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madreJ [45]
3 years ago
8

For the reaction N2(g) + 3H2(g) 2NH3(g), what will happen if hydrogen gas was removed from the reaction mixture? There will be a

shift toward the products. There will not be a shift in equilibrium. There will be a shift toward the reactants. The value for Keq will change.
Chemistry
1 answer:
Hoochie [10]3 years ago
7 0
This uses something called <span>Le Chatelier's principle. It states essentially that any stress put upon a system will be corrected.

In more simple terms, it means that in an equilibrium, such as the equation N2(g) + 3H2(g) <=> 2NH3(g), removing a reactant will cause the system to create more of said reactant to compensate for its loss, or adding excess reactant will cause the system to remove some of the added reactant. For future reference, the same principle applies to products in an equilibrium as well. 

In this case, hydrogen gas is a reactant, and hydrogen is being removed. According to </span><span>Le Chatelier's principle, the system will shift to create more hydrogen gas. In essence, it will shift in the direction of the hydrogen gas, so there will be a shift toward the reactants.

To clear something up, Keq will not change, as it is a constant value with constant conditions (such as temperature, pressure, etc.).</span>
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Phosgene, a poisonous gas, when heated will decompose into carbon monoxide and chlorine in a reversible reaction: COCl2 (g) &lt;
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Answer:

8.08 × 10⁻⁴

Explanation:

Let's consider the following reaction.

COCl₂(g) ⇄ CO (g) + Cl₂(g)

The initial concentration of phosgene is:

M = 2.00 mol / 1.00 L = 2.00 M

We can find the final concentrations using an ICE chart.

     COCl₂(g) ⇄ CO (g) + Cl₂(g)

I       2.00            0            0

C        -x             +x           +x

E    2.00 -x          x             x

The equilibrium concentration of Cl₂, x, is 0.0398 mol / 1.00 L = 0.0398 M.

The concentrations at equilibrium are:

[COCl₂] = 2.00 -x = 1.96 M

[CO] = [Cl₂] = 0.0398 M

The equilibrium constant (Keq) is:

Keq = [CO].[Cl₂]/[COCl₂]

Keq = (0.0398)²/1.96

Keq = 8.08 × 10⁻⁴

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How much pure acid is in 520 milliliters of a 13 ​% ​solution?
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Volume of pure acid = 520*0.13 = 67.6 mL
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