Question

Learning Goal: To understand reaction order and rate constants. For the general equation aA+bB?cC+dD, the rate law is expressed as follows: rate=k[A]m[B]n where m and n indicate the order of the reaction with respect to each reactant and must be determined experimentally and k is the rate constant, which is specific to each reaction. Order For a particular reaction, aA+bB+cC?dD, the rate law was experimentally determined to be rate=k[A]0[B]1[C]2=k[B][C]2 This equation is zero order with respect to A. Therefore, changing the concentration of A has no effect on the rate because [A]0 will always equal 1. This equation is first order with respect to B. This means that if the concentration of B is doubled, the rate will double. If [B] is reduced by half, the rate will be halved. If [B] is tripled, the rate will triple, and so on. This equation is second order with respect to C. This means that if the concentration of C is doubled, the rate will quadruple. If [C] is tripled, the rate will increase by a factor of 9, and so on. . Overall reaction order and rate-constant units The sum of the individual orders gives the overall reaction order. The example equation above is third order overall because 0+1+2=3. For the units of rate to come out to be M/s, the units of the rate constant for third-order reactions must be M?2?s?1 since M/s=(M?2?s?1)(M3) For a second-order reaction, the rate constant has units of M?1?s?1 because M/s=(M?1?s?1)(M2). In a first-order reaction, the rate constant has the units s?1 because M/s=(s?1)(M1). Analyzing a specific reaction Consider the following reaction: NO+O3?NO2+O2,rate=k[NO][O3] Part A What is the overall reaction order? Express your answer as an integer. SubmitHintsMy AnswersGive UpReview Part Part B What are the units of the rate constant k for this reaction? What are the units of the rate constant for this reaction? M?s?1 s?1 M?2?s?1 M?1?s?1 M?3?s?1 SubmitHintsMy AnswersGive UpReview Part Part C What would happen to the rate if [NO] were doubled? The rate would What would happen to the rate if were doubled? stay the same. double. triple. quadruple. SubmitHintsMy AnswersGive UpReview Part Part D What would happen to the rate if [O3] were doubled? The rate would What would happen to the rate if were doubled? stay the same. quadruple. triple. double

Answer #1

given reaction:

NO+O3 ---->NO2 +O2

and ratek[NO][O3]

part A) For any given rxn:

aA+bB ----->cC+dD

ratek[A]^m [B]^n ,where m,n are called or taken as the order of rxn with respect to A and B respectively

And overall order is the sum total of all the orders

ovrall orderm+n

For the given rxn, order with respect to NO is 1 and order with respect to O3 is also 1.

overall order1+12

part B) As the overall order is 2 , the given reaction is second order rxn.Units of rate constant is M^-1s^-1

part C) If [NO] is doubled

new ratek(2*[NO])[O3]2*{k[NO][O3]}2*old rate

Thus, the rate is also doubled

part D) If [O3] is doubled

new ratek[NO](2*[O3])2*{k[NO][O3]}2*old rate

Thus, the rate is also doubled

To understand the relationship between the equilibrium constant
and rate constants.
For a general chemical equation
A+B⇌C+D
the equilibrium constant can be expressed as a ratio of the
concentrations:
Kc=[C][D][A][B]
If this is an elementary chemical reaction, then there is a
single forward rate and a single reverse rate for this reaction,
which can be written as follows:
forward ratereverse rate==kf[A][B]kr[C][D]
where kf and kr are the forward and reverse
rate constants, respectively. When equilibrium is reached, the
forward and...

Item 4
The integrated rate laws for zero-, first-, and second-order
reaction may be arranged such that they resemble the equation for a
straight line,y=mx+b.
Order
Integrated Rate Law
Graph
Slope
0
[A]t=−kt+[A]0
[A]t vs. t
−k
1
ln[A]t=−kt+ln[A]0
ln[A]t vs. t
−k
2
1[A]t= kt+1[A]0
1[A]t vs. t
k
Part A
The reactant concentration in a zero-order reaction was
6.00×10−2 mol L−1 after 140 s and 3.50×10−2
mol L−1 after 400 s . What is the rate constant for...

