Use data from Appendix C to calculate the equilibrium constant, K, at 298 K for each of the fol

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

Use data from Appendix $\mathrm{C}$ to calculate the equilibrium constant, $K,$ at $298 \mathrm{~K}$ for each of the following reactions: (a) $\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{Hl}(g)$ (b) $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)$ (c) $3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{6}(g)$

Use data from Appendix $\mathrm{C}$ to calculate the equilibrium constant, $K,$ at $298 \mathrm{~K}$ for each of the following reactions:
(a) $\mathrm{H}_{2}(g)+\mathrm{I}_{2}(g) \rightleftharpoons 2 \mathrm{Hl}(g)$
(b) $\mathrm{C}_{2} \mathrm{H}_{5} \mathrm{OH}(g) \rightleftharpoons \mathrm{C}_{2} \mathrm{H}_{4}(g)+\mathrm{H}_{2} \mathrm{O}(g)$
(c) $3 \mathrm{C}_{2} \mathrm{H}_{2}(g) \rightleftharpoons \mathrm{C}_{6} \mathrm{H}_{6}(g)$

Key Concepts

Equilibrium Constant

The equilibrium constant, K, quantifies the position of equilibrium for a chemical reaction under a given set of conditions. It is defined based on the law of mass action, where the activities (or concentrations and partial pressures) of the products and reactants are raised to the power of their respective coefficients in the balanced chemical equation.

Gibbs Free Energy

Gibbs free energy is a thermodynamic potential that combines enthalpy and entropy into a single value, and its change (?G) indicates whether a process is spontaneous. The relationship ?G° = -RT ln K connects the standard Gibbs free energy change of a reaction to the equilibrium constant, allowing one to calculate K if ?G° is known.

Thermodynamic Data Utilization

Standard thermodynamic data, such as that found in data tables or appendices, provides essential values like standard enthalpies, entropies, and Gibbs free energies of formation for various substances. This data is crucial for determining the overall ?G° for a reaction, especially when calculating equilibrium constants from fundamental thermodynamic principles.

Reaction Stoichiometry

Accurate interpretation of the balanced chemical equation is vital since the stoichiometric coefficients directly influence both the equilibrium expression and the computation of ?G° from standard data. Each coefficient determines how the thermodynamic properties of the reactants and products contribute to the overall energy change of the reaction.

Transcript

00:01 We can calculate the equilibrium constant for a reaction based on gibbs free energy by using the equation delta g0 is equal to negative rt times the natural log of k, where k is the equilibrium constant.

00:18 And so if we rearrange that equation, we can get the natural log of the equilibrium constant by looking at the gibbs free energy.

00:30 And dividing by rt.

00:35 To get k, we can take both sides and take e to the natural log.

00:40 That cancels out.

00:42 We can take e to delta g not over rt, and that's going to give us the equilibrium constant.

00:51 But in order to do that for chemical reactions, we need to first determine what the change in gibbs free energy is for a particular reaction.

01:01 So let's go ahead and take a look at some reactions.

01:06 So let's start off with h2 as a gas plus i2, gas in equilibrium with 2hi.

01:25 Now remember to figure out what the delta g is, we can also take a look at the equation that the delta g of reaction, is equal to the sum of the number of moles times the gibbs energy of formation for products minus the sum of the number of moles.

01:53 M here is just used to differentiate it from n, but they're both number of moles times the gives energy of formation of reactants.

02:03 So you can look up the gives energy of formation for these particular substances.

02:09 Just remember that elements in their natural state like hydrogen here have a gibbs energy of formation of zero.

02:15 Even though i2 is an element, it's not in its natural state at 298 kelvin.

02:22 So the i2 here will have a value.

02:26 So first we can go ahead and calculate the gibbs free energy of reaction using those formation values.

02:34 And so we've got two moles of h .i, each with a gibbs -free energy of 1 .30 kilojoules per mole.

02:47 That is our only product, so we can subtract now from our reactants.

02:56 And h -2, we've got one mole, but it's gibbs -free energy is zero.

03:05 The sigma here means sum up or add together, so we'll add the i -2 value.

03:11 We have one mole, and its gibbs free energy is 19 .37 kilojoules per mole.

03:22 So the gibbs energy of reaction here is negative 16 .7 kilojoules.

03:33 This is the value that we can plug in to figure out the value of k now.

03:40 So we have k is going to be equal to e to the, and we have our delta g value, negative 16 .77, and we're going to divide that by r and t...

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