Use data from Appendix IVB to calculate the equilibrium...

Use data from Appendix IVB to calculate the equilibrium constant at $25^{\circ} \mathrm{C}$ for each reaction. $\Delta G_{f}^{\circ}$ for $\mathrm{BrCl}(\mathrm{g})$ is $-1.0 \mathrm{kJ} / \mathrm{mol}$ $$\begin{array}{l}{\text { a. } 2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)} \\ {\text { b. } \mathrm{Br}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{BrCl}(g)}\end{array}$$

Use data from Appendix IVB to calculate the equilibrium constant at
$25^{\circ} \mathrm{C}$ for each reaction. $\Delta G_{f}^{\circ}$ for $\mathrm{BrCl}(\mathrm{g})$ is $-1.0 \mathrm{kJ} / \mathrm{mol}$
$$\begin{array}{l}{\text { a. } 2 \mathrm{NO}_{2}(g) \rightleftharpoons \mathrm{N}_{2} \mathrm{O}_{4}(g)} \\ {\text { b. } \mathrm{Br}_{2}(g)+\mathrm{Cl}_{2}(g) \rightleftharpoons 2 \mathrm{BrCl}(g)}\end{array}$$

Key Concepts

Gibbs Free Energy and Equilibrium Relationship

The relationship between the standard Gibbs free energy change (?G°) and the equilibrium constant (K) is given by the equation ?G° = -RT ln K, where R is the universal gas constant and T is the temperature in Kelvin. This equation provides a direct link between thermodynamics and equilibrium, allowing for the calculation of K when ?G° is known, and is central in understanding how energy considerations govern chemical equilibria.

Standard Gibbs Free Energy (?G°)

Standard Gibbs free energy is a thermodynamic quantity that indicates the spontaneity of a reaction under standard conditions, typically 1 bar pressure and a specified temperature (such as 25°C). A negative value implies that the reaction can proceed spontaneously in the forward direction, while a positive value suggests non-spontaneity.

Chemical Equilibrium

Chemical equilibrium refers to the state of a reversible chemical reaction wherein the rates of the forward and reverse processes are equal, leading to constant concentrations of reactants and products over time. Understanding this concept is essential in predicting how changes in conditions, such as temperature or pressure, might shift the equilibrium position to favor either the reactants or the products.

Equilibrium Constant (K)

The equilibrium constant is a dimensionless quantity that provides a measure of the ratio of concentrations (or pressures) of the products to the reactants at equilibrium for a given reaction, each raised to the power of their stoichiometric coefficients. It is critical for quantifying the extent to which a reaction proceeds and is derived under specific standard conditions.

Transcript

00:01 For each reaction that we are given in this problem, we are going to solve for the value of the equilibrium constant k, and we are going to use this equation outlined in red in order to help us determine that value.

00:15 We can use the given stoichiometry of the reaction as well as data in the appendix in the standard change in gibbs -free energy of the overall reaction.

00:27 At a temperature of 298 kelvin and the given value for the gas constant r, we can rearrange that equation to solve directly for the equilibrium constant k.

00:42 In part a, this is the reaction that we are working with.

00:46 We need to begin by using thermodynamic data in the appendix to solve for that change in gibbs -free energy at standard conditions.

00:53 Remember the way that we do that is by looking up the values for the delta g of formation.

00:58 At standard conditions for each one of these chemical species so that we can find the overall delta g of the products and the reactants and then subtract them if we look at this reaction the only product we have is one mole of n204 so when we set up our expression for delta g at standard conditions for this reaction we start with the products and we have one mole of n204 we can look up its delta geof formation value at standard conditions in the appendix and that comes out to 99 .8 kilojoules per mole and that's the only product so for the reactants we now subtract two moles of no2 and now for n02 we look up its delta geo formation value at standard conditions and that will be 51 .3 kilojoules per mole we multiply the delta geof formation values with a number of moles to cancel off units of moles and get total energy units and kilojoules for delta g at standard conditions for this reaction that comes out to negative 2 .8 kilojoules.

02:22 And now if we rearrange this equation to solve for k, we will get the expression that the equilibrium constant k is equal to e to the power of negative delta g at standard conditions divided by r t.

02:44 So k equals e to the power of negative delta g at standard conditions, which we just solve for to be negative 2 .8 kilojoules.

02:57 We divide this by r, 8 .314 joules per mole kelvin...

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