1. Spectroscopic studies of the reaction of azide ion with the cluster [Ru3(CO)12] in acetone solvent confirmed the occurrence of process (i) under a CO atmosphere. Kinetic measurements of this rapid [Ru3(CO)12]+ N3– →[Ru3(NCO)(CO)11]–+ N2(i) reaction revealed the rate law, Rate =k[Ru][N3–], and the activation parameters ΔH‡= 61.6 ± 3.4 kJ mol–1 and ΔS‡= 3.5 ± 11.8 J K–1 mol–1. These results are consistent with the addition of N3– to a CO ligand of [Ru3(CO)12] to form an intermediate under either steady-state or pre-equilibrium conditions etc.

Question

1. Spectroscopic studies of the reaction of azide ion with the cluster [Ru3(CO)12] in acetone solvent confirmed the occurrence of process (i) under a CO atmosphere. Kinetic measurements of this rapid [Ru3(CO)12]+ N3– →[Ru3(NCO)(CO)11]–+ N2(i) reaction revealed the rate law, Rate =k[Ru][N3–], and the activation parameters ΔH‡= 61.6 ± 3.4 kJ mol–1 and ΔS‡= 3.5 ± 11.8 J K–1 mol–1. These results are consistent with the addition of N3– to a CO ligand of [Ru3(CO)12] to form an intermediate under either steady-state or pre-equilibrium conditions etc.

a. What is the rate constant (with units) if the rate of reaction is 1.23 Ms- and the concentration of the Ru cluster is .050 M and the azide concentration is .025 M?

b. Ignoring the uncertainty values, what is the free energy value at 298K? Is the reaction spontaneous?

2. What are the three main intermolecular forces? Which one describes the intermolecular forces in water? In terms of enthalpy and entropy,why does sodium azide dissolve spontaneously in water?

Answer 1

a. To calculate the rate constant, we can use the given rate law:

Rate = k[Ru][N3-]

Given:
Rate = 1.23 Ms-1 (1 Ms-1 = 1 mol/s)
[Ru] = 0.050 M
[N3-] = 0.025 M

From the rate law, we know that the rate of the reaction is directly proportional to the concentration of both the Ru cluster and the azide ion. Therefore, we can rewrite the rate law as:

1.23 Ms-1 = k * (0.050 M) * (0.025 M)

Simplifying, we find:

k = (1.23 Ms-1) / ((0.050 M) * (0.025 M))
k = 984 Ms-1

Therefore, the rate constant is 984 Ms-1.

b. To determine the free energy value at 298K, we need the equation:

ΔG = ΔH - TΔS

Given:
ΔH‡ = 61.6 kJ mol-1 (ignore uncertainty values)
ΔS‡ = 3.5 J K-1 mol-1 (ignore uncertainty values)
T = 298 K

Converting the enthalpy to J/mol:
ΔH‡ = 61.6 kJ mol-1 * 1000 J/kJ = 61,600 J mol-1

Plugging the values into the equation, we find:

ΔG = (61,600 J mol-1) - (298 K * 3.5 J K-1 mol-1)
ΔG = 61,600 J mol-1 - 1,043 J mol-1
ΔG = 60,557 J mol-1

To determine if the reaction is spontaneous, we need to compare ΔG to zero. If ΔG is negative, the reaction is spontaneous. If ΔG is positive, the reaction is non-spontaneous. In this case, ΔG = 60,557 J mol-1, which is positive. Therefore, the reaction is non-spontaneous under these conditions.