14 added 7 characters in body
source | link

where $p$ is the total pressure and $v_i$ is the molar volume of the $i^{\text{th}}$ component in the pure phase. Substituting $v_{i}$ via the ideal gas law and subsequently integrating this equation w.r.t. pressure using the total pressure $p$ as the upper and the partial pressure $p_{i}$ as the lower bound for the integration we get \begin{equation} \int^{\mu^{*}_{i}(p)}_{\mu^{*}_{i}(p_{i})} \mathrm{d} \mu^{*}_{i} = \int_{p_{i}}^{p} \underbrace{v_{i}}_{=\frac{RT}{p}} \mathrm{d} p = R T \int_{p_{i}}^{p} \frac{1}{p} \mathrm{d} p = RT \int_{p_{i}}^{p} \mathrm{d} \ln p \ , \end{equation} so that, introducing the mole fraction $x_{i}$, \begin{equation} \mu_{i}^{*} (p_{i}, T) = \mu_{i}^{*}(p, T) + RT \ln \Bigl(\underbrace{\frac{p_{i}}{p}}_{= x_{i}}\Bigr) = \mu_{i}^{*}(p, T) + RT \ln x_{i} \ . \end{equation} Please, note that there is a dimensionless quantity inside the logarithm. Now, for real gases one has to adjust this equation a little bit: one has to correct the pressure for the errors introduced by the interactions present in real gases. Thus, one introduces the (dimensionless) activity $a_{i}$ by scaling the pressure with the (dimensionless) fugacity coefficient $\varphi_{i}$ \begin{equation} a_{i} = \frac{\varphi_{i} p_{i}}{p^0} \end{equation} where $p^{0}$ is the standard pressure for which $\varphi_{i}=1$ by definition. When this is in turn substituted into the equilibrium equation, whereby the total pressure is chosen to be the standard pressure $p = p^{0}$, the following equation arises \begin{equation} \mu_{i} (p, T) = \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \ln a_{i} \ . \end{equation} Substituting all this togther in our equation for $\Delta G$ and noting that the sum of logarithms can be written as a logarithm of products, $\sum_{i} \ln i = \ln \prod_i i$, one gets \begin{equation} \Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \log \prod [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod [a_{i}]^{\nu_{i}} \ , \end{equation}\begin{equation} \Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \ln \prod_{i} [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod_{i} [a_{i}]^{\nu_{i}} \ , \end{equation} where the standard Gibbs free energy of reaction $\Delta G^{0}$ has been introduced by asserting that the system is under standard pressure. Now, we are nearly finished. One only has to note that $\Delta G = 0$ since the system is in equilibrium and then one can introduce the equilibrium constant $K$, so that \begin{equation} \ln \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \ln K = -\frac{\Delta G^{0}}{RT} \ . \end{equation} So, you see this quantity is dimensionless. The problem is that activities are hard to come by. Concentrations $c_{i}$ or pressures are much easier to measure. So, what one does now, is to introduce a different equilibrium constant \begin{equation} K_{c} = \prod_i [c_{i}]^{\nu_{i}} \ . \end{equation} which is much easier to measure since it depends on concentrations rather than activities. It is not dimensionless but being connected with the "real" dimensionless equilibrium constant via \begin{equation} K = \prod_i [\varphi_{i}]^{\nu_{i}} \left(\frac{RT}{p^{0}}\right)^{\sum_i \nu_{i}} K_{c} \ . \end{equation} it is more or less proportional to $K$ and thus gives qualitatively the same information.

