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Chapter 14: Chemical Equilibrium Notes
Ch. 14 Chemical Equilibrium
14-1
This section introduces the concept of chemical equilibrium.
Key Concepts
Reaction Rates, Equilibrium, Reversible Reactions, Collision theory, Factors affecting reaction rate, Rate laws, Reaction mechanisms, temperature, concentration, Surface area, catalysts, inhibitors, Reaction order, intermediate, Activated complex, Activation energy, Equilibrium constant, Evaluating K_{eq}, Reaction \Delta G^o, Factors affecting equilibrium, LeChatelier’s principle.
Irreversible vs Reversible Reactions
Irreversible Reactions: Reactions that proceed to completion, indicated by a single arrow (→). Examples:
Formation of a precipitate: NaCl(aq) + AgNO3(aq) \rightarrow NaNO3(aq) + AgCl(s)
Formation of a gas: S8(s) + 8O2(g) \rightarrow 8SO_2(g)
Reversible Reactions: Reactions that can proceed in both forward and reverse directions, indicated by a double-sided arrow (⇌). Examples:
Melting ice/refreezing water.
Rechargeable batteries.
Sugar solution.
Writing Reversible Processes
Use a double-headed arrow to indicate reversible processes, e.g., sugar(s) \rightleftharpoons sugar(aq)
Two reactions occur simultaneously:
Sugar(s) \rightleftharpoons sugar(aq)
Sugar(aq) \rightleftharpoons sugar(s)
Both the forward and reverse processes occur simultaneously
Hamilton’s Restaurant Analogy
Illustrates equilibrium with a constant exchange of customers while maintaining a stable population.
Legal occupancy: 10
Equilibrium: 10/3 (3 out, 10 in)
Types of Equilibria
Phase Equilibrium: Example: Solid ⇌ Liquid
Solution Equilibrium: Example: Dissolving a solid in a solvent
Chemical Equilibrium: Example: H2(g) + I2(g) \rightleftharpoons 2HI(g)
Equilibrium Definition
A dynamic condition where two opposing processes occur at equal rates.
The forward and reverse reactions reach a balance point.
Rate of forward reaction equals the rate of the reverse reaction.
Both reactions continue to occur.
No observable, macroscopic changes are visible.
Equilibrium-Reversible Reactions Examples
Reversible reactions can go forwards or backwards
Example 1: 2H2 (g)+ O2 (g) \rightleftharpoons 2H_2O (l)+ energy
Example 2: CO2 + H2O \rightleftharpoons H2CO3
The Equilibrium Constant Expression
Methanol synthesis as a reversible reaction:
CO(g) + 2 H2(g) \rightleftharpoons CH3OH(g)
Forward and reverse rate constants: k1 and k{-1}
Three Approaches to the Equilibrium of Moles of reactants and products
Experiments 1, 2, and 3 demonstrate different starting conditions and the time it takes to reach equilibrium (t_e).
Graphs show the change in moles of CO, H2, and CH3OH over time.
Three Approaches to Equilibrium in the Reaction: CO(g) + 2 H2(g) \rightleftharpoons CH3OH(g)
Table 15.1 provides initial and equilibrium amounts (mol) and concentrations (mol/L) for three experiments conducted in a 10.0-L flask at 483 K.
Experiment 1: Initial amounts of CO and H2 are 1.000 mol each, with no initial CH3OH. At equilibrium, the amounts are 0.911 mol, 0.822 mol, and 0.0892 mol, respectively.
Experiment 2: Initial amount of CH3OH is 1.000 mol, with no initial CO or H2. At equilibrium, the amounts are 0.753 mol, 1.506 mol, and 0.247 mol, respectively.
Experiment 3: Initial amounts of CO, H2, and CH3OH are 1.000 mol each. At equilibrium, the amounts are 1.380 mol, 1.760 mol, and 0.620 mol, respectively.
