Answer:
B) Δ[C]/Δt = 3,60x10⁻² M⁻¹s⁻¹ [A] [B]
Explanation:
For the reaction A + B → C
The formula for rate of reaction is:
Δ[C]/Δt = k [A] [B]
As you have [A], [B] and Δ[C]/Δt information you can multiply [A] times [B] and take this value as X and Δ[C]/Δt as Y. The slope of this lineal regression will be k.
Thus, you must obtain:
y = 3,60x10⁻² X
Thus, rate of reaction is:
B) Δ[C]/Δt = 3,60x10⁻² M⁻¹s⁻¹ [A] [B]
I hope it helps!
Answer:
Correct answer: E) (Δ[C]/Δt) = 8.37 x 10–2 M –2 s –1 [A]2 [B]
Explanation:
To determine the rate expression for the reaction it is necessary to use the initial rate method. In this is necessary to measure the initial rate (Δ[C]/ Δt) of the reaction changing the concentration of the reactants one at each time. If high concentrations of A and B are used experimentally, they will vary little from their initial value, at least during the first minutes of the reaction. Under these conditions, the initial velocity will be approximately a constant.
The general rate expression for the reaction is
(Δ[C]/Δt) = [tex]k.[A]^{\alpha}.[B]^{\beta}[/tex]
where k is the rate constant, [A] and [B] are the concentration of A and B respectively, α and β are the reaction orders of A and B respectively.
To determine the reaction orders is necessary to write the ratio between the first and the second conditions:
[tex]\frac{v_{1}}{v_{2}} =\frac{k.[A]_{1} ^{\alpha}.[B]_{1} ^{\beta}}{k.[A]_{2}^{\alpha}.[B]_{2}^{\beta}} \\ \frac{5.81x10^{-4}}{1.16x10^{-3}} = \frac{k.{(0.215M)} ^{\alpha}.(0.150M) ^{\beta}}{k.{(0.215M)}^{\alpha}.(0.300M)^{\beta}}\\\frac{5.81x10^{-4}}{1.16x10^{-3}} =\frac{(0.150M) ^{\beta}}{(0.300M)^{\beta}}\\0.500 = 0.500^{\beta}[/tex]
β = 1 thus, B has a first order.
Also, is necessary to write the ratio between the first and the third conditions:
[tex]\frac{v_{1}}{v_{3}} =\frac{k.[A]_{1} ^{\alpha}.[B]_{1} ^{\beta}}{k.[A]_{3}^{\alpha}.[B]_{3}^{\beta}} \\ \frac{5.81x10^{-4}}{2.32x10^{-3}} = \frac{k.{(0.215M)} ^{\alpha}.(0.150M) ^{\beta}}{k.{(0.430M)}^{\alpha}.(0.150M)^{\beta}}\\\frac{5.81x10^{-4}}{1.16x10^{-3}} =\frac{(0.215M) ^{\alpha}}{(0.430M)^{\alpha }}\\0.250 = 0.500^{\alpha}[/tex]
α = 2 thus, A has a second order.
Therefore, the rate expression is
(Δ[C]/Δt) = [tex]k.[A]^{2}.[B]^{1}[/tex]
Replacing the data of the first conditions to obtain the rate constant:
[tex]k = \frac{v}{k.[A]^{2}.[B]^{1}} = \frac{5.81x10-4M/s}{(0.215M)^{2} .(0.150M)} =8.37x10^{-2} M^{-2}s^{-1}[/tex]
