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

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

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

If products A and B produce product C, a general equation for the formation of C is of the kind rate = k*[A]^a * [B]^b

[ ] is used for the concentration of each compound.

When k, m and n are unknown, chemists run lab trials to calculate them..

You have these data from 3 trials

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

Trials 1 and 2 are run at constant [A] which permits to calculate the exponent b, in this way

rate 1 = 3.0 * 10^ -3 = k [A1]^a * [B1]^b

rate 2 = 6.0*10^-3 = k [A2]^a * [B2]^b

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

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

6.0 / 3.0  =( 0.020 / 0.010)^b => 2 = 2^b => b = 1

In the same way trials 1 and 3, which were run at constan [B], are used to calculate a

rate 3 / rate 1 = 12 / 3.0   = (1.0)^a / (0.5)^a => 4 = 2^a => 2^2 = 2^a => 2 = a

Now use any of the data to find k

With the second trial: rate = 6*10^-3 m/s = k (0.5)^2 * (0.02) =>

k = 6.0*10^-3 M/s / (0.05 M^3) = 0.12 M^-2 s^-1

With the calcualted values of k, a and b you 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

2.3*10^-2 M/s IS THE ANWSER

The initial rate for the formation of C : v = 2.194.10⁻² M/s

Further explanation

The reaction rate (v) shows the change in the concentration of the substance (changes in addition to concentrations for reaction products or changes in concentration reduction for reactants) per unit time.

Can be formulated:

Reaction: aA ---> bB or A = reagent

B = product

v = reaction rate

t = reaction time

For A + B reactions ---> C + D

Reaction speed can be formulated: where

v = reaction speed, M / s

k = constant, mol¹⁻⁽ᵃ⁺ᵇ⁾. L⁽ᵃ⁺ᵇ⁾⁻¹. S⁻¹

a = reaction order to A

b = reaction order to B

[A] = [B] = concentration of substances

Reactions that occur:

A + 2B --> C

So the reaction speed equation:

v = k. [A]ᵃ [B]ᵇ

1. look for the reaction order a

we look for the same data [B], namely data 1 and 3  4 = 2ᵃ

a = 2

2. look for the reaction order b

we look for the same data [A], namely data 1 and 2  2 = 2ᵇ

b = 1

So the rate reaction

v = k .[A]²[B]

To find K , use data 1

1.4.10⁻³ = k [0.2]²[0.03] k = 1.17 M⁻¹S⁻¹

The initial rate for the formation of c at 25 ∘c, if [A]=0.50 and [B]=0.075

v  = 1.17 [0.5]²[0.075]

v = 2.194.10⁻² M/s

the factor can decrease the rate of a chemical reaction

increase the rate of a chemical reaction

Which of the following does not influence the effectiveness of a detergent

Keywords: reaction rate, reaction order, molar concentration, products, reactants  The initial rate for the formation of C at 25°C, if [A]=0.50M and [B]=0.075M is 0.022 M/s

Further explanation

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50M and [B]=0.075M. Express your answer to two significant figures and include the appropriate units. Consider the reaction

A+2B → C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

Trial                    [A]                    [B]                    Rate

(M)                   (M)                    (M/s)

1                       0.20                 0.030                1.4 x 10 - 3

2                       0.20                 0.060                2.9 x 10      - 3

3                       0.40                 0.030                5.8 x 10       - 3 We will take a look on 1st and 3rd data.  because   have units of so (2 significant values)

Subject:  chemistry

Chapter:  chemical kinetics

Keywords: the initial rate, formation, c, time, M

The question is incomplete, here is the complete question:

Calculate the initial rate for the formation of C at 25°C, if [A]=0.50 M and [B]=0.075 M. Express your answer to two significant figures and include the appropriate units.Consider the reaction

A + 2B ⇔ C

whose rate at 25°C was measured using three different sets of initial concentrations as listed in the following table:

The table is attached below as an image.

The initial rate for the formation of C at 25°C is Explanation:

Rate law is defined as the expression which expresses the rate of the reaction in terms of molar concentration of the reactants with each term raised to the power their stoichiometric coefficient of that reactant in the balanced chemical equation.

For the given chemical equation: Rate law expression for the reaction: where,

a = order with respect to A

b = order with respect to B

Expression for rate law for first trial: ....(1)

Expression for rate law for second trial: ....(2)

Expression for rate law for third trial: ....(3)

Dividing 2 by 1, we get: Dividing 3 by 1, we get: Thus, the rate law becomes: ......(4)

Now, calculating the value of 'k' by using any expression.

Putting values in equation 1, we get: Calculating the initial rate of formation of C by using equation 4, we get: [A] = 0.50 M

[B] = 0.075 M

Putting values in equation 4, we get: Hence, the initial rate for the formation of C at 25°C is  Rate = k * [A]^2 * [B]^1

Use the data from any trial to calculate k.

k = (rate)/([A]^2 * [B]^1)

E.g., for Trial 1, we have

rate = 3.0×10−3 M/s
[A] = 0.50 M
[B] = 0.010 M

Plug those numbers in and crank out the answer.

Now with the calculated value of k, calculate the initial rate for [A] = 0.50 M and [B] = 0.075 M

rate = k * [A]^2 * [B]^1

k = calculated value
[A] = 0.50 M
[B] = 0.075 M

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

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