Three Approaches to Equilibrium Table 15.2
Table 15.2 explores different ways to calculate the equilibrium constant using the data from Table 15.1
Trial 1: \frac{[CH3OH]}{[CO][H2]}
Experiment 1: \frac{0.00892}{0.0911 \times 0.0822} = 1.19
Experiment 2: \frac{0.0247}{0.0753 \times 0.151} = 2.17
Experiment 3: \frac{0.0620}{0.138 \times 0.176} = 2.55
Trial 2: \frac{[CH3OH]}{[CO](2 \times [H2])}
Experiment 1: \frac{0.00892}{0.0911 \times (2 \times 0.0822)} = 0.596
Experiment 2: \frac{0.0247}{0.0753 \times (2 \times 0.151)} = 1.09
Experiment 3: \frac{0.0620}{0.138 \times (2 \times 0.176)} = 1.280
Trial 3: \frac{[CH3OH]}{[CO][H2]^2}
Experiment 1: \frac{0.00892}{0.0911 \times (0.0822)^2} = 14.5
Experiment 2: \frac{0.0247}{0.0753 \times (0.151)^2} = 14.4
Experiment 3: \frac{0.0620}{0.138 \times (0.176)^2} = 14.5
Trial 3, where each concentration is raised to the power of its stoichiometric coefficient, yields essentially the same value for each experiment. This value is the equilibrium constant K_c.
The Equilibrium Constant
No matter the starting composition, the same ratio of concentrations is achieved at equilibrium.
For a general reaction: aA + bB \rightleftharpoons cC + dD
The equilibrium constant expression is: K{eq} = \frac{[C]^c[D]^d}{[A]^a[B]^b}, where K{eq} is the equilibrium constant.
Square brackets [ ] indicate molar concentrations (only for gas and aqueous).
The Magnitude of Equilibrium Constants
The equilibrium constant, K_{eq}, is the ratio of products to reactants.
The larger K_{eq}, the more products are present at equilibrium.
The smaller K_{eq}, the more reactants are present at equilibrium.
If K_{eq} >> 1, then products dominate at equilibrium and equilibrium lies to the right.
If K_{eq} << 1, then reactants dominate at equilibrium and the equilibrium lies to the left.
Factors Affecting K_{eq}
Each reaction has its own characteristic K_{eq} value.
The only factor that affects the value of K_{eq} is temperature.
Many combinations of concentrations can result in the same value for K_{eq} at the same temperature.
Treat K_{eq} as if it were dimensionless (no units).
The Equilibrium Constant - Writing Expressions
To write K_{eq} expressions:
Start with a balanced equation.
Ignore solids and liquids.
In the numerator, the products are multiplied together; coefficients become exponents.
In the denominator, the reactants are multiplied together; coefficients become exponents.
Example 1 - Writing Equilibrium Expression
Write the equilibrium expression for:
4NH3(g) + 7O2(g) \rightleftharpoons 4NO2(g) + 6H2O(g)
K{eq} = \frac{[NO2]^4[H2O]^6}{[NH3]^4[O_2]^7}
Example 2
[NH_3] = 3.1 \times 10^{-2} M
[N_2] = 8.5 \times 10^{-2} M
[H_2] = 3.1 \times 10^{-3} M
N2(g) + 3H2(g) \rightleftharpoons 2NH_3(g)
What is the value of K_{eq} ?
Example 3
[N_2] = 6.4 \times 10^{-3} M
[O_2] = 1.7 \times 10^{-3} M
K_{eq} = 1.1 \times 10^{-3}
N2(g) + O2(g) \rightleftharpoons 2NO (g) at equilibrium
What is the value of [NO] ?
[NO]^2 = K{eq} [N2] [O_2]
Example 4
Write the equilibrium expression for: C(s) + H2O(g) \rightleftharpoons CO(g) + H2(g)
Equilibrium constant expressions do not contain concentrations for solids and liquids unless everything is solid or liquid in a reaction (homogeneous reactions with all solids or liquids).
Example 5
Write the equilibrium expression for: 2H2O(l) \rightleftharpoons H3O^+(aq) + OH^-(aq)
K{eq} = Kw = [H_3O^+][OH^-]
Example 6
Write the equilibrium expression for: CO2(g) + H2(g) \rightleftharpoons CO(g) + H_2O(l)
Equilibrium Understanding Question
Which of the following statements is correct?
At equilibrium the reaction stops.
At equilibrium the rates of the forward and reverse reactions are equal.
At equilibrium the rates of the forward and reverse reactions are zero.
The Equilibrium Constant in Terms of Pressure
For gaseous reactions @ equilibrium, we denote the equilibrium constant as K_p.
aA (g) + bB(g) \rightleftharpoons cC (g)+ dD(g)
Kp = \frac{PC^c PD^d}{PA^a P_B^b}
PA, PB, PC, PD are partial pressures of four gases A, B, C, D at equilibrium.
