Reaction Quotient - Video Tutorials & Practice Problems
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concept
Calculating the Reaction Quotient
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Hey everyone. So in this video, we're gonna take a look at the reaction quotient. Now the reaction quotient itself is represented by the variable Q. It is a ratio of product reacting concentrations at a particular time like the equilibrium constant K, it can be calculated by setting up an expression and ignoring solids and liquids. So our equilibrium constant K looks at concentrations at equilibrium Q looks at a particular time that time could be an equilibrium or could be outside of equilibrium. So that's the distinction between the two, but both of them will ignore solids and liquids when setting up their expressions. Click on the next video. And let's take a look at how to set up one of these expressions using our reaction quotient cue.
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example
Reaction Quotient Example
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1m
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Here in this example, question, it states the formation of gaseous ammonia is displayed by the equation given below. Here, we have one mole of nitrogen gas reacting with one mole of hydrogen gas to produce two moles of ammonia gas. What is the reaction quotient if the following amounts in moles of each component is placed in a 10 L vessel. All right. So here we have the moles of N two H two and NH three respectively. So remember we're going to set up an expression in the same way, we would set up an expression for equilibrium constant K. But here it's gonna be Q equals products over reactants. If we look everything is in the gaseous state, there are no solids or liquids. So we're not going to ignore any of the compounds. So here would be NH three, remember your coefficient becomes your exponent or power. So this would be squared divided by N two times H two. Remember when it comes to these expressions, we typically use the units of molarity or atmospheres. Here, they're giving us moles and they're also giving us leaders with this information, we can figure out our molarity. So you would divide each one of these moles by 10 L. And when you do that, you get our new molarity for each one. So here NH will become 0.0529 molar. It'd still be squared divided by N two, which is 0.0650 times H two, which is 0.0330. When we punch that into our calculators, we'll get 1.3046 as our reaction quotient Q value. If we look at the question, all these numbers have three significant figures. So I'm gonna round this down to 1.30 as our final answer. So this will represent the value of our reaction quotient Q.
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concept
Comparing Q to K
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3m
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So once you've learned how to calculate Q, you can compare it to your equilibrium constant K. Now, here we're gonna say once Q, once you've determined Q compare to K to determine which direction the chemical reaction will shift. And this is important to, will always shift towards K in order to maintain equilibrium. Here, we're going to say the balanced chemical equation will shift in the same direction as Q. So if Q moves in the forward direction to get to K, your chemical reaction moves in the forward direction, if Q moves in the reverse direction to get decay, then your chemical reaction also moves in the reverse direction. Here we're going to say shifting towards a side causes all molecules on that side to increase in amount. So here, let's take a look at this graph. We're doing a Q to K comparison in the first one, Q will be less than K. So here the reaction will shift. So Q is less than K, let's say that Q is equal to 10 and K is equal to 30 Q will always shift to K to get to equilibrium or maintain equilibrium. So Q would move this way. So the reaction will follow suit, the reaction will shift to the right to maintain equilibrium. Here, our chemical equation is two moles of no gas plus BR two liquid gives us two moles of nobr gas. It would also move in the same direction. Now, remember wherever we're heading to the site, we're going to will increase in amount. So our product amount would increase. But there's a balance in terms of this. If my product side is increasing, then my reactant site has to be decreasing. Now, let's flip everything here. Now, let's say that Q is greater than K. So Q is 75 K is still 30. So Q will always shift to K. So Q will go this way. So the reaction will shift in the reverse direction or to the left to maintain equilibrium. So in the same way, it goes this way, we're heading towards the reactant side. Since we're going towards the reactant side, the reactants would be increasing in amount. Things have to balance out. If the reactive side is increasing, then my product side has to be decreasing. Now, finally, let's look at if Q is equal to K. So here, if Q is equal to K, then the reaction is said to be at equilibrium. Since they're both equal to 30 Q can't shift anywhere. It's already at K. And if we're not shifting anywhere, then my chemical reaction wouldn't be shifting anywhere. So he would say that the reactions and product amounts are unchanged or you can say they're constant, they're, they're not increasing or decreasing. They are where they are. OK. So they're remaining unchanged or constant. OK. So that's the way you want to take a look when it comes to your reaction, quotient Q, if you know what it is, you can compare decay and that can help you determine which direction you're going to shift to maintain equilibrium. This can be directly applied, cure chemical reaction.
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example
Reaction Quotient Example
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2m
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Hey, everyone. So for this example question, it says for the reaction, we have four moles of hydrogen bromide gas reacting with one mole of oxygen gas to produce two moles of bromine gas plus two water liquids. We're told that the equilibrium constant is 0.063 at 400. Kelvin. Now, if the reaction quotient is 0.100 which of the following statements is true. All right. So remember our equilibrium constant is K and our reaction quotient is Q. When we have both Q and K, we can compare them to one another to tell which direction our chemical reaction will shift. So do a number line here K represents our equilibrium amount. So put it in the middle now is Q bigger than less than or equal to K. Well, Q is 0.100. So Q is larger. But remember Q will always shift towards K to maintain equilibrium. The direction Q shifts in the same direction that our chemical reaction will shift. So a chemical reaction also shifts in the reverse direction. Now remember the direction we're moving towards will be increasing in amount. Since we're moving towards the reactants, our reactors would be increasing in amount and things must be balanced. If our reaction site is increasing, then our product side has to be decreasing. Based on this information, we have to say which of the following statements is true. So if we take a look here, it says H pr will decrease. Well, H pr is a reactant, we're moving towards the reactant site so it to be increasing, not decreasing. So this is false. 02 will increase. This is true. It's reacting to what should be going up next. BR two will increase. H2O will increase. Both of these are products. We're moving away from the product side. So they should be decreasing in amount, not increasing amount. So out of all the statements only option B is true.
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Problem
Problem
For the reaction: 2 CO2 (g) ⇌ 2 CO (g) + 2 O2 (g), the equilibrium constant is 3.12 × 10−4 at 400 K, while the reaction quotient is 4.18 × 10−4. If initially we have 0.20 atm CO2, 0.30 atm CO and 0.15 atm O2, which of the following statements is not true?
a) The pressure of CO2 will be greater than 0.20 atm.
b) The pressure of CO will be less than 0.30 atm.
c) The pressure of O2 will be greater than 0.15 atm.
d) The pressure of O2 will be less than 0.15 atm.
e) The reaction will favor reactants.
A
The pressure of CO2 will be greater than 0.20 atm.
B
The pressure of CO will be less than 0.30 atm.
C
The pressure of O2 will be greater than 0.15 atm.
D
The pressure of O2 will be less than 0.15 atm.
E
The reaction will favor reactants.
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Problem
Problem
The equilibrium constant for the following gas phase reaction is 0.75 at 750 K. After a short time, analysis of a small amount of the reaction mixture shows the concentrations to be [NOBr] = 1.25 M, [NO] = 0.80 M and [Br2] = 0.50 M. Which of the following statements is/are true?
2 NOBr (g) ⇌ NO (g) + Br2 (g)
a) The reaction mixture is at equilibrium.
b) No further reaction will occur.
c) The partial pressure of NOBr will increase.
d) The partial pressure of NO will increase.
e) The reaction will shift to the left, the reactant side.
A
The reaction mixture is at equilibrium.
B
No further reaction will occur.
C
The partial pressure of NOBr will increase.
D
The partial pressure of NO will increase.
E
The reaction will shift to the left, the reactant side.
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