Question

Given the data calculated in Parts A, B, C, and D, determine the initial rate for a reaction that starts with 0.45 M of reagent A and 0.90 M of reagents B and C?

Trial [A]

(M) [B]

(M) [C]

(M) Initial rate

(M/s)

1 0.20 0.20 0.20 6.0×10−5

2 0.20 0.20 0.60 1.8×10−4

3 0.40 0.20 0.20 2.4×10−4

4 0.40 0.40 0.20 2.4×10−4

Answer 1

`\large\boxed{\large\boxed{0.0014M/s}}`

Step-by-step explanation:

From the table, first find the order of reaction, then find the rate constant, write the rate equation, and, finally, subsititue the data for the reaction that starts with 0.45M of reagent A and 0.90 M of reagents B and C.

1. Table

Trial [A] (M) [B] (M) [C] (M) Initial rate (M/s)

1 0.20 0.20 0.20 6.0×10⁻⁵

2 0.20 0.20 0.60 1.8×10⁻⁴

3 0.40 0.20 0.20 2.4×10⁻⁴

4 0.40 0.40 0.20 2.4×10⁻⁴

2. Orders

a) From trials 3 and 4 you learn that the initial concentration of B does not change the change teh rate of the reaction. Hence the order with respect to [B] is 0.

b) From trials 1 and 2 you learn that when the concentration of C is tripled the rate of reaction is also tripled:

• 0.60 / 0.2 = 3, and

• 1.8×10⁻⁴ / 6.0×10⁻⁵ = 3

Hence, the order with respect to C is 1.

c) From trials 1 and 3 you get:

• 0.40/0.2 = 2

• 2.4×10⁻⁴ / 6.0×10⁻⁵ = 4

Which means that when the concentration of A is doubled, the rate of the reaction is quadruplicated. Hence, the order of the reaction with respect to A is 2.

3. Rate equation

Ther orders are:

`a=2\\\\b=0\\\\c=1`

Hence the rate is:

`rate=k[A]^a{B}^b[C]^c\\ \\ rate=k[A]^2[B]^0[C]^1=k[A]^2C`

4. Rate constant, k

You can use any trial to find the value of the constant, k

Using trial 1:

`6.0* 10^(-5)M/s=k(0.20M)^2(0.20M)\\ \\ k=(6.0* 10^(-5)M/s)/((0.20M)^2(0.20M))=0.0075M^(-2)s^(-1)`

5. Rate law:

`rate=k[A]^2C=0.0075[A]^2[C]`

6. Substitute

Subsititue the data for the reaction that starts with 0.45M of reagent A and 0.90 M of reagents B and C.

`rate=0.0075M^(-2)s^(-1)[A]^2[C]=0.0075M^(-2)s^(-1)[0.45M]^2[0.9M]`

`r=0.00136688M/s\approx 0.0014M/s`