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 answer is b & c i believe : )

Answer from: Quest

chemical digestion occurs in the stomach were hcl and pepsin forms chyme, the mouth in which salivary amylase breaks down starch to form bolus, and the small intestine specifically the duodenum.

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