Gibbs Free Energy And Equilibrium - Video Tutorials & Practice Problems
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1
concept
∆Gº and K
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Now, when we discuss Gibbs free energy and equilibrium, we're really making a comparison between Gibbs free energy and our equilibrium constant. Now, here we're gonna say the relationship between our Gibbs free energy, which is delta G zero and our equilibrium constant, which is K. And we'll say EQ can also be observed in the following formula. Here. We're going to say use this formula when your equilibrium constant to EQ is given or can be calculated here. The formula is that the change in our standard gives free energy equals negative RT LNK. And here we're talking about keq here R would be our gas constant which we've seen in the past. It's equal to 8.314 joules over moles times K. Remember we employ this type of R constant. Whenever we're talking about energy, velocity or speed. In this section, we're not really talking about velocity or speed, we're talking about energy because we're using the units of either Jos or kilojoules. So just keep that in mind when we're talking about joules or kilojoules, then we're using the R value in the form of 8.314 joules over moles times K.
2
example
Gibbs Free Energy And Equilibrium Example
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Here, the example says a certain reaction takes place at 25 °C and has an equilibrium constant of 2.8 times 10 to the four determine the Gibbs free energy of the reaction. All right. So we need to find Gibbs free energy. So delta G zero, they're giving us the equilibrium constant. So the formula that connects them together is negative RT and K, we plug in our, our value which is 8.314 jewels over moles times K temperature needs to be in Kelvin. So at 2 73.15 to this degree Celsius, which gives us 2 98.15 K. And then we have allen of our equilibrium constant, which is 2.8 times 10 of the four. When we multiply everything across the board, we see that Kelvin will cancel out, we'll get at the end negative 2.5 times 10 to the four jewels per mole. So here this will be our final answer for gibbs free energy of this reaction.
3
Problem
Problem
For reaction, Ag2CO3 (s) ⇌ Ag2O (s) + CO2 (g), the ∆Hº = 79.14 kJ/mol, ∆Sº = 167.2 J/K.
Determine the equilibrium constant at which the temperature is 365.1 K.
A
1.90 × 10−9
B
8.23 × 10−21
C
388.1
D
2.58 × 10−3
4
concept
∆G, ∆Gº and Q
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1m
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Here, we're going to say that the following equation relates the change in the Standard Gipps free energy, which deals with standard conditions and the change with gives free energy which is under non-standard conditions. Here, we're gonna say use this formula when given, gives free energy under non standard conditions or dealing with the reaction quotient Q. Here we're going to say that this version of Gibbs free energy formula is the change in the non-standard Gibbs free energy equals the change in the Standard Gibbs free energy plus RT Lnq. Here we're gonna say R is equal to 8.314 joules over moles times K. Remember we use this version of R anytime we're dealing with energy and we know we're dealing with energy whenever we have units of joules or kilojoules involved, right? So keep this in mind, this equation is a way of us connecting the non-standard conditions with the standard conditions in relation to gives free energy.
5
Problem
Problem
Consider a hypothetical reaction at 38 ºC, X2 (aq) + 2 Y (s) ⇌ 3 Z (aq), with a ∆G of −75.8 kJ.
Concentrations of reactants and products: [X2] = 1.4 M, [Y] = 0.34 M, [Z] = 2.6 M. Calculate Keq of the reaction under standard conditions.
A
4.13 × 1015
B
2.67 × 1011
C
9.60 × 1012
D
6.67 × 1013
6
example
Gibbs Free Energy And Equilibrium Example
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2m
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He says the given reaction has a change in the standard Gibbs free energy of negative 374 kilojoules and partial pressures of sulfur tetra fluoride, fluorine gas and sulfur hexafluoride are 0.63 atmospheres, 0.95 atmospheres and 1.7 atmospheres respectively. Here, we need to calculate the change and the non standard gives free energy of the reaction for this reaction. So here is our balanced equation. It's a 1 to 1 relationship with everything here. We're going to say that delta G equals delta G zero or not plus RT LNQ. Here, they're giving us these values. They're gonna help us find cue Q just like an equilibrium constant is equal to products over reactants. It ignores solids and liquids, but everything here is a gas. So it equals the pressure of sulfur hexafluoride divided by the pressure of sulfur tetra fluoride times the pressure of fluorine gas plug in the value. So we have 1.7 divided by 0.63 times 0.95. So Q here equals 2.840. So we have what our Q is, we have what delta G zero is here are uses jewels. So we're gonna convert these kilojoules into jewels by multiplying them by 1000. When we do that, we get negative 37, 4000 jewels plus 8.314 jewels over moles times K. Here temperature, we're not given a temperature. So we're assuming it's under standard temperature which is 2 98.15 Kelvin. And then we have Allen of Q which is 2.840. When we plug all this in, this will give us the change in the non-standard gives free energy which comes out to negative 3.71 times 10 to the five Jules. OK. So this would be our final answer for this given question.
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