Respuesta :
Answer:
(a) Rate = k[A][B]⁰[C]² = k[A][C]².
(b) The rate of the reaction will be doubled as the the rate of the reaction will be doubled.
(c) The rate of the reaction will not be changed as the rate of the reaction is zero order for B and does not depend on the concentration of reactant B.
(d) The rate of the reaction will increase by a factor of 9 when [C] is tripled and the other reactant concentrations are held constant.
(e) The rate of the reaction will increase by a factor of 27 the concentrations of all three reactants are tripled.
(f) The rate of the reaction will decrease by a factor of 8 the concentrations of all three reactants are cut in half.
Explanation:
(a) Write the rate law for the reaction.
The rate of the reaction is the change in the concentration of the reactants (decrease) or the products (increase) with time.
Rate = ΔConcentration/Δt.
The rate of the reaction is directly proportional to the concentration of the reactants.
R ∝ [A][B][C].
Since the mentioned reaction is first order in A, zero order in B, and second order in C.
∴ Rate = k[A][B]⁰[C]² = k[A][C]²,
where, k is the rate constant of the reaction.
(b) How does the rate change when [A] is doubled and the other reactant concentrations are held constant?
If we consider the concentration of [A₁] = [A₀].
As the concentration of [A] is doubled, [A₂] = [2A₀]. [B] and [C] are held constant.
∴ Rate₁ = k[A₁][B₁]⁰[C₁]² = k[A₀][B₀]⁰[C₀]²
Rate₂ = k[A₂][B₂]⁰[C₂]² = k[2A₀][B₀]⁰[C₀]²
∴ Rate₁/Rate₂ = (k[A₀][B₀]⁰[C₀]²)/(k[2A₀][B₀]⁰[C₀]²) = 1/2.
∴ Rate₂ = 2(Rate₁).
So, the rate of the reaction will be doubled.
(c) How does the rate change when [B] is tripled and the other reactant concentrations are held constant?
If we consider the concentration of [B₁] = [B₀].
As the concentration of [A] is doubled, [B₂] = [3B₀]. [A] and [C] are held constant.
∴ Rate₁ = k[A₁][B₁]⁰[C₁]² = k[A₀][B₀]⁰[C₀]²
Rate₂ = k[A₂][B₂]⁰[C₂]² = k[A₀][3B₀]⁰[C₀]²
∴ Rate₁/Rate₂ = (k[A₀][B₀]⁰[C₀]²)/(k[A₀][3B₀]⁰[C₀]²) = 1.
∴ Rate₂ = (Rate₁).
So, the rate of the reaction will not be changed as the rate of the reaction is zero order for B and does not depend on the concentration of reactant B.
(d) How does the rate change when [C] is tripled and the other reactant concentrations are held constant?
If we consider the concentration of [C₁] = [C₀].
As the concentration of [A] is doubled, [C₂] = [3C₀]. [A] and [B] are held constant.
∴ Rate₁ = k[A₁][B₁]⁰[C₁]² = k[A₀][B₀]⁰[C₀]²
Rate₂ = k[A₂][B₂]⁰[C₂]² = k[A₀][B₀]⁰[3C₀]²
∴ Rate₁/Rate₂ = (k[A₀][B₀]⁰[C₀]²)/(k[A₀][B₀]⁰[3C₀]²) = 1/9.
∴ Rate₂ = 9(Rate₁).
So, the rate of the reaction will increase by a factor of 9.
(e) By what factor does the rate change when the concentrations of all three reactants are tripled?
If we consider the concentration of [A₁] = [A₀], [B₁] = [B₀], and [C₁] = [C₀].
As the concentration of all three reactants are tripled:
[A₂] = [3A₀], [B₂] = [3B₀], and [C₂] = [3C₀].
∴ Rate₁ = k[A₁][B₁]⁰[C₁]² = k[A₀][B₀]⁰[C₀]²
Rate₂ = k[A₂][B₂]⁰[C₂]² = k[3A₀][3B₀]⁰[3C₀]²
∴ Rate₁/Rate₂ = (k[A₀][B₀]⁰[C₀]²)/(k[3A₀][3B₀]⁰[3C₀]²) = 1/27.
∴ Rate₂ = 27(Rate₁).
So, the rate of the reaction will increase by a factor of 27.
(f) By what factor does the rate change when the concentrations of all three reactants are cut in half?
If we consider the concentration of [A₁] = [A₀], [B₁] = [B₀], and [C₁] = [C₀].
As the concentration of all three reactants are tripled:
[A₂] = [1/2A₀], [B₂] = [1/2B₀], and [C₂] = [1/2C₀].
∴ Rate₁ = k[A₁][B₁]⁰[C₁]² = k[A₀][B₀]⁰[C₀]²
Rate₂ = k[A₂][B₂]⁰[C₂]² = k[1/2A₀][1/2B₀]⁰[1/2C₀]²
∴ Rate₁/Rate₂ = (k[A₀][B₀]⁰[C₀]²)/(k[1/2A₀][1/2B₀]⁰[1/2C₀]²) = 8.
∴ Rate₂ = 1/8(Rate₁).
So, the rate of the reaction will decrease by a factor of 8.
The rate law for the reaction is = k[A][B]⁰[C]² = k[A][C]².
Calculations and Parameters:
The rate law for the reaction is given mathematically as:
Rate = ΔConcentration/Δt.
Since the rate of change is directly proportional to the reactants, then
R ∝ [A][B][C].
Given that the first order is in A, there is a zero-order in B, and the second order in B, it can be analyzed further as:
Rate = k[A][B]⁰[C]² = k[A][C]²,
Where k is the constant.
To find the factor by which the rate changes when the concentrations of all three reactants are cut in half:
Recall the concentration of [A₁] =
- [A₀], [B₁] = [B₀],
- [C₁] = [C₀].
As the concentration of all three reactants is tripled:
- [A₂] = [1/2A₀],
- [B₂] = [1/2B₀],
- [C₂] = [1/2C₀].
Rate₁ = k[A₁][B₁]⁰[C₁]² = k[A₀][B₀]⁰[C₀]²
Rate₂ = k[A₂][B₂]⁰[C₂]² = k[1/2A₀][1/2B₀]⁰[1/2C₀]²
When we divide the two, Rate₁/Rate₂ = (k[A₀][B₀]⁰[C₀]²)/(k[1/2A₀][1/2B₀]⁰[1/2C₀]²)
= 8.
∴ Rate₂ = 1/8(Rate₁).
Therefore, the rate of the reaction will decrease by a factor of 8.
The other answers are:
- (b) The way in which the rate of the reaction will change is that it would be doubled as the rate of the reaction will be doubled.
- (c) The thing which will occur when B is tripled is that it will remain unchanged as the rate of the reaction is zero-order for B and does not depend on the concentration of reactant B.
- (d) The way in which the rate of the reaction will change when [C] is tripled and the other reactant concentrations are held constant is that it will increase by a factor of 9.
- (e) The rate of the reaction will increase by a factor of 27 as the concentrations of all three reactants are tripled.
- (f) The rate of the reaction will decrease by a factor of 8 if the concentrations of all three reactants are cut in half.
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