Answer:
e- 7.25 x 10³.
Explanation:
∵ ΔG = -RTlnK,
where, ΔG is the free energy change.
R is the general gas constant (R = 8.324 J/mol.K).
K is the equilibrium constant of the reaction.
- For the reaction: <em>N₂(g) + 3H₂(g) → 2NH₃(g),</em>
K = (PNH₃)²/(PN₂)(PH₂)³ = (0.65)²/(1.9)(1.6)³ = 5.43 x 10⁻².
∵ ΔG = -RTlnK.
∴ ΔG = -(8.314 J/mol.K)(298 K) ln(5.43 x 10⁻²) = 7.218 x 10³ J/mol.
Aromatic side chain exhibits an electronic excited state that is closer in energy to the ground state.
- In order to respond to this query, we must decide whether a peptide bond or an aromatic side chain is demonstrating an electronic exited state that is more closely related to the ground state in terms of energy.
- When our energy is as low as possible, we are in the ground state.
- What I want to point out is that if we can choose between the two options—peptide bond or aromatic side chain—without knowing the specific reasons, we can immediately rule out two potential answers.
- Consider what we already know about energy, we have:
E = h x c/λ
- That indicates that when we have more energy, a wavelength decreases. Lower energy corresponds to higher wavelength.
- Aromatic side chains absorb between 250 and 290 nm, while peptide bonds do so between 190 and 250 nm.
- According to our breakdown, we have an electron excited state that is more closely related to the ground state in terms of energy as wavelength increases.
Thus, Aromatic side chain exhibits an electronic excited state that is closer in energy to the ground state.
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brainly.com/question/14483627
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