Consider the sublimation of iodine at 25.0^∘ C : I2(s) ⟶I2(g) a. Find ΔGrxn^∘ at 25.0 ^∘ C. b. F

Consider the sublimation of iodine at 25.0^∘ C : I2(s)...

Consider the sublimation of iodine at $25.0^{\circ} \mathrm{C}$ : $$\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g)$$ a. Find $\Delta G_{\mathrm{rxn}}^{\circ}$ at $25.0{ }^{\circ} \mathrm{C}$. b. Find $\Delta G_{\mathrm{rxn}}$ at $25.0^{\circ} \mathrm{C}$ under the following nonstandard conditions: i $P_{\mathrm{I}_{2}}=1.00 \mathrm{mmHg}$ ii $P_{\mathrm{I}_{2}}=0.100 \mathrm{mmHg}$ c. Explain why iodine spontaneously sublimes in open air at $25.0^{\circ} \mathrm{C}$

Consider the sublimation of iodine at $25.0^{\circ} \mathrm{C}$ :
$$\mathrm{I}_{2}(s) \longrightarrow \mathrm{I}_{2}(g)$$
a. Find $\Delta G_{\mathrm{rxn}}^{\circ}$ at $25.0{ }^{\circ} \mathrm{C}$.
b. Find $\Delta G_{\mathrm{rxn}}$ at $25.0^{\circ} \mathrm{C}$ under the following nonstandard conditions:
i $P_{\mathrm{I}_{2}}=1.00 \mathrm{mmHg}$
ii $P_{\mathrm{I}_{2}}=0.100 \mathrm{mmHg}$
c. Explain why iodine spontaneously sublimes in open air at $25.0^{\circ} \mathrm{C}$

Key Concepts

Phase Transitions and Sublimation

Sublimation is a phase transition in which a substance transitions directly from the solid phase to the gas phase without passing through an intermediate liquid phase. Understanding sublimation involves addressing the energetic and equilibrium factors that govern phase changes, and it serves as a practical example of how Gibbs free energy and pressure conditions influence the spontaneity of phase transitions.

Reaction Quotient (Q) and Nonstandard Conditions

The Reaction Quotient (Q) is a dimensionless number that compares the current state of a reaction system to its equilibrium state by using the activities or concentrations of the reacting species at any given moment. Under nonstandard conditions, the actual free energy change is determined by the equation ?G = ?G° + RT ln Q, which indicates how deviations from standard state affect the spontaneity and direction of a reaction.

Gibbs Free Energy

Gibbs Free Energy is a thermodynamic potential that measures the maximum reversible work obtainable from a system at constant temperature and pressure. It is used to predict the spontaneity of a process; if the change in Gibbs free energy (?G) is negative, the process is spontaneous, while a positive ?G indicates nonspontaneity.

Standard Gibbs Free Energy Change (?G°)

The standard Gibbs free energy change represents the change in free energy when all reactants and products are in their standard states, often defined as 1 atm pressure for gases and 1 M concentration for solutions. This concept is crucial in thermodynamics as it provides a baseline for comparing reactions and is directly linked to the equilibrium constant through the relation ?G° = -RT ln K.

Transcript

00:01 In this question we've been given a system in which we've been told that iodine is subliming.

00:07 And by definition, supplementation is a change of a solid into a gas without first of all person through the liquid phase.

00:16 When we are looking at phase change, we are normally used to a solid that melts into a liquid and the liquid vaporizing into a gas.

00:26 But then when it comes to supplementation, we've got a solid that, is changing into a gas without passing through the liquid phase and we do this by manipulating the temperature or the pressure of the system this is particularly useful in other systems like for example co2 now coming back to what we've been given we've been asked to determine the standard gives energy change of this reaction and if we look at the standard gives the gives energy energy and this is a state function a state function so because of that we can use the principle to say the standard gives energy of the reaction is going to be equal to the sum of the standard gives energy changes of formation of the products minus the summation this will be the stoichiometric coefficients multiplied by and here we are looking at the standard gives energy changes of the reactants so if we are to apply this this is going to be we are going to be getting these from our data booklets where our product is going to be the cuss and our reactant is the solid so remember for the solid this is going to be zero so what we're going to have is one mole of the cuss 19 .3 minus one more multiplied by zero so this is going to be equal to one nine point three and this is in kilojolt now moving on, we know that by definition, delta g, the standard gives energy change.

02:17 This is no longer the standard.

02:18 It is given by the standard gives energy change plus rt, lin.

02:26 And we've already calculated the standard gives energy change as 19 .3.

02:34 So now our goal is to determine the unknown parameters that we have in here.

02:39 Determined that is our r our t and finally our cue for us to add them to this 19 .3 to determine the gibbs energy change of that reaction so moving on we have our temperature is 25 degrees celsius our temperature is 25 degrees celsius we convert this into kelvin to get to 98 .15 kelvin and we have p i2 being equal to 1 millimeter of mercury and this is an equivalent of 1 .3158 times 10 to the power negative 3 atmosphere and we know that q is equal to p i2 and this is equal to this value that we have calculated here so moving on and applying this we have our delta g being equal to 19 .3 times 10 to the power 3 and this will be in jolz plus 8 .314 jols per mole kelvin multiplied by temperature 298 .15 multiplied by lean 1 .3 times 10 to the power negative 3...

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