Answer:
See explanation
Explanation:
The third law of thermodynamics states that "the entropy of a perfect crystal of a pure substance approaches zero as the temperature approaches zero" (Wikipedia).
One example of the third law of thermodynamics has to do with steam. Steam is gaseous water. Since it is a gas, its molecules are free to move around therefore its entropy is high. When the temperature of the steam is decreased below 100 degrees, the molecules of steam loose energy and turn into liquid water and do not move as freely as they did in the gaseous state. If the temperature is further decreased to yield ice at zero degrees, the molecules of water are "frozen" in their positions and the entropy of the system decreases to zero.
Also, the ions in ionic crystal solids move around when the substance is in solution or in molten state hence the substance conducts electricity. When the ionic substance is in solid state, the ions do not move about and the entropy of the solid system tends towards zero.
Answer:
-122 J/K
Explanation:
Let's consider the following balanced reaction.
N₂(g) + 2 O₂(g) ⇒ 2 NO₂(g)
We can calculate the standard reaction entropy (ΔS°) using the following expression.
ΔS° = Σ ηp × Sf°p - Σ ηr × Sf°r
where,
- η: stoichiometric coefficients of products and reactants
- Sf°r: entropies of formation of products and reactants
ΔS° = 2 mol × 240.06 J/K.mol - 1 mol × 191.61 J/K.mol - 2 mol × 205.14 J/K.mol
ΔS° = -121.77 J/K ≈ -122 J/K
<u>Given:</u>
Mass of Ba = 1.50 g
Mass of H2O = 100.0 g
Initial temp T1 = 22 C
Final Temp T2 = 33.1 C
specific heat c = 4.18 J/g c
<u>To determine:</u>
The reaction enthalpy
<u>Explanation:</u>
The heat released during the reaction is:
q = - mc(T2-T1) = - (100+1.5) g *4.18 J/g C * (33.1-22) C = -4709.4 J
# moles of Ba = Mass of Ba/Atomic mass of Ba = 1.5 g/137 g.mol-1 = 0.0109 moles
ΔH = q/mole = - 4709.4 J/0.0109 moles = - 432 kJ/mol
Ans : The enthalpy change for the reaction is -432 kJ/mol
This is the actual answer
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Methane, nitrogen, dark matter, and radiation.