where $p$ is the total pressure and $v_i$ is the molar volume of the $i^{\text{th}}$ component in the pure phase. Substituting $v_{i}$ via the ideal gas law and subsequently integrating this equation w.r.t. pressure using the total pressure $p$ as the upper and the partial pressure $p_{i}$ as the lower bound for the integration we get \begin{equation} \int^{\mu^{*}_{i}(p)}_{\mu^{*}_{i}(p_{i})} \mathrm{d} \mu^{*}_{i} = \int_{p_{i}}^{p} \underbrace{v_{i}}_{=\frac{RT}{p}} \mathrm{d} p = R T \int_{p_{i}}^{p} \frac{1}{p} \mathrm{d} p = RT \int_{p_{i}}^{p} \mathrm{d} \ln p \ , \end{equation} so that, introducing the mole fraction $x_{i}$, \begin{equation} \mu_{i}^{*} (p_{i}, T) = \mu_{i}^{*}(p, T) + RT \ln \Bigl(\underbrace{\frac{p_{i}}{p}}_{= x_{i}}\Bigr) = \mu_{i}^{*}(p, T) + RT \ln x_{i} \ . \end{equation} Please, note that there is a dimensionless quantity inside the logarithm. Now, for real gases one has to adjust this equation a little bit: one has to correct the pressure for the errors introduced by the interactions present in real gases. Thus, one introduces the (dimensionless) activity $a_{i}$ by scaling the pressure with the (dimensionless) fugacity coefficient $\varphi_{i}$ \begin{equation} a_{i} = \frac{\varphi_{i} p_{i}}{p^0} \end{equation} where $p^{0}$ is the standard pressure for which $\varphi_{i}=1$ by definition. When this is in turn substituted into the equilibrium equation, whereby the total pressure is chosen to be the standard pressure $p = p^{0}$, the following equation arises \begin{equation} \mu_{i} (p, T) = \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \ln a_{i} \ . \end{equation} Substituting all this togther in our equation for $\Delta G$ and noting that the sum of logarithms can be written as a logarithm of products, $\sum_{i} \ln i = \ln \prod_i i$, one gets \begin{equation} \Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \log \prod [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod [a_{i}]^{\nu_{i}} \ , \end{equation} where the standard Gibbs free energy of reaction $\Delta G^{0}$ has been introduced by asserting that the system is under standard pressure. Now, we are nearly finished. One only has to note that $\Delta G = 0$ since the system is in equilibrium and then one can introduce the equilibrium constant $K$, so that \begin{equation} \ln \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \ln K = -\frac{\Delta G^{0}}{RT} \ . \end{equation} So, you see this quantity is dimensionless. The problem is that activities are hard to come by. Concentrations $c_{i}$ or pressures are much easier to measure. So, what one does now, is to introduce a different equilibrium constant \begin{equation} K_{c} = \prod_i [c_{i}]^{\nu_{i}} \ . \end{equation} which is much easier to measure since it depends on concentrations rather than activities. It is not dimensionless but being connected with the "real" dimensionless equilibrium constant via \begin{equation} K = \prod_i [\varphi_{i}]^{\nu_{i}} \left(\frac{RT}{p^{0}}\right)^{\sum_i \nu_{i}} K_{c} \ . \end{equation} it is more or less proportional to $K$ and thus gives qualitatively the same information.

where $p$ is the total pressure and $v_i$ is the molar volume of the $i^{\text{th}}$ component in the pure phase. Substituting $v_{i}$ via the ideal gas law and subsequently integrating this equation w.r.t. pressure using the total pressure $p$ as the upper and the partial pressure $p_{i}$ as the lower bound for the integration we get \begin{equation} \int^{\mu^{*}_{i}(p)}_{\mu^{*}_{i}(p_{i})} \mathrm{d} \mu^{*}_{i} = \int_{p_{i}}^{p} \underbrace{v_{i}}_{=\frac{RT}{p}} \mathrm{d} p = R T \int_{p_{i}}^{p} \frac{1}{p} \mathrm{d} p = RT \int_{p_{i}}^{p} \mathrm{d} \ln p \ , \end{equation} so that, introducing the mole fraction $x_{i}$, \begin{equation} \mu_{i}^{*} (p_{i}, T) = \mu_{i}^{*}(p, T) + RT \ln \Bigl(\underbrace{\frac{p_{i}}{p}}_{= x_{i}}\Bigr) = \mu_{i}^{*}(p, T) + RT \ln x_{i} \ . \end{equation} Please, note that there is a dimensionless quantity inside the logarithm. Now, for real gases one has to adjust this equation a little bit: one has to correct the pressure for the errors introduced by the interactions present in real gases. Thus, one introduces the (dimensionless) activity $a_{i}$ by scaling the pressure with the (dimensionless) fugacity coefficient $\varphi_{i}$ \begin{equation} a_{i} = \frac{\varphi_{i} p_{i}}{p^0} \end{equation} where $p^{0}$ is the standard pressure for which $\varphi_{i}=1$ by definition. When this is in turn substituted into the equilibrium equation, whereby the total pressure is chosen to be the standard pressure $p = p^{0}$, the following equation arises \begin{equation} \mu_{i} (p, T) = \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \ln a_{i} \ . \end{equation} Substituting all this togther in our equation for $\Delta G$ and noting that the sum of logarithms can be written as a logarithm of products, $\sum_{i} \ln i = \ln \prod_i i$, one gets \begin{equation} \Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \ln \prod_{i} [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod_{i} [a_{i}]^{\nu_{i}} \ , \end{equation} where the standard Gibbs free energy of reaction $\Delta G^{0}$ has been introduced by asserting that the system is under standard pressure. Now, we are nearly finished. One only has to note that $\Delta G = 0$ since the system is in equilibrium and then one can introduce the equilibrium constant $K$, so that \begin{equation} \ln \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \ln K = -\frac{\Delta G^{0}}{RT} \ . \end{equation} So, you see this quantity is dimensionless. The problem is that activities are hard to come by. Concentrations $c_{i}$ or pressures are much easier to measure. So, what one does now, is to introduce a different equilibrium constant \begin{equation} K_{c} = \prod_i [c_{i}]^{\nu_{i}} \ . \end{equation} which is much easier to measure since it depends on concentrations rather than activities. It is not dimensionless but being connected with the "real" dimensionless equilibrium constant via \begin{equation} K = \prod_i [\varphi_{i}]^{\nu_{i}} \left(\frac{RT}{p^{0}}\right)^{\sum_i \nu_{i}} K_{c} \ . \end{equation} it is more or less proportional to $K$ and thus gives qualitatively the same information.