These partial pressures must be expressed in atmospheres.
Relation Between Both Equilibrium Constants K{eq} and Kp
For gaseous substances, we can use the ideal-gas equation to convert between concentration (in molarity, M) and pressure (in atm):
PV = nRT
P = (\frac{n}{V})RT = MRT where M is concentration in mol/L
For substance “A” we can write: PA=[A]RT, so PA=[A] \cdot constant
Relation Between K{eq} and Kp Continued
Therefore, the equivalence between both equilibrium constants is:
Kp = K{eq}(RT)^{\Delta n}
Where \Delta n = c+d-a-b
Ways Different States of Matter Can Appear in the Equilibrium Constant, K
Molarity | Partial Pressure | ||
|---|---|---|---|
gas, (g) | YES | YES | |
aqueous, (aq) | YES | --- | |
liquid, (l) | --- | --- | |
solid, (s) | --- | --- | |
K_{eq} | K_p | ||
(sometimes expressed as K_c) |
Changing the Chemical Equation
The expression for K depends on the form of the chemical equation written to describe the equilibrium system. Consider the N2O4-NO_2 system:
Many other equations could be written for this system, for example,
Another equation that might be written:
Example 7: Expressing the Equilibrium Constant
At a certain temperature, the value of the equilibrium constant for the reaction is K1. CS2(g) + 3 O2(g) \rightleftharpoons CO2(g) + 2 SO_2(g)
(a) Write an expression for the equilibrium constant, K1. (b) Write an expression for the equilibrium constant, K2, of the reverse reaction.
(c) Write an expression for the related equilibrium constant, K_3
(1/3) CS2(g) + O2(g) = (1/3) CO2(g) + (2/3) SO2(g)
(a) K1 = \frac{[CO2][SO2]^2}{[CS2][O_2]^3}
(b) K2 = \frac{[CS2][O2]^3}{[CO2][SO2]^2} = \frac{1}{K1}
(c) K3 = \frac{[CO2]^{1/3}[SO2]^{2/3}}{[CS2]^{1/3}[O2]} = K1^{1/3}
Adding Chemical Equations
If a reaction can be expressed as the sum of two or more reactions, K for the overall reaction is the product of the equilibrium constants of the individual reactions.
That is, if
reaction 3 = reaction 1 + reaction 2
then,
K(reaction 3) = K(reaction 1) x K(reaction 2)
Adding Chemical Equations Example
To illustrate the application of this rule, consider
SO2(g) + 1/2 O2(g) \rightleftharpoons SO3(g) \quad K1 = 2.2
NO2(g) \rightleftharpoons NO(g) + 1/2 O2(g) \quad K_2= 4.0
Adding these equation eliminates 1/2O2 ; the result is
SO2(g) + NO2(g) \rightleftharpoons SO_3(g) + NO(g)
For this overall reaction, K3 = K1 x K_2 =2.2 x 4.0 = 8.8
Adding Chemical Equations - Validity Demonstration
The validity of this rule can be demonstrated by writing the expression for K for the individual reactions and multiplying:
SO2(g) + 1/2 O2(g) \rightleftharpoons SO3(g) \quad K1 = 2.2
NO2(g) \rightleftharpoons NO(g) + 1/2 O2(g) \quad K_2= 4.0
SO2(g) + NO2(g) \rightleftharpoons SO3(g) + NO(g) \quad K3 = 2.2 x 4.0 = 8.8
Predicting Reaction Direction from a Nonequilibrium Mixture
Calculate Reaction Quotient, Q
The expression for Q is the same as K BUT nonequilibrium concentrations rather than equilibrium concentrations are put into the expression
Compare the value of Q to K.
The Reaction Quotient (Qc)
The reaction quotient (Qc) is calculated by substituting the initial concentrations of the reactants and products into the equilibrium constant (Kc) expression.
IF
Qc > Kc system proceeds from right to left to reach equilibrium
Qc = Kc the system is at equilibrium
Qc < Kc system proceeds from left to right to reach equilibrium
Example 8
At a certain temperature K{eq} = 55 and a reaction vessel contains a mixture with the following concentrations: [SO3] = 0.85 M, [NO] = 1.2 M , [SO2] = 1.5 M and [NO2] = 2.0 M.
