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Why is water not part of the equilibrium constant (acid base, equilibrium, water, chemistry)?
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Lynn Holland
Why is water not part of the equilibrium constant (acid base, equilibrium, water, chemistry)?
For reactions occurring in an aqueous solution, the expression for the equilibrium constant does not contain the active mass of water even if water is a reactant or a product because water is in excess and practically there is no change in its active mass. But, if water is not present in excess then its active mass has to be included in the equilibrium constant expression. For example, for the esterification reaction
CH3COOH + C2H5OH <=> CH3COOC2H5 + H2O
The equilibrium constant K = [CH3COOC2H5]*[H2O]/[CH3COOH]*[C2H5OH] ,
but for the hydrolysis of ester
CH3COOC2H5 + H2O <=> CH3COOH + C2H5OH
K = [CH3COOH]*[C2H5OH]/[CH3COOC2H5]
because water is present in excess and there is practically no change in its active mass.
For the gaseous reaction
2 C2H2(g) + 5 O2 (g) <=> 4 CO2(g) + 2H2O(g)
K= [CO2]^4 *[H2O]^2/[C2H2]^2 *[O2]^5
But if water is present in the liquid state K does not include the active mass of water. This is because the active mass of a pure solid or a liquid is 1. Another way to look at it is that the partial pressure of water vapour remains constant ( because of temperature remaining constant) if liquid water is the product so there is no need of including its active mass in the expression for K.
For reactions occurring in an aqueous solution, the expression for the equilibrium constant does not contain the active mass of water even if water is a reactant or a product because water is in excess and practically there is no change in its active mass. But, if water is not present in excess then its active mass has to be included in the equilibrium constant expression. For example, for the esterification reaction
CH3COOH + C2H5OH <=> CH3COOC2H5 + H2O
The equilibrium constant K = [CH3COOC2H5]*[H2O]/[CH3COOH]*[C2H5OH] ,
but for the hydrolysis of ester
CH3COOC2H5 + H2O <=> CH3COOH + C2H5OH
K = [CH3COOH]*[C2H5OH]/[CH3COOC2H5]
because water is present in excess and there is practically no change in its active mass.
For the gaseous reaction
2 C2H2(g) + 5 O2 (g) <=> 4 CO2(g) + 2H2O(g)
K= [CO2]^4 *[H2O]^2/[C2H2]^2 *[O2]^5
But if water is present in the liquid state K does not include the active mass of water. This is because the active mass of a pure solid or a liquid is 1. Another way to look at it is that the partial pressure of water vapour remains constant ( because of temperature remaining constant) if liquid water is the product so there is no need of including its active mass in the expression for K.
Very good question, I wish I had more students like you! In many reactions in aqueous solution, water is one of the species in the reaction. Indeed, so why is water not included in the expression for the equilibrium constant? You are learning about the equilibrium constant expressed as the product-quotient of the concentrations of the species in the reaction. However, the equilibrium constant is actually the product-quotient of the activities of the species in the reaction. (This won’t make sense until you’ve had thermodynamics, but briefly, the activity of a species is a measure of how the free energy of the system changes with the concentration of that species). The activity of a species is defined relative to a reference state, where the activity is 1. For a dissolved species in an aqueous solution, the reference state is taken as a 1m solution with the properties of infinite dilution. Using this reference state, it follows that we can approximate the activity of a dissolved species by its concentration, provided that the concentration is not too high. This is why we use concentrations when teaching this stuff in freshman chemistry. For some species (e.g., the solvent), it is more convenient to have the reference state be the pure substance. Again, at the reference state, the activity of a species is 1. So, in an aqueous solution that is not too concentrated, the activity of water is approximately 1 and we can ignore it. (Note, if your chemistry teacher says its because the concentration of water is constant at 55.5 moles/liter and that this number is factored in the equilibrium constant, he/she is winging it and isn’t giving the real answer!). This is also why the solid precipitate does not show up in the equilibrium expression for a solubility reaction; the activity of the solid is 1, as it is at its reference state, and can be ignored.
Very good question, I wish I had more students like you! In many reactions in aqueous solution, water is one of the species in the reaction. Indeed, so why is water not included in the expression for the equilibrium constant? You are learning about the equilibrium constant expressed as the product-quotient of the concentrations of the species in the reaction. However, the equilibrium constant is actually the product-quotient of the activities of the species in the reaction. (This won’t make sense until you’ve had thermodynamics, but briefly, the activity of a species is a measure of how the free energy of the system changes with the concentration of that species). The activity of a species is defined relative to a reference state, where the activity is 1. For a dissolved species in an aqueous solution, the reference state is taken as a 1m solution with the properties of infinite dilution. Using this reference state, it follows that we can approximate the activity of a dissolved species by its concentration, provided that the concentration is not too high. This is why we use concentrations when teaching this stuff in freshman chemistry. For some species (e.g., the solvent), it is more convenient to have the reference state be the pure substance. Again, at the reference state, the activity of a species is 1. So, in an aqueous solution that is not too concentrated, the activity of water is approximately 1 and we can ignore it. (Note, if your chemistry teacher says its because the concentration of water is constant at 55.5 moles/liter and that this number is factored in the equilibrium constant, he/she is winging it and isn’t giving the real answer!). This is also why the solid precipitate does not show up in the equilibrium expression for a solubility reaction; the activity of the solid is 1, as it is at its reference state, and can be ignored.
