EQUILIBRIUM REVIEW SHEET chemistry homework help

EQUILIBRIUM REVIEW SHEET 2:
#1) ​Explain about in what conditions a chemical equilibrium occurs. Explain about what it

means. Explain why limiting reactants does not work in a chemical equilibrium..#2) ​The following chemical reaction, N2​ O​ ​4⇌​ 2NO​2​ + 57.2 kJ, is at equilibrium; however, the

N​2O​ ​4​ starts with only N-15 isotope and the NO2​ ​ is N-14 isotope. At equilibrium, the concentrations will not change. What will happen to the isotopes of nitrogen?

#3) ​In the above reaction which way would the reaction shift in the following conditions and how would the equilibrium constant change:

a) NO​2​ is added.
b) N​2O​ ​4​ is added.
c) A catalyst is added.
d) The system is heated.
e) The total pressure is increased. f) The volume is increased.
g) N​2​ gas is added.
h) The temperature is decreased.

#4) ​State five reactions where equilibrium occurs around us. important?

Why is the equilibrium constant

Write the equilibrium constant for #5,6. These are the equilibrium conditions.#5)​ ____ N​2​O​4 ​⇌​ ____ NO​2 [N​2​O​4]​ = 0.65 M, [NO​2]​ = 0.25 M

#6)​ ____ HI + ____ H​2​O​2 ​⇌​ ____ H​2O​ + ____ I​2[HI] = 0.55 M, [H​2​O​2​] = 0.80 M, [H​2​O] = 0.50 M, [I​2​] = 0.75 M

#11) ​H​ O​ (aq) + SO​ (g)⇌H​ SO​ (aq) K​ = 1.5 x 103​ ​ Initial conditions are [H​ O​ ] = 1M, SO​ (g) 2​2​ 2​ 2​4​ c​ 2​2​ 2​

= 0.5M, H​2S​ O​4​ = zero. What are the equilibrium conditions?​ ​Find through​ ​the quadratic equation and through approximation.

#12) ​N​2O​ ​4(​ g) + S​8(​ s) ⇌ SO​2(​ g) +N​2(​ g) The equation is not balanced. If the initial conditions

are [SO​2]​ = 15atm, [N​2]​ =5atm, [N​2O​ ​4]​ =3atm, S8​ =​ 15g. What​ a​ re the equilibrium concentrations for the following constants?
a)Kp=1.4×101​ 2 b)Kp=5.0×10-​14 c)Kp=24 IftheKp=1.4×101​ 0​,​ whatisK?

#13)​ ​2HF(g) ⇌ H​2(​ g) + F​2(​ g). Kp = 16.

following initial conditions?
a) [HF] = 1atm. [H​2]​ =[F​2]​ = 0
b) [HF]= [H​2]​ =[F​2]​ = 1atm
c) [HF]= [H​2]​ =[F​2]​ = 1M
d) [HF] = 1atm. [H​2]​ =[F​2]​ = 0.5atm
e) [HF]=1atm. [H​2]​ =2atm. [F​2]​ =0.5atm

What are the equilibrium concentrations for the

#14) ​Which types of reactions have a VERY HIGH K​c​ value? Why?

#15) ​If A + 2B ​⇌​ C + D. K​c​ = 50 E​2​ + D ​⇌​ F K​c​ = 12

E​2​ + A ​⇌​G K​c=​ 40
2A +2B ​⇌​G K​c=​ 100
What is the value of the following reaction? 2 C + 2F ​⇌​ 4E​2 A + B⇌0.5G?

