Answer:
rate = k[A][B] where k = k₂K
Explanation:
Your mechanism is a slow step with a prior equilibrium:
![\begin{array}{rrcl}\text{Step 1}:& \text{A + B} & \xrightarrow [k_{-1}]{k_{1}} & \text{C}\\\text{Step 2}: & \text{C + A} & \xrightarrow [ ]{k_{2}} & \text{D}\\\text{Overall}: & \text{2A + B} & \longrightarrow \, & \text{D}\\\end{array}](https://tex.z-dn.net/?f=%5Cbegin%7Barray%7D%7Brrcl%7D%5Ctext%7BStep%201%7D%3A%26%20%5Ctext%7BA%20%2B%20B%7D%20%26%20%5Cxrightarrow%20%5Bk_%7B-1%7D%5D%7Bk_%7B1%7D%7D%20%26%20%5Ctext%7BC%7D%5C%5C%5Ctext%7BStep%202%7D%3A%20%26%20%5Ctext%7BC%20%2B%20A%7D%20%26%20%5Cxrightarrow%20%5B%20%5D%7Bk_%7B2%7D%7D%20%26%20%5Ctext%7BD%7D%5C%5C%5Ctext%7BOverall%7D%3A%20%26%20%5Ctext%7B2A%20%2B%20B%7D%20%26%20%5Clongrightarrow%20%5C%2C%20%26%20%5Ctext%7BD%7D%5C%5C%5Cend%7Barray%7D)
(The arrow in Step 1 should be equilibrium arrows).
1. Write the rate equations:
![-\dfrac{\text{d[A]}}{\text{d}t} = -\dfrac{\text{d[B]}}{\text{d}t} = -k_{1}[\text{A}][\text{B}] + k_{1}[\text{C}]\\\\\dfrac{\text{d[C]}}{\text{d}t} = k_{1}[\text{A}][\text{B}] - k_{2}[\text{C}]\\\\\dfrac{\text{d[D]}}{\text{d}t} = k_{2}[\text{C}]](https://tex.z-dn.net/?f=-%5Cdfrac%7B%5Ctext%7Bd%5BA%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%20%3D%20-%5Cdfrac%7B%5Ctext%7Bd%5BB%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%20%3D%20-k_%7B1%7D%5B%5Ctext%7BA%7D%5D%5B%5Ctext%7BB%7D%5D%20%2B%20k_%7B1%7D%5B%5Ctext%7BC%7D%5D%5C%5C%5C%5C%5Cdfrac%7B%5Ctext%7Bd%5BC%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%20%3D%20k_%7B1%7D%5B%5Ctext%7BA%7D%5D%5B%5Ctext%7BB%7D%5D%20-%20k_%7B2%7D%5B%5Ctext%7BC%7D%5D%5C%5C%5C%5C%5Cdfrac%7B%5Ctext%7Bd%5BD%5D%7D%7D%7B%5Ctext%7Bd%7Dt%7D%20%3D%20k_%7B2%7D%5B%5Ctext%7BC%7D%5D)
2. Derive the rate law
Assume k₋₁ ≫ k₂.
Then, in effect, we have an equilibrium that is only slightly disturbed by C slowly reacting to form D.
In an equilibrium, the forward and reverse rates are equal:
k₁[A][B] = k₋₁[C]
[C] = (k₁/k₋₁)[A][B] = K[A][B] (K is the equilibrium constant)
rate = d[D]/dt = k₂[C] = k₂K[A][B] = k[A][B]
The rate law is
rate = k[A][B] where k = k₂K
Answer:
c
Explanation:
you don't think of particles as dense and less dense
dust particles containvenergy but won't release it unless acted upon a force
gas molecules move freely and collide with dust particles which is correct
Answer: Option D is correct.
Explanation: Equation given by de Broglie is:
where, 
= wavelength of the particle
h = Planck's constant
m = mass of the particle
v = velocity of the particle
In option A, football will have some mass and is moving with a velocity of 25 m/s, hence it will have some wavelength.
In Option B, unladen swallow also have some mass and is moving with a velocity of 38 km/hr, hence it will also have some wavelength.
In Option C, a person has some mass and is running with a velocity of 7 m/hr, hence it will also have some wavelength.
As, the momentum of these particles are large, therefore the wavelength will be of small magnitude and hence, is not observable.
From the above, it is clearly visible that all the options are having some wavelength, so option D is correct.
It can be easily judged that only the first reaction is an acid base reaction among the three given in the question. So, we can avoid the other two reactions given in the question. Now let us focus and write down the balanced chemical equation of <span>P-Toluidine + HCl.
</span><span>C7H9N + HCl = C7H10N (+) + Cl (-)
</span>
I hope the answer has come to your help.
