Your equilibrium constant is represented by the variable capital K. Now recall that the equilibrium constant, which can also be represented as K sub EQ can be expressed as KP A or KC. Now KP, we use this version of our equilibrium constant when dealing with gasses in units of atmospheres, KC, we use this when dealing with aqueous solutions which are typically found in molarity. Remember molarity itself is moles per liter. We're gonna say here that KP and KC can be related together through the use of a formula. Now, this formula is KP equals KC RT to the delta N. Now KC and KP again, are just equilibrium constants. They're just equilibrium constant based on the units that we're using, whether that's in atmospheres or molarity R is our gas constant. If you've seen my videos on the ideal gas law, you know that R here equals 0.08206 L times atmospheres over moles times K and the temperature is T in Kelvin. Now here, what is delta N exactly? Well, delta N, here's, here's when we're looking at the moles of gasses within our balanced chemical equation. And here and it's just the gas coefficients of those gaseous compounds here, delta N equals the moles of gas as products minus the moles of gas as reactants. OK. So it's just products minus reactants. We look at our balanced equation, we see which compounds on both sides of our arrows. Our reversible arrows are gasses. We do the number of moles of gasses, products minus the number of moles of gas as reactants. That'll give us our delta. So just remember your equilibrium constant can be expressed as KC and KP. And equation that connects them together is KP equals KC RT to the delta N.
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example
Kp and Kc Example
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Here in this example question, it says calculate KP for the fog reaction with KC equal to 0.77 at 570 degrees Kelvin. All right. So here we're given our equation that two A gas is react with one B solid to produce CD gas plus three E gas. All right. The equation that connects KP to KC. Remember is KP equals KC RT to the delta N. Here we're told that our KC is 0.77 R is our gas constant, which is 0.08206. Temperature needs to be in Kelvin, which it's already given in Kelvin. So we plug that in. Now, delta N, delta N is our moles of gas as products. So how many moles of gas do we have as products? We have one, we have three moles. So that's four minus the moles of gas as reactants. This is a solid. So we're gonna ignore it. Here's a gas, there's two. So delta and here equals two. All right. So then what we're gonna do here is we're gonna say 0.77 when we multiply our R and T together, what's found inside is 46.7742. Remember that's still being squared, they were gonna say 46.7742 squared comes out to 2187.825786. Take that number multiply it by 0.77 gives us 16 84.625855. As our answer, we want to do it in terms of six figs 0.77 has two sick figs, 570 has two sick figs. So we're just gonna round this up two, 1.7 times 10 to the three. So here we have our answer to two significant figures. So this would be our final value for KP.
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Problem
Problem
Partial pressures of the following equilibrium mixture at 955 K are: 130 torr methane, 92 torr hydrogen sulfide, 167 torr hydrogen gas and 532 torr carbon disulfide. What is the value of Kc at 955 K?
CH4(g) + 2 H2S(g) ⇌ 4 H2(g) + CS2(g)
A
1.1×10–4
B
1.3×10–5
C
8.3×10–3
D
1.0×10–2
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Problem
Problem
Consider the hypothetical reaction: ? X(g) + 3 Y(g) ⇌ 3 Z(g), where Kp = 1.16 x 10–3 and Kc = 1.3 at 135°C. Find the value of the coefficient of X.
A
5
B
3
C
4
D
2
5
concept
Value of ∆n
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So when comparing KP to KC, it's important to understand the value of delta N. Here, we're going to say the value of delta N can determine if KC is greater than less than, or equal to KP. Remember the equation is KP equals KC RT to the delta N. Delta N represents the moles of gas as products minus the moles of gas as reactants. We can solve it mathematically to see how KP and KC relate to one another. If we know what DELTA N is or we can just simply say if more moles of gas are being produced. So basically you start out with, let's say three moles of gasses reactants and you end up with four moles of gasses reactants, your number of moles of gas has increased. So if more moles of gas are produced, that would mean that delta N would be equal to or greater than one, then that would mean KP is greater than KC. So here just remember if K, if delta N is greater than, or equal to, to one, then KP is greater than KC or KC is less than KP. If delta N is less than zero then KC is greater than KP. And if it's equal to zero, then they're equal to one another. Now, why is that? Well, if we do KP equals KC RT to the delta N and we make delta N equal to zero, remember anything to the zero power equals one. So KP equals KC times one, anything times one is equal to itself. So KP equals KC. So again, we can mathematically figure out what DELTA N is to see what the relationship between KC and KP is. And remember this relationship, or you can just simply say that if you have a certain number of moles of gasses reactants, and if you go to the product side, there's even more moles of gas of a product. That means that KP is going to be larger than KC. OK. So just you can use that type of logic in order to subvert the math and just look at the question itself in a qualitative manner. All right. So let's put the practice what we just learned in the following example and practice questions.
6
example
Kp and Kc Example
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For the following reactions, identify whether KP is greater than less than or equal to KC. So we're gonna take a look at it in a qualitative as well as a quantitative manner. OK. And see if we get the same answer. So let's just do it for the first one. In the first one, we have four moles of ammonia gas reacting with three moles of oxygen gas to produce two moles of nitrogen gas plus six moles of water vapor. Here, if we were to figure out what delta N is, remember, delta N equals moles of gas as products minus moles of gas as reactants. Our moles of gasses products is two and six. So that's eight minus four and then three, that's seven. So recall that if delta N is equal to or greater than one, that KP would be greater than KC. So that's what this is telling me KP is greater than KC. We could also say that we start out with seven moles of gasses on the reactant side. And now we have eight moles of gasses as products. The number of moles has increased KP is connected to gasses in units of atmospheres, right. So since there's more gas, more moles of gas being produced than we originally had, then KP would be greater than KC. So again, you can solve it mathematically, we could think of it in a qualitative way. More moles of gas on product side means that KP is greater than KC. Less moles of gas on product side means that KP is less than KC. Right? So if we take a look at the next one, we have two moles of gas as reactants and a total of two moles of gas products. So the moles are equal on both sides, there wasn't an increase or decrease. So KP would be equal to KC. If you were to solve this, using delta N, you'll get delta N equal to zero, which remains also KP is equal to KC. And then finally, what do we have for the last one? We have four moles of gas as reactants and we only have two moles of gases products. So what do you think the answer will be here? Did our number of moles increase decrease or stay the same? So we went from four moles of gas to two moles of gas. So the number of moles of gas went down. So that means KP would be smaller KP would be less than KC. So these would be the answers for parts A B and C for the following example, question
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Problem
Problem
For which reaction(s) will Kp = Kc?
a) N2(g) + O2(g) ⇌ 2 NO(g)
b) CaCO3(s) ⇌ CaO(s) + CO2(g)
c) 2 CH4(g) ⇌ C2H2(g) + 3 H2(g)
d) H2O(g) + CO(g) ⇌ H2(g) + CO2(g)
A
a) only
B
b) only
C
c) only
D
d) only
E
a) and d)
F
b) and c)
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Problem
Problem
Select the correct choice below for the reaction: PCl5(g) ⇌ PCl3(g) + Cl2(g)