1answer.
Ask question
Login Signup
Ask question
All categories
  • English
  • Mathematics
  • Social Studies
  • Business
  • History
  • Health
  • Geography
  • Biology
  • Physics
  • Chemistry
  • Computers and Technology
  • Arts
  • World Languages
  • Spanish
  • French
  • German
  • Advanced Placement (AP)
  • SAT
  • Medicine
  • Law
  • Engineering
goblinko [34]
3 years ago
14

For the equilibrium: 2 NO (g) <----> N2(g) + O2 (g), Kp=2400. If initially, only NO is present at a partial pressure o

f 37.30 atm, what will the partial pressures of N2 and O2 be at equilibrium?
1827 atm
38.08 atm
1.725 atm
36.55 atm
Chemistry
2 answers:
nekit [7.7K]3 years ago
7 0

Answer:

p_{N_2}^{eq}=p_{O_2}^{eq}}=18.27atm

Explanation:

Hello,

In this case, for the given chemical reaction, the law of mass action turns out:

Kp=\frac{p_{N_2}^{eq}p_{O_2}^{eq}}{(p_{NO}^{eq})^2}

Whereas the equilibrium pressures, based on the stoichiometry and the change x, changes to:

Kp=\frac{(x)(x)}{(37.30-2x)^2}=2400

Solving for x via quadratic equation, one obtains:

x_1=18.27atm\\x_2=18.84atm

In such a way, the feasible solution is 18.27 atm since the other pressure lead to a negative pressure of NO at the equilibrium, therefore, the equilibrium pressures of nitrogen and oxygen that are equal, result in:

p_{N_2}^{eq}=p_{O_2}^{eq}}=x_1=18.27atm

Best regards.

Thepotemich [5.8K]3 years ago
5 0

The reaction is           2 NO (g) <----> N2(g) + O2

partial pressures

Initial                                 37.30          0          0

Change                            -2p                +p        +p

Equilibrium                          37.30-p        p          p

Kp = pN2 X pO2 / (pNO)^2

2400 = p^2 /  (37.30-p)^2

3339096 - 179040p + 2400p^2 = p^2

2399p^2 + 3339096  -179040 p = 0

On solving

p = 36.55atm

Thus partial pressure of N2 and O2 = 36. 55 atm


You might be interested in
3.<br> КОН<br> +<br> —<br> —<br> Н3РО4<br> 1<br> +<br> K3PO4<br> —<br> Н,0<br> —
expeople1 [14]
What does this mean?
3 0
3 years ago
Why is this image a good model for the Law of Conservation of Mass?
ZanzabumX [31]

Answer:

The Law of Conservation of Mass states that mass is neither created nor destroyed in chemical reactions. Since the number and type of atoms in the reactant side of the chemical equation are the same as on the product side, the Law of Conservation of Mass has been demonstrated.

Explanation:

In the answer.

8 0
3 years ago
Describe how hydrogen and oxygen form water
Gwar [14]

Answer:

The  hydrogen molecules combine with the oxygen molecules, 2 hydrogen molecules, and 1 oxygen molecules is the amount needed to make one water atom or molecule, whatever you want to call it.

7 0
3 years ago
Read 2 more answers
True or false? To prepare 20.0 mL of 2.00 M solution of hydrochloric acid from a 10.0 M stock solution, you should take 4.00 mL
Aleksandr-060686 [28]
Sorry I don’t know but o have to answer for pionts
6 0
3 years ago
If a 3.0 picogram (pg) sample of curium-245 has a half-life of 8500 years, which conclusion can be drawn?
fomenos

Answer:

its A

Explanation:

8 0
2 years ago
Other questions:
  • What is another word for the sound barrier?
    11·2 answers
  • Thermochromic ink, for which new uses are appearing in the fashion industry, is able to do what?
    15·1 answer
  • Did democritus belive atoms retain there identity as a chemical reaction
    6·2 answers
  • What is the order in which electrons start filling the orbitals?
    12·2 answers
  • If you can walk 2.1 miles in 35 minutes, how many days will take you to con
    9·1 answer
  • How do you know if a compound will dissociate or ionize in water?
    11·1 answer
  • how does the percentage by mass of the solute describe the concentration of an aqueous solution os potassium sulfate
    12·1 answer
  • In metallic bonds, electrons are referred to as electrons.​
    7·1 answer
  • PLEASE HELP!!!!!!!!!!!
    12·1 answer
  • In an endothermic reaction, what happens to the surroundings?
    13·1 answer
Add answer
Login
Not registered? Fast signup
Signup
Login Signup
Ask question!