Question

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

[A], M [B], M Δ [C]/ Δ t (initial) M/s 0.215 0.150 5.81 x 10–4 0.215 0.300 1.16 x 10–3 0.430 0.150 2.32 x 10–3
A) (Δ[C]/Δt) = 1.80 x 10–2 M –1 s –1 [A][B]
B) (Δ[C]/Δt) = 3.60 x 10–2 M –1 s –1 [A][B]
C) (Δ[C]/Δt) = 1.20 x 10–1 M –2 s –1 [A][B]2
D) (Δ[C]/Δt) = 5.57 x 10–2 M –3 s –1 [A]2 [B]2
E) (Δ[C]/Δt) = 8.37 x 10–2 M –2 s –1 [A]2 [B]

Answer 1

Correct answer: E) (Δ[C]/Δt) = 8.37 x 10–2 M –2 s –1 [A]2 [B]

Step-by-step 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:

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

`(v_(1))/(v_(3)) =(k.[A]_(1) ^(\alpha).[B]_(1) ^(\beta))/(k.[A]_(3)^(\alpha).[B]_(3)^(\beta)) \\ (5.81x10^(-4))/(2.32x10^(-3)) = \frac{k.{(0.215M)} ^(\alpha).(0.150M) ^(\beta)}{k.{(0.430M)}^(\alpha).(0.150M)^(\beta)}\\(5.81x10^(-4))/(1.16x10^(-3)) =((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 = (v)/(k.[A]^(2).[B]^(1)) = (5.81x10-4M/s)/((0.215M)^(2) .(0.150M)) =8.37x10^(-2)  M^(-2)s^(-1)`

Answer 2

B) Δ[C]/Δt = 3,60x10⁻² M⁻¹s⁻¹ [A] [B]

Step-by-step 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!