Answer:
The vertical line indicating bonds breaking
Explanation:
a. W = 0 J
b. W = - 308.028 J
<h3>Further explanation</h3>
Given
Nitrogen gas expands in volume from 1.6 L to 5.4 L
Required
The work done
Solution
Isothermal :
W = -P . ΔV
Input the value :
a. At a vacuum, P = 0
So W = 0
b. At pressure = 0.8 atm
W = - 0.8 x ( 5.4 - 1.6)
W = -3.04 L.atm ( 1 L.atm = 101.325 J)
W = - 3.04 x 101.325
W = - 308.028 J
Answer is: K <span>be for the reaction at 375 K is 326.
</span>Chemical reaction: N₂(g) + 3H₂(g) ⇌ 2NH₃(g); ΔH = -92,22 kJ/mol.
T₁<span><span> = 298 K
</span>T</span>₂<span><span> = 375 K
</span><span>Δ<span>H = -92,22 kJ/mol = -92220 J/mol.
R = 8,314 J/K</span></span></span>·mol.<span>
K</span>₁ = 6,8·10⁵.<span>
K</span>₂ = ?The van’t Hoff equation: ln(K₂/K₁) = -ΔH/R(1/T₂ - 1/T₁).
ln(K₂/6,8·10⁵) = 92220 J/mol / 8,314 J/K·mol (1/375K - 1/298K).
ln(K₂/6,8·10⁵) = 11092,13 · (0,00266 - 0,00335).
ln(K₂/6,8·10⁵) = -7,64.
K₂/680000= 0,00048
K₂ = 326,4.
To calculate the pH of a solution that has a [H3O+] of 7.22x10^-7. You would do the following
pH=-log[H3O+]
pH=-log[7.22x10^-7]
pH=?
