From the values given for ΔH^∘ and ΔS^∘, calculate ΔG^∘ for each of the following reactions at 2

From the values given for ΔH^∘ and ΔS^∘, calculate ΔG^∘ for...

From the values given for $\Delta H^{\circ}$ and $\Delta S^{\circ},$ calculate $\Delta G^{\circ}$ for each of the following reactions at $298 \mathrm{~K}$. If the reaction is not spontaneous under standard conditions at $298 \mathrm{~K},$ at what temperature (if any) would the reaction become spontaneous? $$ \begin{array}{l} \text { (a) } 2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g) \\ \qquad \begin{array}{c} \Delta H^{\circ}=-844 \mathrm{~kJ} ; \Delta S^{\circ}=-165 \mathrm{~J} / \mathrm{K} \\ \text { (b) } 2 \mathrm{POCl}_{3}(g) \longrightarrow 2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g) \\ \Delta H^{\circ}=572 \mathrm{~kJ} ; \Delta S^{\circ}=179 \mathrm{~J} / \mathrm{K} \end{array} \end{array} $$

From the values given for $\Delta H^{\circ}$ and $\Delta S^{\circ},$ calculate $\Delta G^{\circ}$ for each of the following reactions at $298 \mathrm{~K}$. If the reaction is not spontaneous under standard conditions at $298 \mathrm{~K},$ at what temperature (if any) would the reaction become spontaneous?
$$
\begin{array}{l}
\text { (a) } 2 \mathrm{PbS}(s)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{PbO}(s)+2 \mathrm{SO}_{2}(g) \\
\qquad \begin{array}{c}
\Delta H^{\circ}=-844 \mathrm{~kJ} ; \Delta S^{\circ}=-165 \mathrm{~J} / \mathrm{K} \\
\text { (b) } 2 \mathrm{POCl}_{3}(g) \longrightarrow 2 \mathrm{PCl}_{3}(g)+\mathrm{O}_{2}(g) \\
\Delta H^{\circ}=572 \mathrm{~kJ} ; \Delta S^{\circ}=179 \mathrm{~J} / \mathrm{K}
\end{array}
\end{array}
$$

Key Concepts

Gibbs Free Energy

Gibbs free energy is a thermodynamic function that combines a system's enthalpy and entropy to predict the spontaneity of a reaction. It is defined by the equation ?G = ?H ? T?S, where ?G < 0 indicates a spontaneous process under the given conditions. This concept is central in chemical thermodynamics for determining whether reactions will proceed on their own.

Enthalpy Change

Enthalpy change (?H) refers to the heat absorbed or released by a system during a chemical reaction at constant pressure. It tells us whether a reaction is exothermic (releasing heat, ?H < 0) or endothermic (absorbing heat, ?H > 0), and it contributes directly to the overall energy balance affecting the reaction's spontaneity.

Entropy Change

Entropy change (?S) represents the change in disorder or randomness within a system that occurs during a reaction. A positive ?S indicates increased disorder, which can help drive a reaction toward spontaneity, especially at higher temperatures, while a negative ?S suggests a decrease in disorder, opposing spontaneity.

Temperature Dependence of Spontaneity

Temperature is a critical factor in reaction spontaneity because it affects the T?S term in the Gibbs free energy equation. Even if a reaction appears nonspontaneous under standard conditions, adjusting the temperature can shift the balance between enthalpy and entropy contributions and potentially make the reaction spontaneous.

Standard Reaction Conditions

Standard reaction conditions provide a common baseline for thermodynamic calculations, typically defined as 298 K temperature, 1 atm pressure, and specified reactant and product concentrations. These conditions allow for consistent comparison of thermodynamic data, such as ?G, ?H, and ?S, across different reactions.

Transcript

00:01 To determine whether a reaction is spontaneous or not, we have to know whether the gives free energy of the reaction was positive or negative looking at the equation delta g equals delta h minus t delta s.

00:13 If it's negative, it's spontaneous, and if it's positive, it's not.

00:18 And since we have a combination of enthalpy and entropy, different combinations might give us times where reactions are always spontaneous, not spontaneous.

00:30 Spontaneous ever, or whether there is a high temperature spontaneity or low temperature spontaneity.

00:38 And so if a reaction is shown as non -spontaneous, sometimes we can look to determine the temperature at which the reaction would become spontaneous.

00:51 So let's look at a couple examples.

00:53 We'll figure out whether the reaction is spontaneous or not, if it's non -spontaneous, what temperature could it become spontaneous potentially? so let's start out with a reaction where we have lead sulfide and oxygen, and that becomes lead to oxide and sulfur dioxide.

01:41 All right.

01:42 Now, if we know the delta h for the process occurring at 298 kelvin, we know that because we have the knot.

01:51 It's under standard conditions.

01:54 We have negative 844 kilojoules is our delta h.

01:59 And our delta s is equal to negative 165 joules per kelvin.

02:11 Now be careful when you're using the gibbs free energy equation because we need everything in the same units.

02:17 And delta g is normally given in kilojoules per mole.

02:24 Now, delta h is given, or i should say here in kilojoules.

02:30 We're dealing with one mole of each substance.

02:33 Now here, for our delta h, we're given in kilojoules, but if you notice the s is in joules per kelvin, we're going to have to change that to kilojoules to be able to use it in our equation by dividing by a thousand.

02:53 And now we can just plug it into delta g equals delta h minus t delta s.

03:01 We have our negative 844 kilojoules for delta h.

03:08 It's done at standard conditions so we know 298 kelvin is our temperature and our delta s is negative 0 .165 joules per kelvin, kilovans cancel out and we wind up with the delta g here as negative 795 kilojoules...

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