Our reaction balanced equation at equilibrium N2(g) + 3 H2(g) ↔ 2 NH3(g)
and we have the Kp value at equilibrium = 4.51 X 10^-5
A) 98 atm NH3, 45 atm N2, 55 atm H2
when Kp = [P(NH3)]^2 / [P(N2)] * [P(H2)]^3
= 98^2 / (45 * 55^3) = 1.28 x 10^-3
by comparing the Kp by the Kp at equilibrium(the given value) So,
Kp > Kp equ So the mixture is not equilibrium,
it will shift leftward (to decrease its value) towards the reactants to achieve equilibrium.
B) 57 atm NH3, 143 atm N2, no H2
∴ Kp = [P(NH3)]^2 / [P(N2)]
= 57^2 / 143 = 22.7
∴Kp> Kp equ (the given value)
∴it will shift leftward (to decrease its value) towards reactants to achieve equilibrium.
c) 13 atm NH3, 27 atm N2, 82 atm H2
∴Kp = [P(NH3)]^2 / [P(N2)] * [P(H2)]^3
= 13^2 / (27* 82^3) = 1.14 X 10^-5
∴ Kp< Kp equ (the given value)
∴it will shift rightward (to increase its value) towards porducts to achieve equilibrium.
Answer:
Explanation:
<u>1) Convert the mass of water into number of moles</u>
- Molar mass of water: 18.015 g/mol
- Number of moles, n = mass in grams / molar mass
n = 255 g / 18.015 g/mol = 14.15 mol
<u>2) Use the formula E = n × ΔH vap</u>
This is, you have to multiply the molar ΔH vaporization by the number of moles to find the total energy to boil the given amount of water.
- E = 14.15 mol × 40,650 J/mol = 575,395.5 J
<u>3) Round to the correct number of significant figures.</u>
The mass of water is the measurement with the least number of significant figures (3), so you must report the answer with 3 significant figures,
Answer:
closed container of N2O4 and NO2 is at equilibrium. NO2 is added (increasing the concentration) to the container.
N2O4 (g)
