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

c is having 4 valence electrons and is bonded to three hydrogens and one sulphur atom, therefore, the number of bond pair is equal to four. as all the valence electrons are used in forming the bonds so, there is no lone pair left on carbon.

s is having 6 valence electrons and is bonded to one other sulfur atom and one carbon atom, therefore, the number of bond pair is equal to 2. as only two valence electrons are used in forming the bonds so, four electrons are left. therefore, two lone pairs are present on s.