Question

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

Answer 1

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

Answer: 4.5 * 10^-3 m/s

Answer 2

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

$\large {\boxed {\boxed {\bold {v ~ = ~ - \frac {\Delta A} {\Delta t}}}}$

or

$\large {\boxed {\boxed {\bold {v ~ = ~ + \frac {\Delta B} {\Delta t}}}}$

A = reagent

B = product

v = reaction rate

t = reaction time

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

Reaction speed can be formulated:

$\large {\boxed {\boxed {\bold {v ~ = ~ k. [A] ^ a [B] ^ b}}}$

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

$\frac{v_3}{v_1}=\frac{[B_3]}{[B_1]}$

$\frac{5.8}{1.4}=[\frac{0.4}{0.2}]^x$

4 = 2ᵃ

a = 2

• 2. look for the reaction order b

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

$\frac{v_2}{v_1}=\frac{[B_3]}{[B_1]}$

$\frac{2.9}{1.4}=[\frac{0.06}{0.03}]^b$

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]

$k=\frac{1.4.10^{-3}}{0.04\:.\:0.03}$

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

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Keywords: reaction rate, reaction order, molar concentration, products, reactants