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AP®︎/College Chemistry
Unit 7: Lesson 6
Using the reaction quotientWorked example: Using the reaction quotient to predict a pressure change
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We can predict how the total pressure of a reaction mixture will change as the system moves toward equilibrium by first calculating the reaction quotient, Qₚ, and then comparing Qₚ to Kₚ. Created by Yuki Jung.
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Could we calculate how much the total pressure will decrease?(10 votes)
We would only be able to figure out exactly how much it decreases if we knew the volume and temperature of the system. If we did, though, the Ideal Gas Law (PV = nRT) would come in handy. By simply plugging in the volume for V, the temperature for T, and the number of moles of total gas for n, we could calculate the pressures at non-equilibrium and at equilibrium and simply subtract them from each other to get the total decrease in pressure.(9 votes)
In the above video Sal said that increasing pressure, the reaction favors the side with fewer molecules. in this video is said the opposite(8 votes)
This is a different situation - in the earlier video Sal was talking about the consequences of increasing the pressure on a reaction that was already at equilibrium.
Here Yuki is discussing what happens to the pressure in a reaction that is not yet at equilibrium. Once this reaction reaches equilibrium (which takes into account the difference in pressure) it would then react the same way - i.e. increasing the pressure would further shift the equilibrium towards the products.(16 votes)
At5:31, why the number of gas molecule will decrease in the system? I thought that both sides are gases and since the conservation of molecule, the total number of gas molecule won't be changed.(6 votes)
The three gas molecules on the left will be combined to make just one gas molecule of a "bigger" gas, the conservation of atoms has nothing to do with the number of gas molecules(5 votes)
- At5:23how did she figure out the no. of moles on both sides?(3 votes)
using the coefficients of the gases: on the reactants side you do 1 + 2 = 3 and on the product side you directly get 1(1 vote)
- Hey, guys!
On my textbook when talking about change in temperature and its effect on K, it also introduces the Van't Hoff Equation.
ln( K2/K1)=∆Hº/R * (1/T1-1/T2)
However, I don't think Khan and his colleagues talk about his equation. Can someone please help me to explain it a little bit? Is this a difficult concept to understand?(2 votes)- Well, the Van't Hoff equation shows the dependence of equilibrium constant on the temperature.
ln(K2/K1) = deltaH/R*(1/T1-1/T2)
ln(K2/K1) = deltaH/R * (T2-T1)/T1T2
Now what would happen if the reaction is exothermic (deltaH < 0) and we increase the temperature (T2 > T1). Use the Le Chatelier's principle to figure that out.
The forward reaction will be disfavoured - K2 will be less than K1 (lesser products formed now)
This equation show the same thing.
ln(K2/K1) = delta H (which is -ve) / R * (T2-T1) (which is +ve) /T1/T2
thus, ln (K2/K1) = negative
ln K2 - ln K1 = negative
=> K2 < K1.
Similarly you can figure out the effect of change in temperature (increase and decrease) on endothermic reactions.(3 votes)
Can someone explain to me why the total P is decreasing? I thought that as the moles decrease and the favored reaction is forward, the pressure is supposed to be increasing.(2 votes)
the reaction is favouring the product side,so the number of gas molecules decrease and thus total p decreases(2 votes)
Does the pressure decrease when the amount of gas decrease? I thought it would decrease when there are more gases because the pressure is more evenly distributed.(1 vote)
If volume and temperature remain constant, the pressure decreases when the amount of gas decreases.
However, each gas in a container exerts its own pressure independently of the other gases.
If a gas in a container exerts a pressure of 1 atm, you can pump in other gases, and the first gas will still have a partial pressure of 1 atm even though the total pressure in the container has increased.(4 votes)
Hi, I think I'm getting a little tripped up by the existence of Qc. Is Qc just a different way of saying Keq (aaaaa sorry, my device isn't cooperating with subscripts today but I hope that makes sense!)? Is Qc also the equilibrium coefficient?
Have a wonderful day and to any fellow future students out there, good luck!(1 vote)- The formula looks the same, but Qc is calculated using the concentrations of reactant(s) and product(s) found in the reaction mixture at a given moment.
Keq is determined based on the concentration values found for that reaction when it is at equilibrium.
Does that help?(3 votes)
- Why is the K(c) for CO(g)+2H_2(g)<-->CH_3 OH(g) 14.5? There's 1 mole for CH_3 OH and 1 mole of CO, and 2 moles of 2H_2 so shouldn't it be 1/(1*1^2)=1?(1 vote)
So the actual expression for Kc is given at1:18(it's Qc but they have the same form). This expression incorporates the coefficients from the balanced chemical equation at the top by having them act as exponents for the quantities in the expression.
