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Consider the dissociation of a weak acid HA \(\left(K_{\mathrm{a}}=4.5 \times 10^{-3}\right)\) in water: $$ \mathrm{HA}(a q) \rightleftharpoons \mathrm{H}^{+}(a q)+\mathrm{A}^{-}(a q) $$ Calculate \(\Delta G^{\circ}\) for this reaction at \(25^{\circ} \mathrm{C}\).

Short Answer

Expert verified
The standard Gibbs free energy change for the dissociation of the weak acid HA in water at 25掳C is 螖G掳 = 13.39 kJ/mol.

Step by step solution

01

Convert temperature to Kelvin

First, we need to convert the given temperature from Celsius to Kelvin. The temperature is given as 25掳C. To convert it to Kelvin, we add 273.15: $$ T = 25 + 273.15 = 298.15 \ \mathrm{K} $$
02

Calculate Gibbs free energy change

Now we can use the formula to find 螖G掳, which relates the equilibrium constant, Ka, and the standard Gibbs free energy change, 螖G掳: $$ \Delta G^{\circ} = -RT\ln K $$ Plugging in the known values, R = 8.314 J/mol路K, T = 298.15 K, and Ka = 4.5 x 10鈦宦: $$ \Delta G^{\circ} = -(8.314 \ \mathrm{J/mol \cdot K})(298.15 \ \mathrm{K})\ln(4.5 \times 10^{-3}) $$
03

Solve for 螖G掳

Now, we compute the standard Gibbs free energy change: $$ \Delta G^{\circ} = -(8.314)(298.15) \cdot (-5.41) $$ $$ \Delta G^{\circ} = 13389 \ \mathrm{J/mol} = 13.39 \ \mathrm{kJ/mol} $$ So, the standard Gibbs free energy change for the dissociation of the weak acid HA in water at 25掳C is 13.39 kJ/mol.

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Key Concepts

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

Weak Acid Dissociation
Weak acids are acids that don't completely dissociate in water. This means they release their hydrogen ions, but only to a certain extent. Not every pair of water molecules and acid molecules splits into hydrogen ions and their corresponding negative ions.
Instead, an equilibrium is established between the associated and dissociated forms of the acid in solution.
In the case of a weak acid like HA, the dissociation can be represented by the reaction:

\[ \mathrm{HA} (aq) \rightleftharpoons \mathrm{H}^+ (aq) + \mathrm{A}^- (aq) \]
This partial dissociation is why weak acids have an equilibrium constant, which can be used to measure how far dissociation occurs.
Equilibrium Constant
The equilibrium constant, represented as \( K_a \) for acids, shows the extent to which a weak acid dissociates in water.
For the dissociation of HA, \( K_a \) is defined by the expression:

\[ K_a = \frac{[\mathrm{H}^+][\mathrm{A}^-]}{[\mathrm{HA}]} \]
This numerical value reflects the balance between the amounts of reactants and products at equilibrium.
  • A high \( K_a \) implies significant dissociation, acting more like a strong acid.
  • A low \( K_a \) indicates less dissociation reminiscent of weaker acids.
This value directly affects the Gibbs free energy calculation, linking chemical equilibrium with thermodynamic principles.
Thermodynamics
Thermodynamics in chemistry deals with energy changes during reactions. Gibbs free energy (\( \Delta G \)) is a key part of this.
This value tells us whether a reaction can happen spontaneously. The formula used for calculating \( \Delta G^{\circ} \) is:

\[ \Delta G^{\circ} = -RT\ln K \]
Here,
  • \( R \) is the gas constant, typically 8.314 J/mol路K.
  • \( T \) is the temperature in Kelvin.
  • \( K \) is the equilibrium constant (\( K_a \)).
A negative \( \Delta G^{\circ} \) means the reaction goes by itself without needing extra energy, a positive one shows it's non-spontaneous.
For the weak acid dissociation, \( \Delta G^{\circ} \) gives insight into the energetics of the equilibrium.
Chemical Equilibrium
Chemical equilibrium occurs when the forward and reverse reactions happen at the same rate. In this state, concentrations of reactants and products remain constant over time.
For weak acids, this equilibrium plays a role in determining how much acid dissociates in solution.
While individual molecules may continue to dissociate and recombine, the overall concentration doesn't change.
This balance is what the equilibrium constant \( K_a \) quantifies, allowing for understanding of how system conditions like concentration or temperature might shift the balance.
The calculation of \( \Delta G^{\circ} \) connects these equilibrium concepts with the energy required or released, further bridging thermodynamics with observable chemical balance.

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Most popular questions from this chapter

You remember that \(\Delta G^{\circ}\) is related to \(R T \ln (K)\) but cannot remember if it's \(R T \ln (K)\) or \(-R T \ln (K) .\) Realizing what \(\Delta G^{\circ}\) and \(K\) mean, how can you figure out the correct sign?

The equilibrium constant \(K\) for the reaction $$ 2 \mathrm{Cl}(g) \rightleftharpoons \mathrm{Cl}_{2}(g) $$ was measured as a function of temperature (Kelvin). A graph of \(\ln (K)\) versus \(1 / T\) for this reaction gives a straight line with a slope of \(1.352 \times 10^{4} \mathrm{~K}\) and a \(y\) -intercept of \(-14.51\). Determine the values of \(\Delta H^{\circ}\) and \(\Delta S^{\circ}\) for this reaction. See Exercise 79

Using Appendix 4 and the following data, determine \(S^{\circ}\) for \(\mathrm{Fe}(\mathrm{CO})_{5}(g)\) $$ \begin{aligned} \mathrm{Fe}(s)+5 \mathrm{CO}(g) & \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g) & & \Delta S^{\circ}=? \\ \mathrm{Fe}(\mathrm{CO})_{5}(l) & \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(g) & & \Delta S^{\circ}=107 \mathrm{~J} / \mathrm{K} \\ \mathrm{Fe}(s)+5 \mathrm{CO}(g) & \longrightarrow \mathrm{Fe}(\mathrm{CO})_{5}(l) & & \Delta S^{\circ}=-677 \mathrm{~J} / \mathrm{K} \end{aligned} $$

For mercury, the enthalpy of vaporization is \(58.51 \mathrm{~kJ} / \mathrm{mol}\) and the entropy of vaporization is \(92.92 \mathrm{~J} / \mathrm{K} \cdot \mathrm{mol}\). What is the normal boiling point of mercury?

a. Using the free energy profile for a simple one-step reaction, show that at equilibrium \(K=k_{\mathrm{f}} / k_{\mathrm{r}}\), where \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) are the rate constants for the forward and reverse reactions. Hint: Use the relationship \(\Delta G^{\circ}=-R T \ln (K)\) and represent \(k_{\mathrm{f}}\) and \(k_{\mathrm{r}}\) using the Arrhenius equation \(\left(k=A e^{-E_{\mathrm{a}} / R T}\right) .\) b. Why is the following statement false? "A catalyst can increase the rate of a forward reaction but not the rate of the reverse reaction."

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