Chemistry, 26.08.2019 14:50 kenziedil2637

Calculate the initial rate for the formation of c at 25 ∘c, if [a]=0.50m and [b]=0.075m

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Answer from: nixki333

The rate of formation of a product depends on the the concentrations of the reactants in a variable way.

If two products, call them A and B react together to form product C, a general equation for the formation of C has the form:

rate = k*[A]^m * [B]^n

Where the symbol [ ] is the concentration of each compound.

Then, plus the concentrations of compounds A and B you need k, m and n.

Normally you run controled trials in lab which permit to calculate k, m and n .

Here the data obtained in the lab are:

Trial      [A]     [B]     Rate
            (M)    (M)      (M/s)
1     0.50 0.010      3.0×10−3
2        0.50 0.020       6.0×10−3
3       1.00 0 .010       1.2×10−2

Given that for trials 1 and 2 [A] is the same you can use those values to find n, in this way

rate 1 = 3.0 * 10^ -3 = k [A1]^m * [B1]^n

rate 2 = 6.0*10^-3 = k [A2]^m * [B2]^n

divide rate / rate 1 => 2 = [B1]^n / [B2]^n

[B1] = 0.010 and [B2] = 0.020 =>

6.0 / 3.0 =( 0.020 / 0.010)^n =>

2 = 2^n => n = 1

Given that for data 1 and 3 [B] is the same, you use those data to find m

rate 3 / rate 1 = 12 / 3.0 = (1.0)^m / (0.5)^m =>

4 = 2^m => m = 2

Now use any of the data to find k

With the first trial: rate = 3*10^-3 m/s = k (0.5)^2 * (0.1) =>

k = 3.0*10^-3 m/s / 0.025 m^3 = 0.12 m^-2 s^-1

Now that you have k, m and n you can use the formula of the rate with the concentrations given

rate = k[A]^2*[B] = 0.12 m^-2 s^-1 * (0.50m)^2 * (0.075m) = 0.0045 m/s = 4.5*10^=3 m/s

4.5 * 10^-3 m/s

Answer from: mountainwilson

The initial rate for the formation of 'c' at 25 $\rm ^\circ C$ has been calculated to be 2.25 $\rm \times\;10^-^2$ M/s.

The rate of formation of the product has been dependent on the concentration of reactants.

The given reaction has been:

A + 2B $\rightarrow$ C

The rate of reaction can be given as:

Rate = k $\rm [A]^a\;[B]^b$

By the experimental data:

[A] [B] Rate

0.5 M 0.01 M 3 $\rm \times\;10^-^3$ M/s

0.5 M 0.02 M 6 $\rm \times\;10^-^3$ M/s

1 M 0.0.1 M 1.2 $\rm \times\;10^-^2$ M/s

For the value of [A] = 0.5 M,

The possible rate can be: 3 $\rm \times\;10^-^3$ M/s or 6 $\rm \times\;10^-^3$ M/s

The ratio of two possible rates has been directly proportional to the concentration of b.

$\rm \dfrac{Rate\;2}{Rate\;1}\;=\;\dfrac{[B2]^b}{[B1]^b}$

$\rm \dfrac{6.0\;\times\;10^-^3}{3.0\;\times\;10^-^3}\;=\;(\dfrac{0.02}{0.01})^b$

2 = $\rm 2^b$

b = 1

Since rate 3 and rate 1 have the same concentration of [B], the possible value for a can be calculated with their ratio.

$\rm \dfrac{Rate\;3}{Rate\;1}\;=\;\dfrac{[A3]^a}{[A1]^a}$

$\rm \dfrac{1.2\;\times\;10^-^2}{3.0\;\times\;10^-^3}\;=\;(\dfrac{1}{0.5})^a$

4 = $\rm 2^a$

a = 2

The value of rate constant k can be calculated by substituting the values for the rate 1.

Rate = k $\rm [A]^a\;[B]^b$

3 $\rm \times\;10^-^3$ M/s = k $\rm [0.5]^2\;[0.01]^1$ M

3 $\rm \times\;10^-^3$ = k (0.25) (0.01)

3 $\rm \times\;10^-^3$ = 0.0025 k

k = 1.2 $\rm M^-^2\;s^-^1$

To calculate the initial rate for the formation of c at [A] = 0.5 M, and [B] = 0.075 M.

Rate = 1.2 $\rm [0.5]^2\;[0.075]^1$

Rate = 0.0225 M/s

Rate = 2.25 $\rm \times\;10^-^2$ M/s

The initial rate for the formation of 'c' at 25 $\rm ^\circ C$ has been calculated to be 2.25 $\rm \times\;10^-^2$ M/s.

For more information about the rate of reaction, refer to the link:

link

Answer from: Quest

- "the choice that is not an indicator of a chemical change is "state of matter changes". more common than not, chemical reactions produce energy in the form of light or heat. along with energy, they also produce a new substance called the product that could be in any state of matter (solid, gas, or liquid)."

Answer from: Quest

im pretty sure its b

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Calculate the initial rate for the formation of c at 25 ∘c, if [a]=0.50m and [b]=0.075m...

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