Explain how the number of valence electrons of group a elements are related to the group number

Sulfuric acid is produced in larger amounts by weight than any other chemical. It is used in manufacturing fertilizers, oil refi

Fed [463]

Answer:

A. -166.6 kJ/mol

B. -127.7 kJ/mol

C. -133.9 kJ/mol

Explanation:

Let's consider the oxidation of sulfur dioxide.

2 SO₂(g) + O₂(g) → 2 SO₃(g) ΔG° = -141.8 kJ

The Gibbs free energy (ΔG) can be calculated using the following expression:

ΔG = ΔG° + R.T.lnQ

where,

ΔG° is the standard Gibbs free energy

R is the ideal gas constant

T is the absolute temperature (25 + 273.15 = 298.15 K)

Q is the reaction quotient

The molar concentration of each gas ([]) can be calculated from its pressure (P) using the following expression:

$[]=\frac{P}{R.T}$

Calculate ΔG at 25°C given the following sets of partial pressures.

Part A 130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=\frac{130atm}{(0.08206atm.L/mol.K).298K} =5.32M$

$[SO_{3}]=\frac{2.0atm}{(0.08206atm.L/mol.K).298K} =0.0818M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0818^{2} }{5.32^{3} } =4.44 \times 10^{-5}$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln (4.44 × 10⁻⁵) = -166.6 kJ/mol

Part B 5.0atm SO₂, 3.0atm O₂, 30atm SO₃ Express your answer using four significant figures.

$[SO_{2}]=\frac{5.0atm}{(0.08206atm.L/mol.K).298K}=0.204M$

$[O_{2}]=\frac{3.0atm}{(0.08206atm.L/mol.K).298K}=0.123M$

$[SO_{3}]=\frac{30atm}{(0.08206atm.L/mol.K).298K}=1.23M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{1.23^{2} }{0.204^{2}.0.123 } =296$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol

Part C Each reactant and product at a partial pressure of 1.0 atm. Express your answer using four significant figures.

$[SO_{2}]=[O_{2}]=[SO_{3}]=\frac{1.0atm}{(0.08206atm.L/mol.K).298K}=0.0409M$

$Q=\frac{[SO_3]^{2} }{[SO_{2}]^{2}.[O_{2}] } =\frac{0.0409^{2} }{0.0409^{3}} =24.4$

ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 24.4 = -133.9 kJ/mol

7 0

3 years ago

When the water level of a stream or river exceeds its natural banks, it has reached its

Natasha2012 [34]

Crest, as rivers in the south are currently experiencing large amounts of flooding, they are cresting.

8 0

4 years ago

Using solution theory and the aide of tables and graphs, explain why the Delta H solution for sodium chloride is favorable to th

prisoha [69]

The solution is non-ideal for NaCl soln so ΔH3 is greater than the sum of ΔH1 and ΔH2. this means the forces of attraction between like molecules is greater than the forces of attraction between unlike molecules. NaCl (table salt) dissolves readily in water. In solid NaCl, the positive sodium ions are attracted to the negative chloride ions. The same is true of the solvent, water; the partially positive hydrogen atoms are attracted to the partially negative oxygen atoms. While NaCl dissolves in water, the positive sodium cations and chloride anions are being stabilized by the water molecule electric dipoles. Thus, the intermolecular bonds between NaCl are broken and the salt is dissolved.

4 0

3 years ago

In a cycle of copper experiment, a student first reacts a piece of copper metal with nitric acid to produce copper(II) nitrate (

LuckyWell [14K]

Answer:

0.631 grams is the theoretical yield of solid copper (Cu) that can be recovered at the end of the experiment

Explanation:

The concentration of the solution is given by :

$[C]=\frac{\text{Moles of compound}}{\text{Volume of solution in Liters}}$

We have:

Concentration of copper (II) nitrate solution = $[Cu(NO_3)_2]=2.41 M$

The volume of solution = 4.12 mL

1 mL= 0.001 L

$4.12 mL= 4.12\times 0.001 L= 0.00412 L$

Moles of copper (II) nitrate in solution = n

$2.41=\frac{n}{0.00412 L}=0.0099292 mol$

Moles of copper (II) nitrate in solution = 0.0099292 mol

1 Mole of copper(II) nitrate has 1 mole of copper then 0.0099292 moles of copper(II) nitrate will have :

$1\times 0.0099292 mol= 0.0099292 \text{ mol of Cu}$

Mass of 0.0099292 moles of copper:

$=0.0099292 mol\times 63.55 g/mol=0.63100 g\approx 0.631 g$

This mass of copper present in the solution is the theoretical mass of copper present in the given copper(II) nitrate solution.

0.631 grams is the theoretical yield of solid copper (Cu) that can be recovered at the end of the experiment

7 0

3 years ago

How does the electron structure of each substance affect the properties of compounds that it forms?

VashaNatasha [74]

Answer:

All of these properties are due to the chemical structure of the compound. The chemical structure includes the bonding angle, the type of bonds, the size of the molecule, and the interactions between molecules. Slight changes in the chemical structure can drastically affect the properties of the compound.Some atoms become more stable by gaining or losing an entire electron (or several electrons). When they do so, atoms form ions, or charged particles. Electron gain or loss can give an atom a filled outermost electron shell and make it energetically more stable.

Explanation:

6 0

3 years ago