Consider the balanced chemical reaction when phosphorus and iodine react to produce phosphorus triodide: 2 P(s) + 3 I2(g) → 2 PI
1 answer:
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Answer 1) : According to the complete question attached in the answer,
The radial wave function which is denoted by shown with orange color crosses through zero point. Also, At the the radial nodes, which are spherical shells to some radial distance away from the nucleus there no electron are found.
Also, the radial probability distribution curve denoted as shown in blue color is observed to touch zero, and shows the place of radial node.
Therefore, the total number of nodes will include both the kinds which has radial and angular nodes which will be represented by <em>'n'</em>.
It is observed that for any atomic orbital, the total number of nodes will be n-1 .
Considering the s orbital of the hydrogen, which has zero angular momentum (l); (l=0), as it has zero angular nodes.
Hence, there will be only radial nodes, which is
(n−1 = total number of radial nodes in s orbitals)
According to the image, there are 4 radial nodes shown, so n = 5 (as n-1 = 4; therefore, n = 5)
This represents the 5s orbital.
Answer 2) The radial nodes are observed in I'm seeing radial nodes at
,
,
and
.
where represents the hydorgen bohr atomic radius = 0.0529177 nm
Explanation : It is quite easy to observe the given graph and find out the approximate values of the radial nodes, it does not requires any equation to be solved. Equation can be used to find the radial nodes if it was supplied along with the question. Although by mere speculation one can find out the answer.
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:
<em>Calculate ΔG at 25°C given the following sets of partial pressures.</em>
<em>Part A 130atm SO₂, 130atm O₂, 2.0atm SO₃. Express your answer using four significant figures.</em>
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
<em>Part B 5.0atm SO₂, 3.0atm O₂, 30atm SO₃ Express your answer using four significant figures.</em>
<em />
ΔG = ΔG° + R.T.lnQ = -141.8 kJ/mol + (8.314 × 10⁻³ kJ/mol.K) × 298 K × ln 296 = -127.7 kJ/mol
<em>Part C Each reactant and product at a partial pressure of 1.0 atm. Express your answer using four significant figures.</em>
<em />
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