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steposvetlana [31]
3 years ago
12

The equilibrium constant for the gas phase reaction N2 (g) + O2 (g) ⇌ 2NO (g) is Keq = 4.20 ⋅ 10-31 at 30 °C. At equilibrium, __

______. The equilibrium constant for the gas phase reaction N2 (g) + O2 (g) 2NO (g) is Keq = 4.20 10-31 at 30 °C. At equilibrium, ________. products predominate reactants predominate roughly equal amounts of products and reactants are present only products are present only reactants are present
Chemistry
2 answers:
pychu [463]3 years ago
5 0

Answer:

At equilibrium, reactants predominate.

Explanation:

For every reaction, the equilibrium constant is defined as the ratio between the concentration of products and reactants. Thus, for the reaction N2 (g) + O2 (g) ⇌ 2NO the expression of its equilibrium constant is:

Keq = \frac{[NO]^{2}}{[O_{2} ][N_{2}]}

Since the equilibrium constant is Keq = 4.20x10-31 the concentration of reactants O2 and N2 must be much higher than products to obtain such a small number as  4.20x10-31 at the equilibrium. Hence, at equilibrium reactants predominate.

saveliy_v [14]3 years ago
3 0

Answer:

The equilibrium constant for the gas phase reaction N₂(g) + O₂(g) ⇌ 2NO(g) is Keq = 4.20x10⁻³¹ at 30 °C. At equilibrium, <em>reactants predominate</em>

Explanation:

For the equilibrium

N₂(g) + O₂(g) ⇌ 2NO(g); <em>Keq = 4.20x10⁻³¹</em>

The formula is:

Keq = \frac{[NO]^2}{[N_{2}][O_{2}]}

As keq = 4.20x10⁻³¹

4.20x10⁻³¹ = \frac{[NO]^2}{[N_{2}][O_{2}]}

As keq <<< 1 mathematically [NO]²<<< [N₂][O₂]

That means, <em>reactants predominate. </em>

<em></em>

I hope it helps!

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Answer:

The difference in mass between 3.01×10^24 atoms of gold and a gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm is :

<u>Difference</u>  <u>in mass</u> =<u> 985.32 - 984.5 = 0.82 g</u>

Explanation:

<u>Part I :</u>

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(Note : You can also take n = 5 mole )

Molar mass of gold = 196.96 g/mole

This means, 1 mole of gold(Au) contain = 196.96 grams

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4.99 moles of gold contain = 984.8 g

Mass of {3.01\times 10^{24}} atoms of gold = 984.5 g

<u>Part II :</u>

Density of Gold = 19.32 g/cm^{3}

Volume of the cuboid = length\times breadth\times height

Volume of the gold bar =6.00\times 4.25\times 2.00

Volume of the gold bar = 51cm^{3}

Using formula,

Density = \frac{mass}{Volume}

Mass = Density\times Volume

Mass = 19.32 \times 51

Mass = 985.32 g

So, A  gold bar with the dimensions 6.00 cm X 4.25 cm X 2.00 cm has mass of <u>985.32 g</u>

<u>Difference</u>  <u>in mass</u> =<u> 985.32 - 984.5 = 0.82 g</u>

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