Perhatikan persamaan termokimia berikut.
1 answer:
The chemical reaction is:
HCl (aq) + NaOH (aq) → NaCl (aq) + H₂O (l).
The enthalpy change for this reaction is ΔH = -54 kJ.
This means that reacting one mole of HCl with one mole of NaOH releases 54 kJ of heat energy.
This is an exothermic reaction, and its energy profile resembles that shown in figure (1).
The question asks:
What is the enthalpy change when 10 mL of 1 M HCl reacts with 20 mL of 1 M NaOH?
Calculating moles of HCl: molarity × volume = 1 M × 10 mL = 10 millimoles.
Assuming complete reaction, the moles of NaOH reacted equals moles of HCl.
Therefore, total moles that reacted = 10 millimoles, producing the same amount of water.
Since one mole of acid-base reaction produces one mole of water, formation of 10 millimoles of water releases energy:
54 × 10 × 10⁻³ = 0.54 kJ
Reaction:
HCl (aq) + NaOH (aq) → NaCl (aq) + H₂O (l).
Enthalpy change: ΔH = -54 kJ.
One mole of HCl reacts with one mole of NaOH and liberates 54 kJ heat.
The reaction is exothermic, with the graph similar to figure (1).
Question:
What is the enthalpy change when 10 mL of 1 M HCl reacts with 20 mL of 1 M NaOH?
Moles of HCl = molarity × volume = 1 M × 10 mL = 10 millimoles.
Moles of NaOH equate to moles reacted.
Total acid-base reaction moles = 10 mmol, equating to water produced.
Since one mole acid-base reaction produces 1 mole of water, the energy released for 10 mmol water is:
54 × 10 × 10⁻³ = 0.54 kJ
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Answer:
= - 0.38 eV
Explanation:
According to Bohr's equation for the energy corresponding to an electron in the nth orbital,
E = -13.6
Where E represents energy level in electron volts (eV)
Z symbolizes the atomic number of the atom
n denotes the principal quantum number
For n = 6
⇒ E = -13.6 ×
= - 0.38 eV
I hope this information helps.