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A reaction has \(K=1.9 \times 10^{-14}\) at \(25^{\circ} \mathrm{C}\) and \(K=9.1 \times 10^{3}\) at \(227^{\circ} \mathrm{C}\) . Predict the signs for \(\Delta G^{\circ}, \Delta H^{\circ},\) and \(\Delta S^{\circ}\) for this reaction at \(25^{\circ} \mathrm{C}\) . Assume \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

Short Answer

Expert verified
At 25掳C, the signs for 螖G掳, 螖H掳, and 螖S掳 are as follows: 螖G掳 is negative, indicating a spontaneous reaction. 螖H掳 is positive, indicating an endothermic reaction where energy is absorbed by the system. 螖S掳 is positive, indicating an increase in entropy of the system.

Step by step solution

01

Write down the Van't Hoff equation

The Van't Hoff equation is given by: \[\frac{d(\ln K)}{dT} = \frac{\Delta H^{\circ}}{RT^2}\] We will first use this equation to solve for 螖H掳.
02

Calculate the value of 螖H掳

In order to find 螖H掳, we will take the derivative of ln(K) with respect to T. To do this, we can use the information of the two equilibrium constants K1 at T1 = 25掳C and K2 at T2 = 227掳C. First, convert the temperatures from degree Celsius to Kelvin: \(T_1 = 25 + 273.15 = 298.15 K\) \(T_2 = 227 + 273.15 = 500.15 K\) Now, we can write: \[\frac{d(\ln K)}{dT} = \frac{\ln(K_2) - \ln(K_1)}{T_2 - T_1}\] \[\frac{d(\ln K)}{dT} = \frac{\ln(\frac{K_2}{K_1})}{T_2 - T_1}\] Substitute the given values of K1, K2, T1 and T2: \[\frac{d(\ln K)}{dT} = \frac{\ln(\frac{9.1\times 10^{3}}{1.9 \times 10^{-14}})}{500.15 - 298.15} = 0.08379 K^{-1}\] Using the Van't Hoff equation, we can now find 螖H掳: \[\Delta H^{\circ} = \frac{d(\ln K)}{dT} \cdot RT^2\] \[\Delta H^{\circ} = 0.08379 \times 8.314 \times (298.15)^2 = 65686 J/mol\]
03

Determine the sign of 螖H掳 and 螖S掳

Now that we have the value of 螖H掳, we can determine its sign. Since 螖H掳 is positive (65686 J/mol), this indicates that the reaction is endothermic. This means that energy is absorbed by the system. Now we can determine the sign of 螖S掳 by checking which direction the value of K moves as temperature increases. In this case, at 25掳C, K is very small (1.9 脳 10鈦宦光伌) practically zero. However, at 227掳C, K has increased significantly (9.1 脳 10鲁). Therefore, we can deduce that 螖S掳 is positive, meaning the entropy of the system increases.
04

Calculate 螖G掳 and predict its sign

Since we know the signs of 螖H掳 and 螖S掳, we can determine the sign of 螖G掳 at 25掳C using the Gibbs-Helmholtz equation: \[\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\] We can predict the sign of 螖G掳 without calculating it (although we do not have the value of 螖S掳). Since 螖H掳 is positive and 螖S掳 is positive, at 25掳C (298.15 K), the 螖G掳 will be negative because the T螖S掳 term will be larger than the 螖H掳 term. Therefore, the reaction is spontaneous at 25掳C, and we can conclude that the signs for 螖G掳, 螖H掳, and 螖S掳 at 25掳C are as follows: 螖G掳: Negative 螖H掳: Positive 螖S掳: Positive

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Gibbs Free Energy
Gibbs Free Energy, represented as \(\Delta G^{\circ}\), is an essential concept in thermodynamics that helps us understand whether a reaction is spontaneous. A reaction is considered spontaneous if it occurs naturally without external influence. The formula for Gibbs Free Energy, \(\Delta G^{\circ} = \Delta H^{\circ} - T\Delta S^{\circ}\), clearly shows the relationship between enthalpy \(\Delta H^{\circ}\), temperature \(T\), and entropy \(\Delta S^{\circ}\).

  • \(\Delta G^{\circ}\) Negative: The reaction is spontaneous.
  • \(\Delta G^{\circ}\) Positive: The reaction is non-spontaneous.
  • \(\Delta G^{\circ}\) Zero: The system is at equilibrium.


When both \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) are positive, as in endothermic reactions increasing disorder, \(T\Delta S^{\circ}\) can outweigh \(\Delta H^{\circ}\), making the reaction spontaneous. This is exactly what happens in our exercise, leading to a negative \(\Delta G^{\circ}\), indicating the reaction's spontaneity despite the positive \(\Delta H^{\circ}\).
Van't Hoff Equation
The Van't Hoff equation is a valuable tool for understanding how the equilibrium constant \(K\) of a reaction depends on temperature. It is expressed as \[\frac{d(\ln K)}{dT} = \frac{\Delta H^{\circ}}{RT^2}.\]This equation tells us that the change in the logarithm of the equilibrium constant with temperature gives insights about \(\Delta H^{\circ}\), the reaction's enthalpy change.

  • K increases with temperature if the reaction is endothermic (\(\Delta H^{\circ}\) is positive).
  • K decreases with temperature if the reaction is exothermic (\(\Delta H^{\circ}\) is negative).


