the equation. If [A] remain constant then that mean [A2]= [A1]. If B doubled, then that mean [B2]= 2[B2]. To find what factor does the rate of reaction increases, we need to divide the first reaction rate with the second. The calculation will be: rate2/rate1= k[A2][B2]³ / k[A1][B1]³ rate2/rate1= [A1][2B1]³ / [A1][B1]³ rate2/rate1= A1*8B1³ / A1*B1³ rate2/rate1= 8/1= 8 The rate of reaction will be 8 times faster.
A reaction was shown to have an increasing half-life with increasing concentration of reactant. What order is this reaction?
We know that half life is related to concentration as:t21=[A]1−n where n is the order of the reactionHere n=2t21=[A]1−2=[A]1For n=0, t21 =[A]1−0=[A]If t21 is doubled, [A] is also doubled then the reaction is of zero order.
What order of reaction has a half life equation of t½ = 0.693/k?
first This is the correct expression for half-life for a first order reaction.
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.