13 deleted 4 characters in body
source | link

where $p$ is the total pressure and $v_i$ is the molar volume of the $i^{\text{th}}$ component in the pure phase. Substituting $v_{i}$ via the ideal gas law and subsequently integrating this equation w.r.t. pressure using the total pressure $p$ as the upper and the partial pressure $p_{i}$ as the lower bound for the integration we get \begin{equation} \int^{\mu^{*}_{i}(p)}_{\mu^{*}_{i}(p_{i})} \mathrm{d} \mu^{*}_{i} = \int_{p_{i}}^{p} \underbrace{v_{i}}_{=\frac{RT}{p}} \mathrm{d} p = R T \int_{p_{i}}^{p} \frac{1}{p} \mathrm{d} p = RT \int_{p_{i}}^{p} \mathrm{d} \ln p \ , \end{equation} so that, introducing the mole fraction $x_{i}$, \begin{equation} \mu_{i}^{*} (p_{i}, T) = \mu_{i}^{*}(p, T) + RT \ln \Bigl(\underbrace{\frac{p_{i}}{p}}_{= x_{i}}\Bigr) = \mu_{i}^{*}(p, T) + RT \ln x_{i} \ . \end{equation} Please, note that there is a dimensionless quantity inside the logarithm. Now, for real gases one has to adjust this equation a little bit: one has to correct the pressure for the errors introduced by the interactions present in real gases. Thus, one introduces the (dimensionless) activity $a_{i}$ by scaling the pressure with the (dimensionless) fugacity coefficient $\varphi_{i}$ \begin{equation} a_{i} = \frac{\varphi_{i} p_{i}}{p^0} \end{equation} where $p^{0}$ is the standard pressure for which $\varphi_{i}=1$ by definition. When this is in turn substituted into the equilibrium equation, whereby the total pressure is chosen to be the standard pressure $p = p^{0}$, the following equation arises \begin{equation} \mu_{i} (p_{i}, T) = \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \ln a_{i} \ . \end{equation}\begin{equation} \mu_{i} (p, T) = \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \ln a_{i} \ . \end{equation} Substituting all this togther in our equation for $\Delta G$ and noting that the sum of logarithms can be written as a logarithm of products, $\sum_{i} \ln i = \ln \prod_i i$, one gets \begin{equation} \Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \log \prod [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod [a_{i}]^{\nu_{i}} \ , \end{equation} where the standard Gibbs free energy of reaction $\Delta G^{0}$ has been introduced by asserting that the system is under standard pressure. Now, we are nearly finished. One only has to note that $\Delta G = 0$ since the system is in equilibrium and then one can introduce the equilibrium constant $K$, so that \begin{equation} \ln \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \ln K = -\frac{\Delta G^{0}}{RT} \ . \end{equation} So, you see this quantity is dimensionless. The problem is that activities are hard to come by. Concentrations $c_{i}$ or pressures are much easier to measure. So, what one does now, is to introduce a different equilibrium constant \begin{equation} K_{c} = \prod_i [c_{i}]^{\nu_{i}} \ . \end{equation} which is much easier to measure since it depends on concentrations rather than activities. It is not dimensionless but being connected with the "real" dimensionless equilibrium constant via \begin{equation} K = \prod_i [\varphi_{i}]^{\nu_{i}} \left(\frac{RT}{p^{0}}\right)^{\sum_i \nu_{i}} K_{c} \ . \end{equation} it is more or less proportional to $K$ and thus gives qualitatively the same information.

