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3 years ago

What stress would shift the equilibrium position of the following system to the right?

Chemistry

2 answers:

MakcuM [25]3 years ago

6 0

Answer:

Decreasing the concentration of $N_2O_3$ => To the left

Heating the system => To the left

Adding a catalyst => No change

Increasing the concentration of NO => To the left

Explanation:

We have to analyze each case:

Decreasing the concentration of N2O3 

The compound $N_2O_3$ is a reactive. When we remove the concentration of this compound the reaction has to go to the left in order to restore the initial equilibrium.

Heating the system 

In this case, the heat (energy) can be considered as a product because it has a negative sign. Therefore, if we add energy the reaction will move to the opposite side. The reactive side

Adding a catalyst

The addition of a catalyst doesn't change the equilibrium of any reaction. The catalyst addition affects the velocity of the reaction but not the equilibrium.

Increasing the concentration of NO

The NO is a product therefore if we increase the concentration of NO will displace the reaction to the opposite side. To the reactants side

yanalaym [24]3 years ago

4 0

Answer:

Decreasing the concentration of N2O3

Explanation:

This is because the products on the right of the reaction occupy more space. One (1) mole of NO and another mole of NO2 will occupy more space than the one (1) mole of N2O3. Therefore decreasing the concentration of N2O3 will shift the reaction to the right because the products will have more space to occupy – hence favoring equilibrium.

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Explanation:

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3 years ago

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Answer:

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2 years ago

Ni-cad (nickel–cadmium) batteries have a slightly lower cell potential than the common alkaline value of 1.5 v normally used in

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The half-reaction are:

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2 years ago

Some fuel cells are powered by hydrogen. Scientists are looking into the decomposition of water (H2O) to make hydrogen fuel with

skad [1K]

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Explanation:

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Source of energy brainly.com/question/2948717

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2 years ago

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what i

Ostrovityanka [42]

Consider the isomerization of butane with equilibrium constant is 2.5 .The system is originally at equilibrium with :

[butane]=1.0 M , [isobutane]=2.5 M

If 0.50 mol/L of butane is added to the original equilibrium mixture and the system shifts to a new equilibrium position, what is the equilibrium concentration of each gas?

Answer:

The equilibrium concentration of each gas:

[Butane] = 1.14 M

[isobutane] = 2.86 M

Explanation:

Butane ⇄ Isobutane

At equilibrium

1.0 M 2.5 M

After addition of 0.50 M of butane:

(1.0 + 0.50) M -

After equilibrium reestablishes:

(1.50-x)M (2.5+x)

The equilibrium expression will wriiten as:

$K_c=\frac{[Isobutane]}{[Butane]}$

$2.5=\frac{2.5+x}{(1.50-x)}$

x = 0.36 M

The equilibrium concentration of each gas:

[Butane]= (1.50-x) = 1.50 M - 0.36M = 1.14 M

[isobutane]= (2.5+x) = 2.50 M + 0.36 M = 2.86 M

3 0

3 years ago