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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Trimix 10/50 is a gas mixture that contians 10% oxygen and 50% helium, and the rest is nitrogen. If a tank of trimix 10/50 has a

Nataly_w [17]

Answer: 1.61 x 10⁴ kPa

Dalton's law states that the sum of the partial pressures of each gas equals the total pressure of the gas mixture. According to this law,

Pi = xi P

where Pi is the partial pressure of the gas i, xi is the mole fraction of the gas i in the gas mixture and P is the total pressure.

The mole fraction is defined as the quotient between the moles of solute (ni) and the total moles of the mixture (nt), which is calculated by adding the moles of all its components:

xi = $\frac{n_{i} }{n_{t} }$

In the Trimix 10/50 mix you have 10% oxygen, 50% helium and 40% nitrogen.

To calculate the total number of moles of the mixture and thus determine the molar fraction of helium, we consider 100 g and calculate the number of moles that represent 10 g of O₂ (n₁), 50 g of He (n₂) and 40 g of N₂ (n₃):

n₁ = 10 g x $\frac{1 mol}{31.998 g}$ = 0.313 mol