Equilibrium Constants of Multiple Reactions Finding #1) ​At 900 ̊C, Kp =1.04 for the following reaction

CaCO​3(​ s) ⇌CaO(s) + CO​2​(g).​At 900 ̊C, Kp = 0.05 for the following reaction 2FeO(s) + CO​2(​ g) ⇌ Fe​2O​ ​3(​ s) + CO(g)What is the equilibrium constant forCO(g) + Fe​2O​ ​3(​ s) + CaO(s) ⇌ 2FeO(s)​ ​+ CaCO​3(​ s)

#2) ​N​2(​ g) + 3H​2(​ g)⇌2NH​3(​ g) has a Kc​ ​ = 1.3 x 10​-2At the same temperature, 2H​​2(​ g) + O​2(​ g) ⇌ 2H​2O​ (g), K​c​ = 3.4 x 103​

What is the equilibrium constant for

2NH​3(​ g) + 1.5O​2(​ g) ⇌ 3H​2O​ (g) + N​2​(g)

#3) ​N​2(​ g) + 3H​2(​ g) ßà 2NH​3​(g) has a K = 1.3 x 10-​ 2At the same temperature, H​2(​ g) + Br​2(​ g) ßà 2HBr (g), K = 3.5 x 104​

What is the equilibrium constant for

1.5 Br​2(​ g) + NH​3(​ g) ⇌ 3HBr(g) + 0.5N​2(​ g)

LOOK AT THE FOLLOWING REACTIONS. CAN THEY BE SOLVED EASILY? ​IF THEY CAN, SOLVE THEM. FIND THE EQUILIBRIUM CONCENTRATIONS.

#23) ​CaSO​4(​ s) ↔ CaO(s) + SO​3(​ g) K = 10
Start with 10 grams of calcium sulfate, 5 grams of calcium oxide, 2M of SO​3.​

#24) ​8Na​2S​ ​2O​ ​3(​ aq) ↔ 8Na​2S​ O​3(​ aq) + S​8(​ s) K = 10,000 [Na​2S​ ​2O​ ​3]​ = 1M [NaSO3​ ]​ = 1M

#25) ​2NO​2(​ g) ↔ N​2O​ ​4(​ g) K​p​ = 100 [NO​2]​ = 2atm [N​2O​ ​4]​ = 5atm

#26) ​CH​4(​ g) + 2O​2(​ g) ↔CO​2​(g) + 2H2​ ​O(l) K = 10,000

[CH​4]​ = 1.5M#27​) ​8O​2​ + S​8(​ s) ↔ 8SO​2​(g)

[O​2]​ = 1M Products = zero. K=256

[SO​2(​ g)] = 0.5M
Al = 100g HCl(aq) = 1.000M Products = zero.

#29​) ​Na​2S​ O​4(​ aq) + 2HCl(g) ↔ NaCl(aq) + 2NaHSO4​ (​ aq) [Na​2S​ O​4]​ = 1M [HCl] = 1M Products = zero.

#30​) ​H​ O+ (aq) + OH- (aq) ↔H​ O(l) K = 101​ 4 3​ 2​

[H​3O​ +] = 1.5 x 10-​8 [OH-] = 1.5 x 10-​6#31​) ​NH​ (g) + H​ O(l) ↔ NH​ +​(aq) + OH-​​(aq) ​ ​ K = 1.5 x 10-​5

[O​2(​ g)] = 0.5M S​8(​ s) = 100g#28​) ​2Al(s) + 6HCl(aq) ↔2AlCl​3(​ aq) + 3H2​ (​ g) K = 10,000

3​ 2​ 4​[NH​3]​ = 0.05M H​2O​ = 1 liter. [NH​4+​ ] = 1.5M [OH-] = 1M

#32​) ​OF​ (g) + H​ O (g) ↔ 2HF(g) + O​ (g) K = 1.6 x 106​ 2​2​ 2​

Initial conditions, reactants = zero. [HF] = 0.8M [O​2]​ = 1.5M

#8) Lead (II) iodide has a Ksp of 1.4 x 10-​ 8​. If 1g of Pb(II) iodide is put into 1 liter of water, what percent of it will dissolve? (3)

K = 8

#9) 0.5g of NaOH is dissolved in 1 liter of water. 0.5g of Ca(NO​3​)​2​ is also dissolved in the same water. Sodium and nitrate ions are completely soluble in water. Ca(OH)​2​ has a Ksp of 1.3 x 10-​6​ at 25°C. Will there be a precipitate? Show your work. (3)