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Scorpion4ik [409]
3 years ago
15

For which of the following equilibria does `"K"_("eq") = ["O"_2]`? A. O2(l) O2(g) B. 2O3(g) 3O2(g) C. 2H2O(l) 2H2(g) + O2(g) D.

2Hg(s) +O2(g) 2HgO(s)
Chemistry
2 answers:
Feliz [49]3 years ago
8 0

For plato users

the answer is a. O2(l) O2(g)

hope this helps!

svp [43]3 years ago
3 0

For a hypothetical reaction:

aA ↔ bB

the equilibrium constant is given as:

Keq = [B]^b/[A]^a

activity of pure solids and liquids = 1

For the given reactions:

A) O2(l) ↔ O2(g)

Keq = [O₂]

B) 2O3(g) ↔ 3O2(g)

Keq = [O₂]³/[O₃]²

C) 2H2O(l) ↔2H2(g) + O2(g)

Keq = [H₂]²[O₂]

D) 2Hg(s) + O2(g) ↔ 2HgO(s)

Keq = [Hg]²[O₂]

Ans: A



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2 years ago
Suppose a 250. mL flask is filled with 0.30 mol of N_2 and 0.70 mol of NO. The following reaction becomes possible:N_2(g) +O2 →
Inessa [10]

Answer:

0.4 M

Explanation:

Equilibrium occurs when the velocity of the formation of the products is equal to the velocity of the formation of the reactants. It can be described by the equilibrium constant, which is the multiplication of the concentration of the products elevated by their coefficients divided by the multiplication of the concentration of the reactants elevated by their coefficients. So, let's do an equilibrium chart for the reaction.

Because there's no O₂ in the beginning, the NO will decompose:

N₂(g) + O₂(g) ⇄ 2NO(g)

0.30 0 0.70 Initial

+x +x -2x Reacts (the stoichiometry is 1:1:2)

0.30+x x 0.70-2x Equilibrium

The equilibrium concentrations are the number of moles divided by the volume (0.250 L):

[N₂] = (0.30 + x)/0.250

[O₂] = x/0.25

[NO] = (0.70 - 2x)/0.250

K = [NO]²/([N₂]*[O₂])

K = \frac{(\frac{0.70 -2x}{0.250})^2 }{\frac{0.30+x}{0.250}*\frac{x}{0.250} }

7.70 = (0.70-2x)²/[(0.30+x)*x]

7.70 = (0.49 - 2.80x + 4x²)/(0.30x + x²)

4x² - 2.80x + 0.49 = 2.31x + 7.70x²

3.7x² + 5.11x - 0.49 = 0

Solving in a graphical calculator (or by Bhaskara's equation), x>0 and x<0.70

x = 0.09 mol

Thus,

[O₂] = 0.09/0.250 = 0.36 M ≅ 0.4 M

3 0
3 years ago
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