Consider the evaporation of methanol at 25.0^∘ C : CH3 OH(l) ⟶CH3 OH(g) a. Find ΔGr^∘ at 25.0 ^∘

Consider the evaporation of methanol at 25.0^∘ C : CH3 OH(l)...

Consider the evaporation of methanol at $25.0^{\circ} \mathrm{C}$ : $$\mathrm{CH}_{3} \mathrm{OH}(l) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g)$$ a. Find $\Delta G_{\mathrm{r}}^{\circ}$ at $25.0{ }^{\circ} \mathrm{C}$. b. Find $\Delta G_{\mathrm{r}}$ at $25.0{ }^{\circ} \mathrm{C}$ under the following nonstandard conditions: i $P_{\mathrm{CH}_{3} \mathrm{OH}}=150.0 \mathrm{mmHg}$ ii $P_{\mathrm{CH}_{3} \mathrm{OH}}=100.0 \mathrm{mmHg}$ iii $P_{\mathrm{CH}_{3} \mathrm{OH}}=10.0 \mathrm{mmHg}$ c. Explain why methanol spontaneously evaporates in open air at $25.0{ }^{\circ} \mathrm{C}$.

Consider the evaporation of methanol at $25.0^{\circ} \mathrm{C}$ :
$$\mathrm{CH}_{3} \mathrm{OH}(l) \longrightarrow \mathrm{CH}_{3} \mathrm{OH}(g)$$
a. Find $\Delta G_{\mathrm{r}}^{\circ}$ at $25.0{ }^{\circ} \mathrm{C}$.
b. Find $\Delta G_{\mathrm{r}}$ at $25.0{ }^{\circ} \mathrm{C}$ under the following nonstandard conditions:
i $P_{\mathrm{CH}_{3} \mathrm{OH}}=150.0 \mathrm{mmHg}$
ii $P_{\mathrm{CH}_{3} \mathrm{OH}}=100.0 \mathrm{mmHg}$
iii $P_{\mathrm{CH}_{3} \mathrm{OH}}=10.0 \mathrm{mmHg}$
c. Explain why methanol spontaneously evaporates in open air at $25.0{ }^{\circ} \mathrm{C}$.

Key Concepts

Gibbs Free Energy and Spontaneity

The sign of the Gibbs free energy change determines whether a process is spontaneous. A negative free energy change indicates that a process will proceed spontaneously under the given conditions. This concept connects the thermodynamic calculations to observable physical behavior, such as phase transitions.

Vapor Pressure and Evaporation

Vapor pressure is the pressure exerted by a vapor in equilibrium with its liquid phase at a given temperature. It is a critical concept in understanding phase change processes like evaporation. When the ambient pressure of the vapor is lower than its equilibrium vapor pressure, evaporation occurs spontaneously as the system moves towards equilibrium.

Nonstandard Conditions and Reaction Quotient

Under nonstandard conditions, the Gibbs free energy change can be adjusted using the reaction quotient (Q), which reflects the ratio of the activities (or partial pressures/concentrations) of products to reactants. The relationship ?G = ?G° + RT ln Q allows prediction of the directionality of a reaction when the system is not at equilibrium.

Standard Gibbs Free Energy Change

This concept involves calculating the change in Gibbs free energy under standard state conditions (usually 1 bar for gases and pure substances for liquids and solids). It provides a baseline reference for comparing how far a process is from equilibrium and is crucial for determining the spontaneity of a chemical reaction under defined conditions.

Transcript

00:01 In this question we've been given the evaporation of methanol, ch3oh from a liquid phase into ch3 or h in the cassius phase.

00:14 So this is the reaction that we're looking at and we've been asked to determine the standard gives energy change of this reaction and eventually it gives energy change.

00:26 And we know the gibbs energy change by definition this is equal to this.

00:31 Gives energy change that will calculate in the first step, plug it in here, and then plus rt, lynn, q, and this qp is going to be partial pressure or the pressure of ch3 oh.

00:55 Now looking at the standard gives energy change, this is going to be, it is a state function, so we are allowed to use the standard gives energy changes from our data booklets to say this is the gibbs energy change of the formation of the products products minus that of the reactants of the reactants and if we are to look at this this is going to be our product this is our product and this is our reactant and remember we'll always have to multiply by this documentary coefficient because these values are given in per more.

01:38 So if we are to make that calculation, the standard gives energy change is going to be 1 multiplied by negative 162 .3 minus the reactant, which is the liquid methanol, which is negative 1, 1 more minus 166 .6 .6.

01:57 So at the end of the day, our standard gives energy change is going to be equal to 4.

02:05 3 and this is in kilojolts.

02:08 Now moving on to applying this formula, we are going to apply this in 2 forms because we've been given two pressures.

02:15 So first of all we have to convert our temperature 25 degrees celsius into 27, 298 .15 kelvin.

02:26 We've also been given to say p, ch3oh is equal to 1 .50 in millimeters per mercury.

02:35 It will do as good to convert this into atmosphere which is 0 .197 -37 atmosphere because usually all these parameters are evaluated against the standard condition of 25 degrees celsius and one atmosphere...

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