SO3(g) + NO(g) \rightleftharpoons SO2(g) + NO_2(g)
Is the reaction at equilibrium and if not which direction will the rxn proceed?
Example 8 - Solution
Solve just as if you were solving for the equilibrium constant.
Then analyze the resulting quotient with the given K_{eq}
Q = \frac{[SO2][NO2]}{[SO_3][NO]} = \frac{[1.5][2.0]}{[0.85][1.2]} = 2.94
Q = 2.94 which is < K_{eq} (55)
If Q < K_{eq} then the numerator of our quotient must increase
therefore the rxn continues in order to increase [products] until it reaches K_{eq}
Calculating Equilibrium Concentrations
Using initial concentrations, stoichiometry and Kc, equilibrium concentrations of all components can be determined.
Example 9
The equilibrium constant (K{eq}) for the reaction is 1.1 x 10^{-3}. If the initial concentrations are [Br2] = 0.063 M and [Br] = 0.012 M, calculate the concentrations of these species at equilibrium.
Br_2 (g) \rightleftharpoons 2Br (g)
Let x be the change in concentration of Br_2
Br_2 (g) | 2Br (g) | |
|---|---|---|
Initial (M) | 0.063 | 0.012 |
Change (M) | -x | +2x |
Equilibrium (M) | 0.063 - x | 0.012 + 2x |
K{eq} = \frac{[Br]^2}{[Br2]} = \frac{(0.012 + 2x)^2}{0.063 - x} = 1.1 x 10^{-3}
Solve for x
Example 9 - Solution Continued
K_c = \frac{(0.012 + 2x)^2}{0.063 - x} = 1.1 x 10^{-3}
4x^2 + 0.048x + 0.000144 = 0.0000693 – 0.0011x
4x^2 + 0.0491x + 0.0000747 = 0
ax^2 + bx + c =0
x=\frac{-b \pm \sqrt{b^2 – 4ac}}{2a}
Br_2 (g) | 2Br (g) | |
|---|---|---|
Initial (M) | 0.063 | 0.012 |
Change (M) | -x | +2x |
Equilibrium (M) | 0.063 - x | 0.012 + 2x |
x = -0.0105
x = -0.00178
At equilibrium, [Br] = 0.012 + 2x = -0.00900 M or 0.00844 M
At equilibrium, [Br_2] = 0.063 – x = 0.063- (-0.00178) = 0.0648 M
Example 10
The decomposition of HI at low temperature was studied by injecting 2.50 mol of HI into a 10.32 L vessel at 25°C. What is [H2] at equilibrium for the reaction
2HI(g) \rightleftharpoons H2(g) + I2 (g); K_c = 1.26 x 10^{-3} ?
Note the moles into a 10.32 L vessel stuff … calculate molarity. Starting concentration of HI: 2.5 mol/10.32 L = 0.242 M
Example 10 - ICE Table
2 HI | H2 | I2 | |
|---|---|---|---|
Initial: | 0.242 M | 0 | 0 |
Change: | -2x | +x | +x |
Equil: | 0.242-2x | x | x |
What we are asked for here is the equilibrium concentration of H2 …
… otherwise known as x. So, we need to solve this beast for x.
Example 10 - Solution
And yes, it’s a quadratic equation. Doing a bit of rearranging:
Kc= \frac{[H2][I_2]}{[HI]^2} = \frac{x^2}{(0.242 -2x)^2}
x = 0.00802 or –0.00925
Since we are using this to model a real, physical system, we reject the negative root. The [H2] at equil. is 0.00802 M.
Solubility and Solubility product Ksp
As a salt dissolves in water and ions are released, they can collide and re-form the solid
Equilibrium is reached when the rate of dissolution equals the rate of recrystallization
CaF_2(s) \leftrightarrow Ca^{2+}(aq) + 2F^-(aq)
Solubility product Ksp is the same as Keq when you dissolve a salt in water (do not include solid in formula
Solubility refers to x
Solubility Product
Solubility product: Ksp
CaF_2(s) \leftrightarrow Ca^{2+}(aq) + 2F^-(aq)
K_{sp} = [Ca^{2+}][F^-]^2
Leave out the CaF_2 because solid
Adding more solid will not effect the amount of solid that can dissolve at a certain temperature
***K_{sp} in your reference sheet (#7)
K_{sp} Values
TABLE 15.4 K_{sp} Values at 25°C for Common Ionic Solids
(List of various ionic solids and their K_{sp} values at 25°C)
Example 11
CuBr has a solubility of 2.0x10^{-4} mol/L at 25°C. Find the K_{sp} value.
