Initial rate of the reaction PCl5-->PCl3 + Cl2 is increased by a factor of 4 when concentration of PCl5 is doubled, therefore the rate is?

[匿名]

2017-10-20 11:00 pm

multiple choice answers are:
a. first order with respect to PCl5
b. fourth order with respect to PCl5
c. second order with respect to PCl5
d. first order with respect to PCl3

更新1:

part 2: A certain first order reaction is 46 % complete in 68 min at 25 C. What is its rate constant? a. 9.1x10^-3 min^-1 b. 1.1x10^-2 min^-1 c. 31 min^-1 d. 51 min^-1 part 3: reaction A-> products. Which is consistent with second order reaction? a. ln[A] v. time gives + slope b. ln[A] v.t time gives - slope c. 1/[A] v. time gives + slope d. 1/[A] v. time gives - slope PART 4: constant of a reaction is 0.025 M-1 s-1. [A]0=8.00 M what's the concentration of A after 25 s. Show work.

回答 (1)

胡雪8°

2017-10-20 11:32 pm

1.
PCl₅ → PCl₃ + Cl₂
Rate = k [PCl₅]ⁿ

When [PCl₅] = [PCl₅]ₒ, Rate = Rateₒ :
Rateₒ = k [PCl₅]ₒⁿ …… [1]

When [PCl₅] = 2 [PCl₅]ₒ, Rate = 4 Rateₒ :
4 Rateₒ = k (2 [PCl₅]ₒ)ⁿ …… [2]

[2]/[1] :
4 = 2ⁿ
n = 2

The answer : c. second order with respect to PCl₅

====
2.
The integrated form of rate law for first order reaction :
ln([A]/[A]ₒ) = -kt
ln(100% - 46%) = -k (68 min)
k = -[ln(0.54)]/68 min⁻¹
k = 9.1 × 10⁻³ min⁻¹

The answer : a. 9.1 × 10⁻³ min⁻¹

====
3.
The integrated form of rate law for a second order reaction :
1/[A] = kt + (1/[A]ₒ)
where 1/[A] and t are variables, while k and 1/[A]ₒ are constants.

1/[A] = kt + (1/[A]ₒ) is a linear equation in the form y = mx + c.
Ploting 1/[A] vs. t, the slope is k (k > 0) and the y-intercept is 1/[A]ₒ.

The answer : c. 1/[A] vs. time gives + slope

====
4.
Refer to the unit of the rate constant.
The reaction is second-order, i.e. Rate = k [A]²

The integration form of rate law for second-order reaction:
1/[A] = kt + (1/[A]ₒ)
1/[A] = (0.025×25) + (1/8.00) M⁻¹
[A] = 1.33 M

👍 2 👎 0