For the reaction a+b+c→d+e, the initial reaction rate was measured for various initial concentrations of reactants. the following data were collected: trial [a] (m) [b] (m) [c] (m) initial rate (m/s) 1 0.10 0.10 0.10 3.0×10−5 2 0.10 0.10 0.30 9.0×10−5 3 0.20 0.10 0.10 1.2×10−4 4 0.20 0.20 0.10 1.2×10−4what is the value of the rate constant k for this reaction? when entering compound units, indicate multiplication of units explicitly using a multiplication dot (multiplication dot in the menu). for example, m−1⋅s−1. express your answer to two significant figures and include the appropriate units. indicate the multiplication of units explicitly either with a multiplication dot or a dash.

k = 0.030 M⁻²· s⁻¹

Explanation:

Trial    [A] (M)        [B] (M)     [C]( M)    Rate (M/s)

1           0.10           0.10         0.10          3.0 x 10-5

2          0.10           0.10         0.30         9.0 x 10-5

3          0.20          0.10         0.10           1.2 x 10-4

4          0.20          0.20        0.10           1.2 x 10-4

The rate law for the reaction can be written as:

rate = k[A]ᵃ[B]ᵇ[C]ⁿ

To obtain k and the exponents, let´s first write the rates for trials 1 and 2:

rate 1 =  3.0 ×10⁻⁵ = k (0.10)ᵃ·(0.10)ᵇ·(0.10)ⁿ

rate 2 = 9.0 ×10⁻⁵ = k (0.10)ᵃ·(0.10)ᵇ·(0.30)ⁿ     (notice that 0.30 = 3·0.10) Then:

rate 2= 9.0 ×10⁻⁵ = k (0.10)ᵃ·(0.10)ᵇ·(3·0.10)ⁿ

Distributing the exponent:

rate 2 = 9.0 ×10⁻⁵ = k (0.10)ᵃ·(0.10)ᵇ·3ⁿ·(0.10)ⁿ

If we divide rate 2 by rate 1:

rate 2 / rate 1 = 9.0 ×10⁻⁵/ 3.0 ×10⁻⁵ = 3

3 = k (0.10)ᵃ·(0.10)ᵇ·3ⁿ·(0.10)ⁿ / k (0.10)ᵃ·(0.10)ᵇ·(0.10)ⁿ

3 = 3ⁿ

n = 1

In the same way, we can proceed with rate 3 and rate 1 to obtain the exponent "a". Notice that (0.20)ᵃ = 2ᵃ·(0.10)ᵃ

rate 3 / rate 1 =   1.2 x 10⁻⁴ / 3.0 x 10⁻⁵ = 4

4 = k·2ᵃ·(0.10)ᵃ·(0.10)ᵇ·(0.10) /  k (0.10)ᵃ·(0.10)ᵇ·(0.10)

4 = 2ᵃ

a = 2

To obtain "b" we use trial 4 and trial and trial 3:

rate 4 / rate 3 = 1.2 x 10⁻⁴ / 1.2 x 10⁻⁴  = 1

1 = k·(0.2)²·2ᵇ·(0.1)ᵇ·(0.1) / k·(0.2)²·(0.1)ᵇ·(0.1)

1 = 2ᵇ

b = 0

The rate law will be:

rate = k[A]²[B]⁰[C]

Solving for k:

k = rate / [A]²[B]⁰[C]

Now, let´s calculate k for each trial. The value of k will be the average value:

k1 = 3.0 ×10⁻⁵ M/s / (0.10 M)²·(0.10 M)⁰·(0.10 M)

k1 = 3.0 ×10⁻⁵ M/s/ 1.0 × 10⁻³ M³ = 0.030 M⁻²· s⁻¹

k2 = 9.0 ×10⁻⁵ M/s / (0.10 M)²·(0.10 M)⁰·(0.30 M) = 0.030 M⁻²· s⁻¹

k3 = 1.2 x 10⁻⁴ M/s / (0.20 M)²·(0.10 M)⁰·(0.10 M) = 0.030 M⁻²· s⁻¹

k4 =  1.2 x 10⁻⁴ M/s / (0.20 M)²·(0.20 M)⁰·(0.10 M) = 0.030 M⁻²· s⁻¹

The value of k is  0.030 M⁻²· s⁻¹

assuming that your periodic table has eighteen columns, you should highlight   the four elements in the third column from the left of the table.

neutral atoms of elements after calcium contain d electrons. the energy of d orbitals of a particular main shell lies between the energy of the s and p orbitals of the next main shell. for example, electrons in the have energy higher than those in the orbital while lower than those of the orbital.

by the aufbau principle, electrons fill orbitals one at a time, starting from those of the lowest energy. the orbital is therefore filled right after orbitals and before orbitals.  

potassium has electron configuration and calcium . the s orbital of a main shell holds a maximum of two electrons. the next element would thus have the electron configuration and satisfy the requirements.

a neutral atom of the element contains three electrons on top of the most recent noble gas element. it should thus be placed in the third column of the periodic table. both scandium and yttrium would satisfy the conditions. an aufbau diagram indicates that the 4f orbital is filled before the 5d. however, it appears that exceptions exists for both lanthanum and actinium. neutral atoms of the two d-block elements also demonstrate the and should be highlighted as well.