What Is the Value of K for This Aqueous Reaction at 298 K?
The question “what is the value of K for this aqueous reaction at 298 k?” is generally followed by a reaction in which the thermodynamic equilibrium constant is to be determined. Under the equilibrium state, the ratio of the reactant’s concentration and that of the reaction result depends on the temperature and type of the equilibrium reaction. The Norwegian chemists Cato Maximilian Guldberg and Peter Waage stated that in an equilibrium reaction, the law of equilibrium applies.
Free Energy and Thermodynamic Equilibrium
To determine the value of the thermodynamic equilibrium constant, it is important to understand the relationship between thermodynamic equilibrium and the standard Gibbs free energy.
What is Gibbs free energy? Free energy is the maximum amount of energy released in a certain process that occurs at constant temperature and pressure. Free energy is symbolized by ΔG. At a constant temperature and pressure, the chemical reaction will take place spontaneously in the direction of a condition with lower free energy changes, until finally reaching a state of equilibrium. Thus, the state of equilibrium determines the lowest free energy value in a reaction system. If a chemical reaction reaches a thermodynamic equilibrium, ΔG=0.
Changes in the free energy in a reaction are the result of changes in pressure or changes in the concentration of substances involved in the reaction. Thus, there is a relationship between changes of free energy and pressure or the concentration of the reaction system.
Free Energy and Pressure
The effect of pressure on the thermodynamic function in the Gibbs free energy equation takes the form of enthalpy and entropy.
Enthalpy is the ability of the system to provide heat energy from the system to the environment. at constant pressure, the enthalpy change in a system shows the amount of heat given. this change is measured using an open calorie meter (system air pressure = outside air pressure).
The enthalpy of an ideal gas, solid, and liquid does not depend on pressure. Real objects at room temperature and pressure usually more or less have these properties so as to simplify the calculation of enthalpy.
In classical thermodynamics, the concept of entropy is defined in the second law of thermodynamics, which states that the entropy of an isolated system always increases or remains constant. Thus, entropy can also be a measurement of the tendency of a process, whether the process tends to be “affected by entropy” or will go to a certain direction. Entropy also shows that heat energy always flows spontaneously from an area of higher temperature to an area of lower temperature.
Gibbs Free Energy Formula to Determine the Value of K
The formula for Gibbs free energy is ΔG⁰ = – RT lnK, with R referring to the universal or ideal gas constant (0.008314 kj/mol-K) and T referring to temperature in Kelvin. Based on this formula, what is the value of K for this aqueous reaction at 298 k? Based on the aforementioned formula, you can reverse the equation to discover the value of K by raising e to the power of each side of the equation because the natural logarithm is the logarithm to the base of the mathematical constant e. The value of e is 2.7182818284591.
Therefore, in order to figure out the value of K, you can inverse the equation to K = e-ΔG/RT
The question that you are attempting to solve is “what is the value of K for this aqueous reaction at 298 k?” Based on this question, you should already get all values needed for solving the equation if the ΔG of the reaction is determined in the question. So, for instance, if you are to solve the value of K in the reaction of αA + βB ↔ γC + δD, with the ΔG determined at 11.49 kJ/mol, you can put all of the known values in the formula to find the value of K.
Let’s start by listing all known values as follows.
e = 2.72
ΔG = 11.49 kJ/mol
R = 0.008314 kj/mol-K
T = 298 K
Then let us put all of the known values in the inverted formula to determine the value of K, which is K = e-ΔG/RT
K = e-ΔG/RT
K = 2.72-11.49 / 0.008314 · 298
K = 2.72-11.49 / 2.48
K = 2.72−4.64
K = 0.0096
K = 9.6 × 10-3
This equation is valid to all problems in which the question “what is the value of K for this aqueous reaction at 298 k?” is given, provided that the value of ΔG is also already clarified.