The equilibrium constant for the reaction, I2(s)+Br2(g) ⇌2 IBr(g) is 0.164 at 25^∘ C. (a) Calcul

The equilibrium constant for the reaction, I2(s)+Br2(g) ⇌2...

The equilibrium constant for the reaction, $\mathrm{I}_{2}(\mathrm{s})+\mathrm{Br}_{2}(\mathrm{g}) \rightleftharpoons 2 \mathrm{IBr}(\mathrm{g})$ is 0.164 at $25^{\circ} \mathrm{C}$. (a) Calculate $\Delta_{r} G^{\ominus}$ for this reaction. (b) Bromine gas is introduced into a container with excess solid iodine. The pressure and temperature are held at 0.164 atm and $25^{\circ} \mathrm{C}$, respectively. Find the partial pressure of IBr(g) at equilibrium. Assume that all the bromine is in the gaseous form and that the vapour pressure of iodine is negligible. (c) In fact, solid iodine has a measurable vapour pressure at $25^{\circ} \mathrm{C}$. In this case, how would the calculation have to be modified?

Key Concepts

Equilibrium Constant (K)

The equilibrium constant is a ratio that describes the state of balance in a chemical reaction by relating the activities (or pressures/concentrations) of products and reactants at equilibrium. In heterogeneous equilibria, pure solids or liquids have an activity of 1, so only the gaseous or aqueous species are considered in forming the equilibrium expression.

Gibbs Free Energy and Its Relation to Equilibrium (?rG° = -RT ln K)

This concept links the standard change in Gibbs free energy (?rG°) to the equilibrium constant via the equation ?rG° = -RT ln K, where R is the gas constant and T is the temperature in Kelvin. It quantifies how far a reaction is from equilibrium, with a more negative ?rG° indicating a reaction that favors product formation.

Partial Pressure Calculations in Gaseous Equilibria

For reactions involving gases, equilibrium can be analyzed using partial pressures. The equilibrium constant expressed in terms of partial pressures allows one to calculate the required pressure of a gaseous product or reactant when the system reaches equilibrium, as only the gaseous species contribute to the equilibrium expression in a heterogeneous system.

Heterogeneous Equilibrium Adjustments (Inclusion of Vapor Pressure)

In systems involving multiple phases, if a substance that is normally considered inactive (like a solid) exhibits a measurable vapor pressure, it introduces additional gaseous species into the equilibrium. The equilibrium calculation must then be modified to incorporate this vapor pressure account, thereby adjusting the equilibrium condition to reflect the actual contributions of all gaseous components.

Transcript

00:01 Hello students, in this question we are given a equilibrium condition.

00:06 I2 in solid state plus bromine in gaseous state gives us 2 times ibr in gaseous.

00:18 All right? so it's equilibrium constant is given k is equal to 0 .164.

00:25 In the first part of the question, we have to calculate del gnor.

00:32 Del r g nor for given reaction and we can do it by using the expression del g nor equal to minus r t ln of k equal to minus value of r we know it's 8 .314 into temperature is given to be 298 into ln of k and the value of k is also given so upon putting the values and calculating it we get it equal to 4479 .17 jule mole inverse, which is equal to plus minus, sorry, which is equal to positive 4 .48 kilojoule inverse.

01:23 In the b part of this question, we are asked, we have to calculate the partial pressure of ibr, gas if we know that bromine gas is introduced into a container with solid iodine and that pressure and temperature held it in this case pressure is equal to 0 .164 atmosphere and the temperature t is equal to 298 kelvin all right so for this case our reaction will be i2 in solid state plus br2 in gaseous gives us 2 times ibr in gaseous state so amounts of the product produced will be it will be alpha n alpha n 0 and then it will be minus 1 minus alpha n to alpha n 1 minus alpha n and 2 alpha n 1 minus alpha n 2 alpha n and this will be the amount and the mole fraction mole fraction will be equal to 1 minus alpha into 2 alpha divided by 1 plus alpha into 1 plus alpha and the partial pressure, partial fraction, will be given by 1 minus alpha into p to alpha divided by 1 plus alpha into 1 plus alpha.

03:25 So, k is equal to partial pressure of ibr divided by p nor raised to 3 to 3 .3...

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