Question

Given the initial rate data for the reaction A + B –––> C, determine the rate expression for the reaction.

Answer 1

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 2

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) = $k.[A]^{\alpha}.[B]^{\beta}$

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:

$\frac{v_{1}}{v_{2}} =\frac{k.[A]_{1} ^{\alpha}.[B]_{1} ^{\beta}}{k.[A]_{2}^{\alpha}.[B]_{2}^{\beta}} \\ \frac{5.81x10^{-4}}{1.16x10^{-3}} = \frac{k.{(0.215M)} ^{\alpha}.(0.150M) ^{\beta}}{k.{(0.215M)}^{\alpha}.(0.300M)^{\beta}}\\\frac{5.81x10^{-4}}{1.16x10^{-3}} =\frac{(0.150M) ^{\beta}}{(0.300M)^{\beta}}\\0.500 = 0.500^{\beta}$

β = 1 thus, B has a first order.

Also, is necessary to write the ratio between the first and the third conditions:

$\frac{v_{1}}{v_{3}} =\frac{k.[A]_{1} ^{\alpha}.[B]_{1} ^{\beta}}{k.[A]_{3}^{\alpha}.[B]_{3}^{\beta}} \\ \frac{5.81x10^{-4}}{2.32x10^{-3}} = \frac{k.{(0.215M)} ^{\alpha}.(0.150M) ^{\beta}}{k.{(0.430M)}^{\alpha}.(0.150M)^{\beta}}\\\frac{5.81x10^{-4}}{1.16x10^{-3}} =\frac{(0.215M) ^{\alpha}}{(0.430M)^{\alpha }}\\0.250 = 0.500^{\alpha}$

α = 2 thus, A has a second order.

Therefore, the rate expression is

(Δ[C]/Δt) = $k.[A]^{2}.[B]^{1}$

Replacing the data of the first conditions to obtain the rate constant:

$k = \frac{v}{k.[A]^{2}.[B]^{1}} = \frac{5.81x10-4M/s}{(0.215M)^{2} .(0.150M)} =8.37x10^{-2} M^{-2}s^{-1}$