The solubility is x. It tells us the amount of solute that can dissolves in 1 L of water
Use ICE chart: solubility tells you x value
Ksp=[Cu^+][Br^-]= x^2 = (2.0x10^{-4} M)^2 = 4.0x10^{-8}
CuBr(s) | Cu+(aq) | Br-(aq) | |
|---|---|---|---|
I | Not imp. | 0 | 0 |
C | - x | + x | +x |
E | Not imp | x | x |
Example 12
Ag2S has a solubility of 3.4x10^{-17} mol/L at 25°C. Find the K{sp} value.
The solubility tells us the amount of solute that can dissolves in 1 L of water
Use ICE chart: solubility tells you x value
Ksp=[Ag^+]^2[S^{2-}]= (2x)^2(x)= 4x^3 = 1.6x10^{-49}
| | Ag_2S(s)
↔ | 2Ag+(aq) | S2-(aq) |
| :---------- | :------------------------ | :------- | :------- |
| I | Not imp. | 0 | 0 |
| C | - x | +2 x | + x |
| E | Not imp | 2x | x |
Example 13
The K{sp} value for Cu(IO3)_2 is 1.4x10^{-7} at 25°C. Calculate its solubility.
Solve for solubility = x value using ICE chart
K{sp}=[Cu^{2+}][IO3^-]^2= (x)(2x)^2 = 1.4x10^{-7} = 4x^3
X = (3.5x10^{-8})^{1/3} = 3.3x10^{-3}
| | Cu(IO3)2(s)
↔ | Cu^{2+}(aq)
| 2IO_3^-(aq) |
| :---------- | :--------------------------- | :-------------------+ | :---------------------- |
| I | Not imp. | 0 | 0 |
| C | -x | +x | +2x |
| E | Not imp | x | 2x |
Example 14
The K{sp} value for BaCO3 is 1.6x10^{-9} at 25°C. Calculate its solubility.
Solve for solubility = x value using ICE chart
K{sp}=[Ba^{2+}][CO3^{2-}]= (x)(x) = 1.6x10^{-9} = x^2
x = 4.0x10^{-5}
| | BaCO_3(s)
↔ | Ba^{2+}(aq)
| CO_3^{2-}(aq) |
| :---------- | :--------------------------- | :-------------------+ | :---------------------- |
| I | Not imp. | 0 | 0 |
| C | -x | +x | +x |
| E | Not imp | x | x |
Le Chatelier’s Principle
Le Chatelier’s Principle: if you disturb an equilibrium, it will shift to undo the disturbance. In a system at equilibrium, the concentrations will always arrange themselves to multiply and divide in the K_{eq} equation to give the same number (at constant temperature).
Le Châtlier’s Principle - Catalyst
can predict how certain changes in a reaction will affect the position of equilibrium
catalyst
speeds up a reaction rate- the forward and reverse
does not effect equilibrium position
Changing Concentration
system will shift away from the added component or towards a removed component
Ex: N2 + 3H2 \rightleftharpoons 2NH_3
if more N_2 is added, then equilibrium position shifts to right
if some NH_3 is removed, then equilibrium position shifts to right
Change in Pressure - Gases
adding or removing gaseous reactant or product is same as changing conc.
adding inert or uninvolved gas
increase the total pressure
doesn’t effect the equilibrium position
N2 +3 H2 \rightleftharpoons 2NH_3
Change in Pressure - Volume
changing the volume
decrease V of the container
Increase the pressure
shifts towards the side of the reaction with less gas molecules
increase V of the container
Decrease the pressure
shifts towards the side of the reaction with more gas molecules
Change in Temperature
all other changes alter the concentration at equilibrium position but don’t actually change value of K
value of K does change with temperature
As the temperature increases, do more reactants or products exist at equilibrium?