By observing how \(K\) changes, we can deduce whether a reaction absorbs or releases heat. In the exercise, the significant increase in \(K\) with temperature implies a positive \(\Delta H^{\circ}\), confirming the reaction is endothermic.
Equilibrium Constant
The equilibrium constant, \(K\), is a vital concept indicating the extent to which a reaction proceeds at a certain temperature. A large \(K\) means the products are favored, while a small \(K\) means the reactants are favored.

  • If \(K >> 1\), the reaction mostly yields products.
  • If \(K << 1\), the reaction remains mostly in the reactants form.


In our exercise, \(K\) changes dramatically from \(1.9 \times 10^{-14}\) at 25掳C to \(9.1 \times 10^3\) at 227掳C. This increase indicates that the reaction shifts significantly towards the products as the temperature rises. This behavior helps us understand the reaction's dynamics and its spontaneity as temperature changes. Larger \(K\) at higher temperatures suggests less reactant and more product formation.
Endothermic Reactions
Endothermic reactions are processes where heat is absorbed from the surroundings, indicated by a positive \(\Delta H^{\circ}\). When a reaction is endothermic:

  • The system absorbs energy.
  • \(\Delta H^{\circ}\) is greater than zero.
  • Heat is required for the reaction to proceed.


In the provided exercise, the positive \(\Delta H^{\circ}\) calculated indicates an endothermic process. This means that the reaction absorbs heat, making it more product-favorable at higher temperatures. As temperature increases, the equilibrium shifts towards the formation of products, correlating with the increase in \(K\). Understanding such reactions helps in predicting how a reaction behaves as conditions change and designing systems that rely on temperature-sensitive reactions.

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Most popular questions from this chapter

Consider the reaction: $$\mathrm{H}_{2} \mathrm{S}(g)+\mathrm{SO}_{2}(g) \longrightarrow 3 \mathrm{S}(g)+2 \mathrm{H}_{2} \mathrm{O}(l)$$ for which \(\Delta H\) is \(-233 \mathrm{kJ}\) and \(\Delta S\) is \(-424 \mathrm{J} / \mathrm{K}\) . a. Calculate the free energy change for the reaction \((\Delta G)\) at 393 \(\mathrm{K} .\) b. Assuming \(\Delta H\) and \(\Delta S\) do not depend on temperature, at what temperatures is this reaction spontaneous?

In the text, the equation $$\Delta G=\Delta G^{\circ}+R T \ln (Q)$$ was derived for gaseous reactions where the quantities in \(Q\) were expressed in units of pressure. We also can use units of mol/L for the quantities in \(Q\)specifically for aqueous reactions. With this in mind, consider the reaction $$\mathrm{HF}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{F}^{-}(a q)$$ for which \(K_{\mathrm{a}}=7.2 \times 10^{-4}\) at \(25^{\circ} \mathrm{C}\) . Calculate \(\Delta G\) for the reaction under the following conditions at \(25^{\circ} \mathrm{C} .\) a. \([\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \mathrm{M}\) b. \([\mathrm{HF}]=0.98 M,\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=2.7 \times 10^{-2} M\) c. \([\mathrm{HF}]=\left[\mathrm{H}^{+}\right]=\left[\mathrm{F}^{-}\right]=1.0 \times 10^{-5} \mathrm{M}\) d. \([\mathrm{HF}]=\left[\mathrm{F}^{-}\right]=0.27 M,\left[\mathrm{H}^{+}\right]=7.2 \times 10^{-4} M\) e. \([\mathrm{HF}]=0.52 M,\left[\mathrm{F}^{-}\right]=0.67 M,\left[\mathrm{H}^{+}\right]=1.0 \times 10^{-3} \mathrm{M}\) Based on the calculated DG values, in what direction will the reaction shift to reach equilibrium for each of the five sets of conditions?

Consider the reaction $$\begin{array}{l}{2 \mathrm{O}(g) \longrightarrow \mathrm{O}_{2}(g)} \\\ {\text { sof } \Delta H \text { and } \Delta S}\end{array}$$ a. Predict the signs of \(\Delta H\) and \(\Delta S .\) b. Would the reaction be more spontaneous at high or low temperatures?

For each of the following pairs, which substance has the greater value of S? a. \(\mathrm{N}_{2} \mathrm{O}(\text { at } 0 \mathrm{K})\) or He (at 10 \(\mathrm{K} )\) b. \(\mathrm{N}_{2} \mathrm{O}(g)\) (at \(1 \mathrm{atm}, 25^{\circ} \mathrm{C} )\) or He(g) (at 1 atm, \(25^{\circ} \mathrm{C} )\) c. \(\mathrm{NH}_{3}(s)\) (at 196 \(\mathrm{K} ) \longrightarrow \mathrm{NH}_{3}(l)(\text { at } 196 \mathrm{K})\)

Consider the reaction $$\mathrm{Fe}_{2} \mathrm{O}_{3}(s)+3 \mathrm{H}_{2}(g) \longrightarrow 2 \mathrm{Fe}(s)+3 \mathrm{H}_{2} \mathrm{O}(g)$$ a. Use \(\Delta G_{f}^{\circ}\) values in Appendix 4 to calculate \(\Delta G^{\circ}\) for this reaction. b. Is this reaction spontaneous under standard conditions at 298 \(\mathrm{K} ?\) c. The value of \(\Delta H^{\circ}\) for this reaction is 100 . kJ. At what temperatures is this reaction spontaneous at standard conditions? Assume that \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) do not depend on temperature.

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