Answer:
Kc = 8.05x10⁻³
Explanation:
This is the equilibrium:
2NH₃(g) ⇄ N₂(g) + 3H₂(g)
Initially 0.0733
React 0.0733α α/2 3/2α
Eq 0.0733 - 0.0733α α/2 0.103
We introduced 0.0733 moles of ammonia, initially. So in the reaction "α" amount react, as the ratio is 2:1, and 2:3, we can know the moles that formed products.
Now we were told that in equilibrum we have a [H₂] of 0.103, so this data can help us to calculate α.
3/2α = 0.103
α = 0.103 . 2/3 ⇒ 0.0686
So, concentration in equilibrium are
NH₃ = 0.0733 - 0.0733 . 0.0686 = 0.0682
N₂ = 0.0686/2 = 0.0343
So this moles, are in a volume of 1L, so they are molar concentrations.
Let's make Kc expression:
Kc= [N₂] . [H₂]³ / [NH₃]²
Kc = 0.0343 . 0.103³ / 0.0682² = 8.05x10⁻³
This problem is to use the Claussius-Clapeyron Equation, which is:
ln [p2 / p1] = ΔH/R [1/T2 - 1/T1]
Where p2 and p1 and vapor pressure at estates 2 and 1
ΔH is the enthalpy of vaporization
R is the universal constant of gases = 8.314 J / mol*K
T2 and T1 are the temperatures at the estates 2 and 1.
The normal boiling point => 1 atm (the pressure of the atmosphere at sea level) = 101,325 kPa
Then p2 = 101.325 kPa
T2 = ?
p1 = 54.0 kPa
T1 = 57.8 °C + 273.15K = 330.95 K
ΔH = 33.05 kJ/mol = 33,050 J/mol
=> ln [101.325/54.0] = [ (33,050 J/mol) / (8.314 J/mol*K) ] * [1/x - 1/330.95]
=> 0.629349 = 3975.22 [1/x - 1/330.95] = > 1/x = 0.000157 + 1/330.95 = 0.003179
=> x = 314.6 K => 314.6 - 273.15 = 41.5°C
Answer: 41.5 °C
Concentration of the reactant,pressure,surface
area of the reactant and temperatur
Decrease because loss of electrons.
