# TBR General Chemistry - Example 3.9

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#### Shjanzey

##### Full Member
7+ Year Member

I have been reviewing this one and I am confused about one specific thing they state in the explanation.

This problem deals with equilibrium and they state that since the Keq is < 1.0 and the reaction starts with all products, then the value of x is going to be significant (e.g. the amount of product shifting towards the reactant side). This part makes sense to me, however they go on to say "more than half shifts over".

I don't understand how they can confidently state that more than half is going to shift. Does this mean that when a significant shift in x occurs during an equilibrium problem that more than half will always shift over? Can I safely make this assumption every single time a large shift will occur?

##### MD
Moderator Emeritus
10+ Year Member
Keq = [products] / [reactants]

If [products] = [reactants] then Keq = 1. Therefore, if Keq < 1 we must have [reactants] > [products].

Assuming we start with [products] = X and [reactants] = 0, to reach a Keq of 1 we would need values of [reactants] = X/2 and [products] = X/2, such that Keq = [X/2] / [X/2] = 1

In this case, half of the products would have converted to reactants.

If we want Keq < 1, then we need [reactants] > X/2, and thus [products] < X/2, meaning we need more than half of the initial products to convert to reactants.

#### Shjanzey

##### Full Member
7+ Year Member
Fantastic explanation, Thanks

#### BerkReviewTeach

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Keq = [products] / [reactants]

If [products] = [reactants] then Keq = 1. Therefore, if Keq < 1 we must have [reactants] > [products].

Assuming we start with [products] = X and [reactants] = 0, to reach a Keq of 1 we would need values of [reactants] = X/2 and [products] = X/2, such that Keq = [X/2] / [X/2] = 1

In this case, half of the products would have converted to reactants.

If we want Keq < 1, then we need [reactants] > X/2, and thus [products] < X/2, meaning we need more than half of the initial products to convert to reactants.

You explain things so well!

##### MD
Moderator Emeritus
10+ Year Member
Keq = [products] / [reactants]

If [products] = [reactants] then Keq = 1. Therefore, if Keq < 1 we must have [reactants] > [products].

Assuming we start with [products] = X and [reactants] = 0, to reach a Keq of 1 we would need values of [reactants] = X/2 and [products] = X/2, such that Keq = [X/2] / [X/2] = 1

In this case, half of the products would have converted to reactants.

If we want Keq < 1, then we need [reactants] > X/2, and thus [products] < X/2, meaning we need more than half of the initial products to convert to reactants.

Okay I'm having a serious issue with this problem. The explanation I've given here works in the case of a 1:1 mole ratio between reactants and products it works, but not for this problem.

Now that said, it happens to hold true that for mole ratios up to around 1:7, a Keq on the other side of 1 (from either starting with only reactants or products) means that more than half shifts over, but how can that be known for this problem?

I really don't see it. BRT can you or someone else explain how, without looking at anything more than the stoichiometric ratio between NO2 and N2O4, we can start the problem with the assumption that more than half will shift over simply because the Keq is less than 1? Because that doesn't hold true for a reaction like 8X <-> Y if you start with only Y.

#### Gauss44

##### Full Member
5+ Year Member
Example 3.9:

At 92.2 degrees celcius, the ksp for the following reaction is .2 atm -1. If you were to place exactly .2 atm of N2O4g into a 1 liter vessel, what would the partial pressure of NO2g be once equilibrium was established?

2NO2g (equilibrium symbol here) N2O4g

############################

Could the combo of temperature and coefficients explain the shift?

##### MD
Moderator Emeritus
10+ Year Member
Example 3.9:

At 92.2 degrees celcius, the ksp for the following reaction is .2 atm -1. If you were to place exactly .2 atm of N2O4g into a 1 liter vessel, what would the partial pressure of NO2g be once equilibrium was established?

2NO2g (equilibrium symbol here) N2O4g

############################

Could the combo of temperature and coefficients explain the shift?

In a way that we're supposed to recognize?

#### Gauss44

##### Full Member
5+ Year Member
In a way that we're supposed to recognize?

Not sure this is correct, but I was thinking: Volume of gas is proportional to number of moles per ideal gas law. Thus, in the absence of any other factors, twice the volume of gas on the left side (which suggests more pressure on the left side at equilibrium). And high temperature usually shifts equilibrium further toward the side with more moles, which in this case, is also the left side.

TBR Tech, or anyone, am I right?

##### MD
Moderator Emeritus
10+ Year Member
Not sure this is correct, but I was thinking: Volume of gas is proportional to number of moles per ideal gas law. Thus, in the absence of any other factors, twice the volume of gas on the left side (which suggests more pressure on the left side at equilibrium). And high temperature usually shifts equilibrium further toward the side with more moles, which in this case, is also the left side.

TBR Tech, or anyone, am I right?

I don't understand what you're saying at all. Your second sentence is totally lost on me because its a fragment, and increasing temperature has no relation to shifting toward more or less moles of involved substance as far as I know; it would depend on whether the reaction is exo/endothermic.

Edit: And furthermore, temperature isn't even changing in the problem. Yeah it takes place at a relatively high temp, so what? Kp is still .2

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#### Gauss44

##### Full Member
5+ Year Member
I don't understand what you're saying at all. Your second sentence is totally lost on me because its a fragment, and increasing temperature has no relation to shifting toward more or less moles of involved substance as far as I know; it would depend on whether the reaction is exo/endothermic.

Edit: And furthermore, temperature isn't even changing in the problem. Yeah it takes place at a relatively high temp, so what? Kp is still .2

Here's what I was trying to say in the second sentence:

volume &#8733; coefficients

I'm not sure if this applies. Chad describes it in his video titled, 2012 3.4 Stoichiometry a little after the 20:00 mark, if you are signed up for those. I know it's for gases only, but other than that, I don't know much about this rule. It comes from the ideal gas law. Then I took it a step further and tried to apply this logic to pressure.

Then regarding the temperature (in the second to last sentence), I'm wondering if applying heat to a gas generally favors the side of an equilibrium equation with more coefficients or more moles of gas, even if the temperature is constant. I'm thinking this because heat usually makes a gas expand. And heating N2O4 would break a bond, resulting in 2NO2. (I doubt we're suppose to know that last sentence though.)

Again, most of this is speculation.

Last edited:

##### MD
Moderator Emeritus
10+ Year Member
Here's what I was trying to say in the second sentence:

volume &#8733; coefficients

I'm not sure if this applies. Chad describes it in his video titled, 2012 3.4 Stoichiometry a little after the 20:00 mark, if you are signed up for those. I know it's for gases only, but other than that, I don't know much about this rule. It comes from the ideal gas law. Then I took it a step further and tried to apply this logic to pressure.

Then regarding the temperature (in the second to last sentence), I'm wondering if applying heat to a gas generally favors the side of an equilibrium equation with more coefficients or more moles of gas, even if the temperature is constant. I'm thinking this because heat usually makes a gas expand. And heating N2O4 would break a bond, resulting in 2NO2. (I doubt we're suppose to know that last sentence though.)

Again, most of this is speculation.

I'm not convinced the temperature matters here simply because the interaction of temperature with number of moles of gas is, as you said, used in the context of the ideal gas law, specifically when there is change in one of those parameters (temp, pressure, etc.).

Anybody else have any ideas?

#### popchap

##### New Member
I'm having trouble understanding the part about partial pressure. The solution says that half of 0.200 atm of N2O4 shifts over then the partial pressure of NO2 is 0.200 atm.

I'm not understanding how the partial pressure of NO2 is 0.200 atm. Can someone please elaborate on this part? Thanks!