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3 years ago

At equilibrium, the concentrations in this system were found to be [N2]=[O2]=0.200 M and [NO]=0.400 M.

Chemistry

1 answer:

Kipish [7]3 years ago

4 0

Answer:

The concentration in equilibrium of NO is 0,550M.

Explanation:

For the reaction:

N₂(g) + O₂(g) ⇄ 2NO

The equilibrium constant is defined as:

k = [NO]² / [N₂][O₂] (1)

Replacing for the concentrations in equilibrium:

k = (0,400M)² / (0,200M)(0,200M)

k = 4,000

If you add more NO until 0,700M, the equilibrium concentrations will be:

[NO] = 0,700M-2x

[N₂] = 0,200M+x

[O₂] = 0,200M+x

Replacing in (1)

4,000 = (0,700M-2x)² / (0,200M+x)²

4,000 = 4x²- 2,8x + 0,49 / x² + 0,4x + 0,04

4x² + 1,6x + 0,16 = 4x²- 2,8x + 0,49

4,4x = 0,33

x = 0,075M

That means that concentration in equilibrium of NO is:

[NO] = 0,700M - 2×0,075M = 0,550M

I hope it helps!

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20.0 g of Nitrogen is produced when oxygen gas reacts with NO gas. 29.8 L of oxygen is required at STP to produce this 20.0 g of

fiasKO [112]

The statement is false as 85675 grams of nitrogen dioxide is produced instead of 20 grams.

Explanation:

Balance equation for the reaction:

2NO + $O_{2}$ ⇒ 2N $O_{2}$

Data given:

mass of nitrogen produced = 20 gram

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0xygen is the limiting reagent so,

number of moles of oxygen: (atomic mass of 1 mole 32 grams/mole

number of moles = $\frac{mass}{atomic mass of 1 mole}$

putting the values in the equation:

= $\frac{29800}{32}$

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931.5 moles of oxygen will give x moles

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1862.5 moles of NO2 produced

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The statement is false as 85675 grams of nitrogen dioxide is produced instead of 20 grams as said in question.

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