Under special conditions, sulfur reacts with anhydrous liquid ammonia to form a binary compound of sulfur and nitrogen. The compound is found to consist of \(69.6 \% \mathrm{~S}\) and \(30.4 \% \mathrm{~N}\). Measurements of its molecular mass yield a value of \(184.3 \mathrm{~g} \mathrm{~mol}^{-1}\). The compound occasionally detonates on being struck or when heated rapidly. The sulfur and nitrogen atoms of the molecule are joined in a ring. All the bonds in the ring are of the same length. (a) Calculate the empirical and molecular formulas for the substance. (b) Write Lewis structures for the molecule, based on the information you are given. (Hint: You should find a relatively small number of dominant Lewis structures.) (c) Predict the bond distances between the atoms in the ring. (Note: The \(\mathrm{S}-\mathrm{S}\) distance in the \(\mathrm{S}_{8}\) ring is \(2.05 \AA\). \()\) (d) The enthalpy of formation of the compound is estimated to be \(480 \mathrm{~kJ} \mathrm{~mol}^{-1} . \Delta H_{f}^{\circ}\) of \(\mathrm{S}(g)\) is \(222.8 \mathrm{~kJ} \mathrm{~mol}^{-1}\). Estimate the average bond enthalpy in the compound.

Step by step solution

01

Calculate the empirical formula

To find the empirical formula, assume 100 g of the compound, which means that we have 69.6 g of sulfur and 30.4 g of nitrogen. Then, convert these masses into moles using the molar masses of sulfur and nitrogen. The molar mass of sulfur is 32.07 g/mol, and the molar mass of nitrogen is 14.01 g/mol. Moles of sulfur = mass / molar mass Moles of sulfur = \(69.6\mathrm{~g}/ 32.07\mathrm{~g/mol} = 2.17\) moles Moles of nitrogen = mass / molar mass Moles of nitrogen = \(30.4\mathrm{~g}/14.01\mathrm{~g/mol} = 2.17\) moles Now, divide both the moles of sulfur and nitrogen by their least value to find their ratio. Since we already have their ratio as 1:1, the empirical formula would be \(\mathrm{SN}\).

02

Calculate the molecular formula

Now we have to find the molecular formula. We will use the molecular mass given in the problem: 184.3 g/mol. Divide the molecular mass by the mass of the empirical formula: \(\frac{184.3\mathrm{~g/mol}}{(32.07+14.01)\mathrm{~g/mol}} = \frac{184.3}{46.08} \approx 4\) The molecular formula is four times the empirical formula, which means the molecular formula is \(\mathrm{S_4N_4}\).

03

Write Lewis structures

To create the Lewis structure for \(\mathrm{S_4N_4}\), start by counting the valence electrons. Sulfur has six valence electrons, and nitrogen has five valence electrons. Total valence electrons = \((6 * 4) + (5 * 4) = 44\) The given hint is that there are a relatively small number of dominant Lewis structures. We'll also consider that all the bonds in the ring are of the same length. Consider a single nitrogen atom bonded to two sulfur atoms and another nitrogen atom. This means it forms a double bond with one sulfur and a single bond with the other sulfur and the other nitrogen. The sulfur atom forms single bonds with two nitrogen atoms. This ring structure would have eight pairs of electrons. Two dominant Lewis structures can be drawn for the molecule by switching the double and single bonds in the ring.

04

Predict the bond distances

Predict the bond distances between the atoms in the ring by averaging the known bond distances. The S-S distance in S8 ring is given as 2.05 Å. We will average the distances between S-S, S-N, and N-N bonds. Since we do not have the exact bond distance values for S-N and N-N, the average bond distance could be calculated approximately. The resulting average bond distance is around \(\textbf{2 Å}\).

05

Estimate the average bond enthalpy in the compound

To estimate the average bond enthalpy, we are given the enthalpy of formation and the ΔH for S(g). We can use the energy change equation to find the energy change for S and N in the compound: ΔH = Σ(Product side bond enthalpy - Reactant side bond enthalpy) The atoms form 4 S-N bonds in the molecule. In S8, S atoms form 16 S-S bonds: ΔH = 480 kJ/mol - (4 * Average bond enthalpy) + 222.8 kJ/mol * (4 * S - 16 * S/8) Now, solve for the average bond enthalpy: \(Average\ bond\ enthalpy = \frac{(480-889.6)}{4} \approx \textbf{-102.4 kJ/mol}\)

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