Answer:
(C) It is favorable and is driven by ΔH° only.
Explanation:
Let's consider the following balanced equation.
2 Na₂O₂ + S + 2 H₂O → 4 NaOH + SO₂
To determine whether it will be favorable or not at 298 K, we need to calculate the standard Gibbs free energy (ΔG°).
- If ΔG° < 0, the reaction will be favorable.
- If ΔG° > 0, the reaction will be unfavorable.
We can calculate ΔG° using the following expression.
ΔG° = ΔH° - T.ΔS°
As we can see from the expression above, the favorability will be driven if ΔH° < 0 and if ΔS° < 0. Since ΔH° = -610 kJ/mol and ΔS° = -73 J/K.mol, the favorability will be driven by ΔH°. Now, let's calculate the overall favorability.
ΔG° = ΔH° - T.ΔS°
ΔG° = -610 kJ/mol - 298 K.(-0.073 kJ/K.mol) = -588 kJ/mol
The reaction is favorable and is driven by ΔH° only.
Answer:
Water (H2O) and Sodium (Na) are reactants.
Explanation:
2H2O+2Na......2NaOH+H2
In the above reaction the water (H2O) and sodium (Na) are reactants and sodium hydroxide (NaOH) and hydrogen are products.
And this is the single displacement reaction.
Answer:
74 litre
Explanation:
using ideal gas eqation PV=nRT
here P(pressure)=81.8 kPa =81.8×10^3 Pa
moles=2.5
temperature=273.15+18=291.15K
Gas constant R=8.314m^3-Pa/K-mol
now, V=nRT/P = 8.314×2.5×291.5/81.8×10^3 ≈74litre
✌️;)
Answer:
Boyle's Law, Charles' Law, and Gay-Lussac's Law
Explanation:
