
What are the standard conditions?
For example, in the equation \[\Delta {G^o} = \,\Delta {H^o}\, - \,\,T\,\Delta {S^o}\]
\[\Delta {G^o}\] seems to be temperature dependent, so does that mean \[\Delta {G^o}\] only applies at a specific pressure?
Answer
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Hint: We have determined the standard condition for the equation \[\Delta {G^o} = \,\Delta {H^o}\, - \,\,T\,\Delta {S^o}\] . Here, the standard condition indicates the standard temperature, standard pressure. We can also use it as STP which means standard temperature and pressure.
Complete answer:
The above equation given in the above question i.e. \[\Delta {G^o} = \,\Delta {H^o}\, - \,\,T\,\Delta {S^o}\] is the equation for standard Gibbs free energy change. The Gibbs free energy is defined in terms of the three functions i.e. temperature, enthalpy and entropy. It is a state function which means it depends on the initial and final state of the reaction. It is used to predict whether the reaction is spontaneous or not at constant temperature and pressure. It is also the chemical potential that is minimized when a system reaches equilibrium at constant pressure and temperature.
The equation for change in Gibbs free energy at constant temperature and pressure where all the thermodynamic quantity are those of the system is:
\[\Delta G = \,\Delta H\, - \,\,T\,\Delta S\]
At standard conditions of temperature and pressure; it is written as:
\[\Delta {G^o} = \,\Delta {H^o}\, - \,\,T\,\Delta {S^o}\]
The standard conditions of temperature and pressure (or STP) are as follows:
Temperature = \[273.15\] K
Pressure = \[1\] bar or \[1\] atm
Also, at STP
\[1\] mol of ideal gas = \[22.711\] L at \[1\] bar
\[1\] mol of ideal gas = \[22.414\] L at \[1\] atm
The value of \[\Delta {G^o}\] calculated under the standard conditions characterizes the “driving force” of the reaction toward equilibrium.
\[\Delta {G^o}\] \[ < \] \[0\] the reaction is spontaneous i.e. reaction is exergonic
\[\Delta {G^o}\] \[ = \] \[0\] the system is at equilibrium.
\[\Delta {G^o}\] \[ > \] \[0\] the reaction is not spontaneous i.e. reaction is endergonic.
Note:
Do not get confused with thermodynamic standard conditions with the “standard temperature and pressure” (STP) widely employed in gas law calculations. It is similar but Temperature is equal to \[298.15\] K and the pressure is equal to \[1\] bar.
Complete answer:
The above equation given in the above question i.e. \[\Delta {G^o} = \,\Delta {H^o}\, - \,\,T\,\Delta {S^o}\] is the equation for standard Gibbs free energy change. The Gibbs free energy is defined in terms of the three functions i.e. temperature, enthalpy and entropy. It is a state function which means it depends on the initial and final state of the reaction. It is used to predict whether the reaction is spontaneous or not at constant temperature and pressure. It is also the chemical potential that is minimized when a system reaches equilibrium at constant pressure and temperature.
The equation for change in Gibbs free energy at constant temperature and pressure where all the thermodynamic quantity are those of the system is:
\[\Delta G = \,\Delta H\, - \,\,T\,\Delta S\]
At standard conditions of temperature and pressure; it is written as:
\[\Delta {G^o} = \,\Delta {H^o}\, - \,\,T\,\Delta {S^o}\]
The standard conditions of temperature and pressure (or STP) are as follows:
Temperature = \[273.15\] K
Pressure = \[1\] bar or \[1\] atm
Also, at STP
\[1\] mol of ideal gas = \[22.711\] L at \[1\] bar
\[1\] mol of ideal gas = \[22.414\] L at \[1\] atm
The value of \[\Delta {G^o}\] calculated under the standard conditions characterizes the “driving force” of the reaction toward equilibrium.
\[\Delta {G^o}\] \[ < \] \[0\] the reaction is spontaneous i.e. reaction is exergonic
\[\Delta {G^o}\] \[ = \] \[0\] the system is at equilibrium.
\[\Delta {G^o}\] \[ > \] \[0\] the reaction is not spontaneous i.e. reaction is endergonic.
Note:
Do not get confused with thermodynamic standard conditions with the “standard temperature and pressure” (STP) widely employed in gas law calculations. It is similar but Temperature is equal to \[298.15\] K and the pressure is equal to \[1\] bar.
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