N2(g) + 3 H2(g) \rightleftharpoons 2 NH_3(g) + heat
Change in Temperature - Energy
if energy is added, the reaction will shift in direction that consumes energy
treat energy as a
reactant: for endothermic reactions
product: for exothermic reactions
N2(g) + 3 H2(g) \rightleftharpoons 2 NH_3(g) + heat
Le Chatelier’s Principle - Practice Questions
As4O6(s) + 6C(s) \rightleftharpoons As_4(g) + 6CO(g)
add CO: to left
add C: No shift
remove As4O6: no shift
remove As_4: to right
decrease volume: to left
add Ne gas: no shift
Le Chatelier’s Principle - Practice Questions
P4(s) + 6Cl2(g) \rightleftharpoons 4PCl_3(l)
decrease volume: to right
increase volume: to left
add P_4: no shift
remove Cl_2: to left
add Kr gas: no shift
add PCl_3: no shift
Le Chatelier’s Principle - Practice Questions
NRG + N2(g) + O2(g) \rightleftharpoons 2NO(g)
endo or exo?
\Delta H=181 kJ
endothermic
increase temp: to right
increase volume: no shift
decrease temp: to left
Free Energy and Equilibrium Constants
For a system at equilibrium the standard free energy change (\Delta G^o) for a reaction is directly related to the equilibrium constant for the reaction
\Delta G^o = -RT \ln K
where
\Delta G^o = std. free energy change
R = gas constant (8.314 J) mol.K
T = temp. in Kelvin
K = equilibrium constant
Free Energy and Equilibrium Constants - Calculations
This equation can be used to calculate \Delta G^o for a reaction (and determine its spontaneity) when the equilibrium constant (or the equilibrium concentrations) is known.
The equation can also be rearranged and used to find the value of the equilibrium constant if \Delta G^o for the reaction is known:
K = e^{-\frac{\Delta G^o}{RT}}
Free Energy and Equilibrium Constants Example
Find \Delta G^o for the following reaction at 25oC if K_{eq} = 7.00 x 10^5.
N2 (g) + 3 H2 (g) \rightleftharpoons 2 NH_3 (g)
\Delta G^o =
Formulas and Variables in Chemical Equilibrium
Here's a compilation of the formulas from the chapter, along with the meanings of each variable:
Equilibrium Constant Expression:
K_{eq} = \frac{[C]^c[D]^d}{[A]^a[B]^b}
K_{eq}: Equilibrium constant
[A], [B]: Molar concentrations of reactants
[C], [D]: Molar concentrations of products
a, b, c, d: Stoichiometric coefficients of reactants and products in the balanced chemical equation
Equilibrium Constant in Terms of Pressure:
Kp = \frac{PC^c PD^d}{PA^a P_B^b}
K_p: Equilibrium constant in terms of pressure
PA, PB: Partial pressures of gaseous reactants at equilibrium
PC, PD: Partial pressures of gaseous products at equilibrium
a, b, c, d: Stoichiometric coefficients of reactants and products in the balanced chemical equation
Relation Between K{eq} and Kp:
Kp = K{eq}(RT)^{\Delta n}
K_p: Equilibrium constant in terms of pressure
K_{eq}: Equilibrium constant
R: Ideal gas constant (0.0821 L·atm/mol·K)
T: Temperature in Kelvin
\Delta n: Change in the number of moles of gas (moles of gaseous products - moles of gaseous reactants)
Reaction Quotient:
Q_c = \frac{[C]^c[D]^d}{[A]^a[B]^b}
Q_c: Reaction quotient at any given time
[A], [B]: Initial or non-equilibrium concentrations of reactants
[C], [D]: Initial or non-equilibrium concentrations of products
a, b, c, d: Stoichiometric coefficients of reactants and products in the balanced chemical equation
Solubility Product Constant:
K_{sp} = [M^{m+}]^x[X^{x-}]^y
K_{sp}: Solubility product constant
[M^{m+}]: Concentration of the cation
[X^{x-}]: Concentration of the anion
x, y: Stoichiometric coefficients of the ions in the dissolution equilibrium
Quadratic Formula:
x = \frac{-b \pm \sqrt{b^2 - 4ac}}{2a}
Used to solve quadratic equations of the form ax^2 + bx + c = 0
Free Energy and Equilibrium Constant:
\Delta G^o = -RT \ln K
\Delta G^o: Standard Gibbs free energy change
R: Ideal gas constant (8.314 J/mol·K)
T: Temperature in Kelvin
K: Equilibrium constant
K = e^{-\frac{\Delta G^o}{RT}}
K: Equilibrium constant
\Delta G^o: Standard Gibbs free energy change
R: Ideal gas constant (8.314 J/mol·K)
T$$: Temperature in Kelvin