Nitrogen (N2) and hydrogen (H2) gases react to form ammonia, which requires -99.4 J/K of standard entropy (ΔS°).
What is standard entropy?
The difference between the total standard entropies of the reaction mixture and the summation of the standard entropies of the outputs is the standard entropy change. Each entropy in the balanced equation needs to be compounded by its coefficient, as shown by the letter "n."
Calculation:
Balancing the given reaction following-
1/2 N₂(g) + 3/2 H₂ (g)→ NH₃ (g)
ΔS° = [1 mol x S° (NH₃)g] - [1/2 mol x S° (N₂)g] - [3/2 mol x S°(H₂)g]
Here S° = standard entropy of the system
Insert into the aforementioned equation all the typical entropy values found in the literature:
ΔS° = [1 mol x 192.45 J/mol.K] - [1/2 mol x 191.61 J/mol.K] - [3/2 mol x 130.684 J/mol.K]
⇒ΔS° = - 99.4 J/K
Therefore, the standard entropy, ΔS° is -99.4 J/K.
Learn more about standard entropy here:
brainly.com/question/14356933
#SPJ4
Explanation:
A. Hydrogen bonding is present in CS2 but not in CO2.
B. CS2 has greater dipole moment than CO2 and thus the dipole-dipole forces in CS2 are stronger.
C. CS2 partly dissociates to form ions and CO2 does not. Therefore, ion-dipole interactions are present in CS2 but not in CO2.
D. The dispersion forces are greater in CS2 than in CO2.
<u><em>PLS MARK BRAINLIEST :D</em></u>
Answer:
✅
Explanation:
DON'T USE DRUGS BECAUSE IT IS GOING DEATH
That would be phosphorus. It’s electron configuration is 1s^2 2s^2 2p^6 3s^2 3p^3
The final volume V₂=4.962 L
<h3>Further explanation</h3>
Given
T₁=20 + 273 = 293 K
P₁= 1 atm
V₁ = 4 L
T₂=100+273 = 373 K
P₂=780 torr=1,02632 atm
Required
The final volume
Solution
Combined gas law :
P₁V₁/T₁=P₂V₂/T₂
Input the value :
V₂=(P₁V₁T₂)/(P₂T₁)
V₂=(1 x 4 x 373)/(1.02632 x 293)
V₂=4.962 L
