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:
0.25M HCl
Explanation:
The reaction of HCl with NaOH is:
HCl + NaOH ⇄ H₂O + NaCl
<em>Where 1 mole of HCl reacts per mole of NaOH</em>
The end point was reached when the student added:
0.0500L × (0.1mol / L) = 0.00500 moles of NaOH
As 1 mole of HCl reacted per mole of NaOH, moles of HCl present are:
<em>0.00500 moles HCl</em>
The volume of the sample of hydrochloric acid was 20.0mL = 0.0200L, and concentration of the sample is:
0.00500 mol HCl / 0.0200L = <em>0.25M HCl</em>
A. electron. The nucleus has protons and neutrons, quark is the particle which forms protons and neutrons.
<span>30.0 ml of 0.15 m K2CrO4 solution will have more potassium ions.
Let's see the relative number of potassium ions for each solution. Since all the measurements are the same, the real difference is the K2CrO4 will only have 2 potassium ions per molecule while the K3PO4 solution will have 3 potassium ions per molecule.
K2CrO4 solution
30.0 * 0.15 * 2 = 9
K3PO4 solution
25.0 * 0.080 * 3 = 6
Since 9 is greater than 6, the K2CrO4 solution will have more potassium ions.</span>
