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Rate Expression In Third Order

What is the rate expression of a second order reaction having different reactants?

Checkout this to get your solution :))

What is the expression of a second-order reaction with the same and different initial concentrations?

I am going to assume that the actual question here is “What is the integrated rate law for a second-order reaction with the same and different initial conditions?” because the question, as written, has a trivial answer:Same concentrations:rate = k [A][math]^2[/math]Different concentrations:rate = k [A] [B]Now, assuming that these were NOT the answers you were seeking, the integrated rate laws for each condition are:Same concentrations:[math]\dfrac{1}{[A_f]}\,- \,\dfrac{1}{[A_i]} = k\Delta t[/math]Different concentrations:[math]\dfrac{1}{([B_i]\,-\,[A_i])}\,ln\dfrac{[A_i][B_f]}{[B_i][A_f]} = k\Delta t[/math]

The reaction 2A + 5B → products is third order in A and third order in B. What is the rate law for this reacti?

The reaction 2A + 5B → products is third order in A and third order in B. What is the rate law for this reaction?
1. rate = k[A]2[B]5

2. rate = k[A]3[B]3

3. rate = k[A]2/7[B]5/7

4. rate = k[A]5[B]2

5. rate = k[A]6[B]3

What is the integrate rate law of a third order reaction?

The integrated rate law of a nth order reaction .i.e.
rate = -d[A]/dt = k∙[A]ⁿ
1/[A]ⁿ⁻¹ = 1/[A]₀ⁿ⁻¹ + (n-1)∙k∙t
([A]₀ denotes initial concentration)

So for a third order reaction, i.e. n=3
1/[A]² = 2∙k∙t + 1/[A]₀²
[A] = 1/√[ 2∙k∙t + 1/[A]₀²] = [A]₀/√[2∙[A]₀∙k∙t + 1]

Consider the following rate law expression:rate=k[A]^2[B].?

Which of the following is NOT true about the reaction having this expression? and explain.
a)the reaction is first order in B
b)the reaction is overall third order
c)the reaction is second order in A
d)Doubling the concentration of A doubles the rate.

How to derive the half life expression from the integrated rate law of a third order reaction?

-dA/dt = k*A³

dA/A³ = -k*dt

∫dA/A³ = -∫k*dt

-1/(2*A²) + C = -k*t

at t = 0 A = A0

-1/(2*A0²) + C = 0

C = 1/(2*A0²)

-1/(2*A²) + 1/(2*A0²) = -k*t
-1/(2*A²) = -k*t - 1/(2*A0²)

1/A² = 2*k*t +1/A0²

at t = th, A = A0/2;

4/A0² = 2*k*th + 1/A0²

3/A0² = 2*k*th

th = 3/(2*k*A0²)

What is an example of a third order reaction with two reactants that behaves as a first order reaction?

Any reaction that has this rate formula k=A^2*B is third order. Say C=2A+B is the stoichiometry (but the kinetics are not necessarily what we have posited, but let's say they are). It doesn't matter what compounds or elements really are A, B and C for this argument. If initial amount of B is very much greater than initial amount of A, say B = 1000*A, then it is effectively constant over the course of the reaction, i.e., it would change only by 0.1% at complete reaction. So then k' = A'*B where A' is the effectively constant value of A over the reaction. So the reaction appears to be first order in B. This is what is called a pseudo-first order rate equation, but remember it is an approximation only when one reactant in the kinetic equation is in large excess.

What is the derivation of third order reaction and its half life?

t 1/2 is inversely proportional to [A]^(n-1) where n is the order of the reaction. So; half lie time for the third order reaction is proportional to 1/[A]^2.

Third order reaction when 2 reactants are same?

A reaction like 2A + B → P is not necessarily 3rd order. The mechanism might be something like:A + B → CC + A → P(or some other variant)3rd order reactions require 3 molecules being present at the same spot, at the same time. Moreover, they must collide at the right orientation to make the reaction happen, with enough energy to overcome the activation energy. Such reaction are extremely rare.