The actual amounts of our chemical species are not the same as the coefficients from the chemical equation. Instead these are given in the problem; 1.2 mol of CO, 1.5 mol of H2, and 2.0 mol of CH3OH. It is these numbers which we plug into the expression to get the reaction quotient, or a ratio of the products and reactants not at equilibrium.
The actual value of the equilibrium constant, Kc, is determined from the concentrations of the chemical species when the reaction is at equilibrium. So that 14.5 number is determined experimentally.
Hope that helps.(1 vote)
- Hi, if the reaction is moving in the forward direction would that not infer a decrease in volume as the stoichiometric coefficient on the left adds to 3 and on the right is 1. As volume is inversely proportional the pressure, shouldn't pressure increase as the reaction moves from left to right? Thanks in advances.(1 vote)
- As @Asif mentioned, we're not talking about any change in volume. You have to assume the volume remains constant and any change in internal pressure is unable to affect the container's volume.(1 vote)
5:31Video transcript
- [Voiceover] A one litre reaction vessel contains 1.2 moles of carbon monoxide, 1.5 moles of hydrogen gas, and 2.0 moles of methanol gas. How will the total pressure change as the system approaches equilibrium at constant temperature? So, our carbon monoxide is
reacting with our hydrogen in a one to two ratio to give us methanol. And this reaction is reversible. We also know the equilibrium
constant for this reaction is 14.5 at some temperature. And we know that the
temperature is staying constant. So, we are going to break this
problem up into two parts. In part one, we're
gonna try to figure out, using the reaction quotient, whether our system is
at equilibrium or not. So, for this reaction, our reaction quotient 'Q' is the product concentration. "C-H-3-O-H" for methanol. Divided by the concentration
of our hydrogen gas; to the second power, because of that
stoichiometrical efficient. And then also in the denominator, we have our carbon monoxide concentration. We can calculate 'Q' by plugging in the
concentrations of these, at this particular moment in time. And we can calculate the concentrations using the volume of the vessel, which is one liter and the mole quantities. We know that concentration is just moles divided by volume. And since we're dividing
everything by one, the initial concentrations will be the same as the number of moles. So, if we write that out, for carbon monoxide, the initial concentration is 1.2 molar. For hydrogen, it's 1.5 molar. And for methanol, it is 2.0 molar. So now we can plug these
concentrations into our expression for 'Q'. And then we get, in our numerator, 2.0. And our denominator, is 1.5 squared times 1.2. So if we plug this all
into our calculators, what I got is at our 'Q', for this particular moment in time, with these concentrations is 0.74. So this tells us, first of all, we know that 'Q' is not equal to 'KC'. So that means we are not at equilibrium. "Not at equilibrium". Which means that our pressures are indeed going to change because the system is going
to try to reach equilibrium. The second thing we can do, using the reaction quotient to, is figure out how the
concentrations will change. Now that we know our
reaction quotient 'Q', we know that our reaction quotient 'QC' is less than 'K'. We can visualize this on a number line. So if we look at all
possible values of 'Q', we know that when 'Q' is zero, we have all reactants. When 'Q' is infinitely large, we have all reactants, we have all products. And then we have all of the
possible values in between. What we're really worried about here, is just looking at the
relative value of 'Q' and 'K'. And seeing how reaction
concentrations are going to shift. So 'Q', we can put on our number line, is somewhere around here. And 'K' is 14.5, so
we'll say it's somewhere around here. So this is our 'Q' and this is our 'K'. We can see that 'Q' is less
than 'K' on our number line. So what's gonna happen is, in order to reach equilibrium, our concentrations are
going to shift to the right to get 'Q' closer to 'K'. Which means, what's going to happen is, the reaction is going to shift to favor making more products. So if we look back at
the balance reaction, what's going to happen here is, it's going to shift to favor the products. So I'm making that top arrow
a little bit more bold. And to tie this into what the problem wants to know, we can figure out how the shift to make more products will
affect the total pressure. So the total pressure for a system that has a
bunch of gas molecules in it, we know that total pressure is related to the moles of gas in the system. So since we're shifting
to favor the reactants. And on the reactants side, we have, we're making one mole of gas. And we're starting with
three moles of reactant gas. We're favoring the side that
has fewer gas molecules. So that means as we shift
to favor the products, we're going to reduce number
of gas molecules in the system, and that's gonna reduce our 'P' total. So the answer is, that P total is going to decrease as our reaction approaches equilibrium and that is because our reaction quotient
'Q' is less than 'K'.