where $p$ is the total pressure and $v_i$ is the molar volume of the $i^{\text{th}}$ component in the pure phase. Substituting $v_{i}$ via the ideal gas law and subsequently integrating this equation w.r.t. pressure using the total pressure $p$ as the upper and the partial pressure $p_{i}$ as the lower bound for the integration we get \begin{equation} \int^{\mu^{*}_{i}(p)}_{\mu^{*}_{i}(p_{i})} \mathrm{d} \mu^{*}_{i} = \int_{p_{i}}^{p} \underbrace{v_{i}}_{=\frac{RT}{p}} \mathrm{d} p = R T \int_{p_{i}}^{p} \frac{1}{p} \mathrm{d} p = RT \int_{p_{i}}^{p} \mathrm{d} \ln p \ , \end{equation} so that, introducing the mole fraction $x_{i}$, \begin{equation} \mu_{i}^{*} (p_{i}, T) = \mu_{i}^{*}(p, T) + RT \ln \Bigl(\underbrace{\frac{p_{i}}{p}}_{= x_{i}}\Bigr) = \mu_{i}^{*}(p, T) + RT \ln x_{i} \ . \end{equation} Please, note that there is a dimensionless quantity inside the logarithm. Now, for real gases one has to adjust this equation a little bit: one has to correct the pressure for the errors introduced by the interactions present in real gases. Thus, one introduces the (dimensionless) activity $a_{i}$ by scaling the pressure with the (dimensionless) fugacity coefficient $\varphi_{i}$ \begin{equation} a_{i} = \frac{\varphi_{i} p_{i}}{p^0} \end{equation} where $p^{0}$ is the standard pressure for which $\varphi_{i}=1$ by definition. When this is in turn substituted into the equilibrium equation, whereby the total pressure is chosen to be the standard pressure $p = p^{0}$, the following equation arises \begin{equation} \mu_{i} (p_{i}, T) = \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \ln a_{i} \ . \end{equation} Substituting all this togther in our equation for $\Delta G$ and noting that the sum of logarithms can be written as a logarithm of products, $\sum_{i} \ln i = \ln \prod_i i$, one gets \begin{equation} \Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \log \prod [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod [a_{i}]^{\nu_{i}} \ , \end{equation} where the standard Gibbs free energy of reaction $\Delta G^{0}$ has been introduced by asserting that the system is under standard pressure. Now, we are nearly finished. One only has to note that $\Delta G = 0$ since the system is in equilibrium and then one can introduce the equilibrium constant $K$, so that \begin{equation} \ln \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \ln K = -\frac{\Delta G^{0}}{RT} \ . \end{equation} So, you see this quantity is dimensionless. The problem is that activities are hard to come by. Concentrations $c_{i}$ or pressures are much easier to measure. So, what one does now, is to introduce a different equilibrium constant \begin{equation} K_{c} = \prod_i [c_{i}]^{\nu_{i}} \ . \end{equation} which is much easier to measure since it depends on concentrations rather than activities. It is not dimensionless but being connected with the "real" dimensionless equilibrium constant via \begin{equation} K = \prod_i [\varphi_{i}]^{\nu_{i}} \left(\frac{RT}{p^{0}}\right)^{\sum_i \nu_{i}} K_{c} \ . \end{equation} it is more or less proportional to $K$ and thus gives qualitatively the same information.

