Answer:
a. Rate = k×[A]
b. k = 0.213s⁻¹
Explanation:
a. When you are studying the kinetics of a reaction such as:
A + B → Products.
General rate law must be like:
Rate = k×[A]ᵃ[B]ᵇ
You must make experiments change initial concentrations of A and B trying to find k, a and b parameters.
If you see experiments 1 and 3, concentration of A is doubled and the Rate of the reaction is doubled to. That means a = 1
Rate = k×[A]¹[B]ᵇ
In experiment 1 and to the concentration of B change from 1.50M to 2.50M but rate maintains the same. That is only possible if b = 0. (The kinetics of the reaction is indepent to [B]
Rate = k×[A][B]⁰
<h3>Rate = k×[A]</h3>
b. Replacing with values of experiment 1 (You can do the same with experiment 3 obtaining the same) k is:
Rate = k×[A]
0.320M/s = k×[1.50M]
<h3>k = 0.213s⁻¹</h3>
Answer:
The steps with correct mechanism are given below:
C
1) CH₄(g) + Cl(g) → CH₃(g) + HCl(g) : This is a slow step.
The rate is given as: R1 = k₁[CH₄][Cl]
2) CH₃(g) + Cl₂(g) → CH₃Cl(g) + Cl(g): This is a fast step.
The rate is given as: Rate = k₂[CH₃][Cl₂]
∴ CH₄(g) + Cl₂(g) → CH₃Cl(g) + HCl(g)
Here, the slowest step will be the rate-determining step.
Answer:
0.143 g of KCl.
Explanation:
Equation of the reaction:
AgNO3(aq) + KCl(aq) --> AgCl(s) + KNO3(aq)
Molar concentration = mass/volume
= 0.16 * 0.012
= 0.00192 mol AgNO3.
By stoichiometry, 1 mole of AgNO3 reacts with 1 mole of KCl to form a precipitate.
Number of moles of KCl = 0.00192 mol.
Molar mass of KCl = 39 + 35.5
= 74.5 g/mol
Mass = molar mass * number of moles
= 74.5 * 0.00192
= 0.143 g of KCl.