The following data were obtained at 25C for a reaction:
aA + bB = cC
Initial Concentrations Initial Reaction Rate
Reaction [A] [B] M/s or mol/Ls
1 0.100 M 0.200 M 5.00x10^-4
2 0.150 M 0.200 M 7.50x10^-4
3 0.150 M 0.600 M 6.75x10^-3
What is the order of the reaction with respect to A?
What is the order of the reaction with respect to B?
What is the rate constant, k, for the reaction?

In a generic chemical reaction involving reactants A and B and
products C and D, aA+bB→cC+dD,
the standard enthalpy ΔrH∘ of the reaction is given by
ΔrH∘=cΔfH∘(C)+dΔfH∘(D)
−aΔfH∘(A)−bΔfH∘(B)
Notice that the stoichiometric coefficients, a,
b, c, d, are an important part of this
equation. This formula is often generalized as follows, where the
first sum on the right-hand side of the equation is a sum over the
products and the second sum is over the reactants:
ΔrH∘=∑productsnΔfH∘−∑reactantsmΔfH∘
where m and...

In a generic chemical reaction involving reactants A and B and
products C and D, aA+bB→cC+dD, the standard enthalpy ΔH∘rxn of the
reaction is given by ΔH∘rxn=cΔH∘f(C)+dΔH∘f(D) −aΔH∘f(A)−bΔH∘f(B)
Notice that the stoichiometric coefficients, a, b, c, d, are an
important part of this equation. This formula is often generalized
as follows, where the first sum on the right-hand side of the
equation is a sum over the products and the second sum is over the
reactants: ΔH∘rxn=∑productsnΔH∘f−∑reactantsmΔH∘f where m and...

Learning Goal: To understand how to use integrated rate laws to
solve for concentration. A car starts at mile marker 145 on a
highway and drives at 55 mi/hr in the direction of decreasing
marker numbers. What mile marker will the car reach after 2 hours?
This problem can easily be solved by calculating how far the car
travels and subtracting that distance from the starting marker of
145. 55 mi/hr×2 hr=110 miles traveled milemarker 145−110
miles=milemarker 35 If we...

The integrated rate laws for zero-, first-, and second-order
reaction may be arranged such that they resemble the equation for a
straight line,y=mx+b.
Order
Integrated Rate Law
Graph
Slope
0
[A]=−kt+[A]0
[A] vs. t
−k
1
ln[A]=−kt+ln[A]0
ln[A] vs. t
−k
2
1[A]= kt+1[A]0
1[A] vs. t
k
Part A
The reactant concentration in a zero-order reaction was
8.00×10−2M after 200 s and
2.50×10−2Mafter 390 s . What is the rate
constant for this reaction?
Express your answer with the...

The integrated rate laws for zero-, first-, and second-order
reaction may be arranged such that they resemble the equation for a
straight line,y=mx+b.
Order
Integrated Rate Law
Graph
Slope
0
[A]=−kt+[A]0
[A] vs. t
−k
1
ln[A]=−kt+ln[A]0
ln[A] vs. t
−k
2
1[A]= kt+1[A]0
1[A] vs. t
k
------------
Part A
The reactant concentration in a zero-order reaction was
5.00×10−2M after 110 s and
4.00×10−2M after 375 s . What is the rate
constant for this reaction?
----------
Part B...

Part A A certain first-order reaction has a rate constant of
2.40×10−2 s−1 at 21 ∘C. What is the value of k at 61 ∘C if Ea =
90.0 kJ/mol ? Express your answer using two significant figures. k
= s−1 SubmitMy AnswersGive Up
Part B A certain first-order reaction has a rate constant of
2.40×10−2 s−1 at 21 ∘C. What is the value of k at 61 ∘C if Ea = 104
kJ/mol ? Express your answer using two...

The integrated rate laws for zero-, first-, and second-order
reaction may be arranged such that they resemble the equation for a
straight line,y=mx+b.
Order
Integrated Rate Law
Graph
Slope
0
[A]=−kt+[A]0
[A] vs. t
−k
1
ln[A]=−kt+ln[A]0
ln[A] vs. t
−k
2
1[A]= kt+1[A]0
1[A] vs. t
k
Part A
The reactant concentration in a zero-order reaction was
5.00×10−2M after 200 s and
2.50×10−2M after 310 s . What is the rate
constant for this reaction?
Express your answer with...

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