For reactions occurring in an aqueous solution, the expression for the equilibrium constant does not contain the active mass of water even if water is a reactant or a product because water is in excess and practically there is no change in its active mass. But, if water is not present in excess then its active mass has to be included in the equilibrium constant expression. For example, for the esterification reaction
CH3COOH + C2H5OH <=> CH3COOC2H5 + H2O
The equilibrium constant K = [CH3COOC2H5]*[H2O]/[CH3COOH]*[C2H5OH] ,
but for the hydrolysis of ester
CH3COOC2H5 + H2O <=> CH3COOH + C2H5OH
K = [CH3COOH]*[C2H5OH]/[CH3COOC2H5]
because water is present in excess and there is practically no change in its active mass.
For the gaseous reaction
2 C2H2(g) + 5 O2 (g) <=> 4 CO2(g) + 2H2O(g)
K= [CO2]^4 *[H2O]^2/[C2H2]^2 *[O2]^5
But if water is present in the liquid state K does not include the active mass of water. This is because the active mass of a pure solid or a liquid is 1. Another way to look at it is that the partial pressure of water vapour remains constant ( because of temperature remaining constant) if liquid water is the product so there is no need of including its active mass in the expression for K.
For reactions occurring in an aqueous solution, the expression for the equilibrium constant does not contain the active mass of water even if water is a reactant or a product because water is in excess and practically there is no change in its active mass. But, if water is not present in excess then its active mass has to be included in the equilibrium constant expression. For example, for the esterification reaction
CH3COOH + C2H5OH <=> CH3COOC2H5 + H2O
The equilibrium constant K = [CH3COOC2H5]*[H2O]/[CH3COOH]*[C2H5OH] ,
but for the hydrolysis of ester
CH3COOC2H5 + H2O <=> CH3COOH + C2H5OH
K = [CH3COOH]*[C2H5OH]/[CH3COOC2H5]
because water is present in excess and there is practically no change in its active mass.
For the gaseous reaction
2 C2H2(g) + 5 O2 (g) <=> 4 CO2(g) + 2H2O(g)
K= [CO2]^4 *[H2O]^2/[C2H2]^2 *[O2]^5
But if water is present in the liquid state K does not include the active mass of water. This is because the active mass of a pure solid or a liquid is 1. Another way to look at it is that the partial pressure of water vapour remains constant ( because of temperature remaining constant) if liquid water is the product so there is no need of including its active mass in the expression for K.
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Very good question, I wish I had more students like you! In many reactions in aqueous solution, water is one of the species in the reaction. Indeed, so why is water not included in the expression for the equilibrium constant? You are learning about the equilibrium constant expressed as the product-quotient of the concentrations of the species in the reaction. However, the equilibrium constant is actually the product-quotient of the activities of the species in the reaction. (This won’t make sense until you’ve had thermodynamics, but briefly, the activity of a species is a measure of how the free energy of the system changes with the concentration of that species). The activity of a species is defined relative to a reference state, where the activity is 1. For a dissolved species in an aqueous solution, the reference state is taken as a 1m solution with the properties of infinite dilution. Using this reference state, it follows that we can approximate the activity of a dissolved species by its concentration, provided that the concentration is not too high. This is why we use concentrations when teaching this stuff in freshman chemistry. For some species (e.g., the solvent), it is more convenient to have the reference state be the pure substance. Again, at the reference state, the activity of a species is 1. So, in an aqueous solution that is not too concentrated, the activity of water is approximately 1 and we can ignore it. (Note, if your chemistry teacher says its because the concentration of water is constant at 55.5 moles/liter and that this number is factored in the equilibrium constant, he/she is winging it and isn’t giving the real answer!). This is also why the solid precipitate does not show up in the equilibrium expression for a solubility reaction; the activity of the solid is 1, as it is at its reference state, and can be ignored.
Very good question, I wish I had more students like you! In many reactions in aqueous solution, water is one of the species in the reaction. Indeed, so why is water not included in the expression for the equilibrium constant? You are learning about the equilibrium constant expressed as the product-quotient of the concentrations of the species in the reaction. However, the equilibrium constant is actually the product-quotient of the activities of the species in the reaction. (This won’t make sense until you’ve had thermodynamics, but briefly, the activity of a species is a measure of how the free energy of the system changes with the concentration of that species). The activity of a species is defined relative to a reference state, where the activity is 1. For a dissolved species in an aqueous solution, the reference state is taken as a 1m solution with the properties of infinite dilution. Using this reference state, it follows that we can approximate the activity of a dissolved species by its concentration, provided that the concentration is not too high. This is why we use concentrations when teaching this stuff in freshman chemistry. For some species (e.g., the solvent), it is more convenient to have the reference state be the pure substance. Again, at the reference state, the activity of a species is 1. So, in an aqueous solution that is not too concentrated, the activity of water is approximately 1 and we can ignore it. (Note, if your chemistry teacher says its because the concentration of water is constant at 55.5 moles/liter and that this number is factored in the equilibrium constant, he/she is winging it and isn’t giving the real answer!). This is also why the solid precipitate does not show up in the equilibrium expression for a solubility reaction; the activity of the solid is 1, as it is at its reference state, and can be ignored.
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