where $p$ is the total pressure and $v_i$ is the molar volume of the $i^{\text{th}}$ component in the pure phase. Substituting $v_{i}$ via the ideal gas law and subsequently integrating this equation w.r.t. pressure using the total pressure $p$ as the upper and the partial pressure $p_{i}$ as the lower bound for the integration we get \begin{equation} \int^{\mu^{*}_{i}(p)}_{\mu^{*}_{i}(p_{i})} \mathrm{d} \mu^{*}_{i} = \int_{p_{i}}^{p} \underbrace{v_{i}}_{=\frac{RT}{p}} \mathrm{d} p = R T \int_{p_{i}}^{p} \frac{1}{p} \mathrm{d} p = RT \int_{p_{i}}^{p} \mathrm{d} \ln p \ , \end{equation} so that, introducing the mole fraction $x_{i}$, \begin{equation} \mu_{i}^{*} (p_{i}, T) = \mu_{i}^{*}(p, T) + RT \ln \Bigl(\underbrace{\frac{p_{i}}{p}}_{= x_{i}}\Bigr) = \mu_{i}^{*}(p, T) + RT \ln x_{i} \ . \end{equation} Please, note that there is a dimensionless quantity inside the logarithm. Now, for real gases one has to adjust this equation a little bit: one has to correct the pressure for the errors introduced by the interactions present in real gases. Thus, one introduces the (dimensionless) activity $a_{i}$ by scaling the pressure with the (dimensionless) fugacity coefficient $\varphi_{i}$ \begin{equation} a_{i} = \frac{\varphi_{i} p_{i}}{p^0} \end{equation} where $p^{0}$ is the standard pressure for which $\varphi_{i}=1$ by definition. When this is in turn substituted into the equilibrium equation, whereby the total pressure is chosen to be the standard pressure $p = p^{0}$, the following equation arises \begin{equation} \mu_{i} (p, T) = \underbrace{\mu_{i}^{*}(p^{0}, T)}_{= \, \mu_{i}^{0}} + RT \ln a_{i} \ . \end{equation} Substituting all this togther in our equation for $\Delta G$ and noting that the sum of logarithms can be written as a logarithm of products, $\sum_{i} \ln i = \ln \prod_i i$, one gets \begin{equation} \Delta G = \underbrace{\sum_i \nu_{i} \mu_{i}^{0}}_{= \, \Delta G^{0}} + RT \underbrace{\sum_i \nu_{i} \ln a_{i}}_{= \, \log \prod [a_{i}]^{\nu_{i}}} = \Delta G^{0} + RT \ln \prod [a_{i}]^{\nu_{i}} \ , \end{equation} where the standard Gibbs free energy of reaction $\Delta G^{0}$ has been introduced by asserting that the system is under standard pressure. Now, we are nearly finished. One only has to note that $\Delta G = 0$ since the system is in equilibrium and then one can introduce the equilibrium constant $K$, so that \begin{equation} \ln \underbrace{\prod_i [a_{i}]^{\nu_{i}}}_{= \, K} = -\frac{\Delta G^{0}}{RT} \qquad \Rightarrow \qquad \ln K = -\frac{\Delta G^{0}}{RT} \ . \end{equation} So, you see this quantity is dimensionless. The problem is that activities are hard to come by. Concentrations $c_{i}$ or pressures are much easier to measure. So, what one does now, is to introduce a different equilibrium constant \begin{equation} K_{c} = \prod_i [c_{i}]^{\nu_{i}} \ . \end{equation} which is much easier to measure since it depends on concentrations rather than activities. It is not dimensionless but being connected with the "real" dimensionless equilibrium constant via \begin{equation} K = \prod_i [\varphi_{i}]^{\nu_{i}} \left(\frac{RT}{p^{0}}\right)^{\sum_i \nu_{i}} K_{c} \ . \end{equation} it is more or less proportional to $K$ and thus gives qualitatively the same information.

12 added 1 character in body
source | link

but since $\mu^{*}_{i}$ is associated with a pure facephase, $\left( \frac{\partial V_{i}}{\partial n_{i}} \right)_{T}$ can be simplified to \begin{equation} \left(\frac{\partial V_{i}}{\partial n_{i}} \right)_{T} = \frac{V_{i}}{n_{i}} = v_{i} \end{equation}

but since $\mu^{*}_{i}$ is associated with a pure face, $\left( \frac{\partial V_{i}}{\partial n_{i}} \right)_{T}$ can be simplified to \begin{equation} \left(\frac{\partial V_{i}}{\partial n_{i}} \right)_{T} = \frac{V_{i}}{n_{i}} = v_{i} \end{equation}

but since $\mu^{*}_{i}$ is associated with a pure phase, $\left( \frac{\partial V_{i}}{\partial n_{i}} \right)_{T}$ can be simplified to \begin{equation} \left(\frac{\partial V_{i}}{\partial n_{i}} \right)_{T} = \frac{V_{i}}{n_{i}} = v_{i} \end{equation}

11 deleted 2 characters in body
source | link
10 edited body
source | link
9 changed log to ln in order to avoid confusion with the decadic logarithm
source | link
8 added details for the Maxwell relation used, corrected spelling
source | link
7 added 24 characters in body
source | link
6 added 791 characters in body
source | link
5 added 791 characters in body
source | link
4 added 147 characters in body
source | link
3 added 105 characters in body
source | link
2 added 295 characters in body
source | link
1
source | link