Question

The standard Gibbs energy change for the reaction \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{aq})+\mathrm{H}_{2} \mathrm{O}(\mathrm{l}) \rightleftharpoons$$$ \mathrm{CH}_{3} \mathrm{CO}_{2}^{-}(\mathrm{aq})+\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})$$is \)27.07 \mathrm{kJmol}^{-1}\( at 298 K. Use this thermodynamic quantity to decide in which direction the reaction is spontaneous when the concentrations of \)\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{aq}), \mathrm{CH}_{3} \mathrm{CO}_{2}^{-}(\mathrm{aq}),\( and \)\mathrm{H}_{3} \mathrm{O}^{+}(\mathrm{aq})\( are \)0.10 \mathrm{M}, 1.0 \times 10^{-3} \mathrm{M},\( and \)1.0 \times 10^{-3} \mathrm{M},$ respectively.

Short answer

Given the concentrations provided, the reaction \(\mathrm{CH}_{3} \mathrm{CO}_{2} \mathrm{H}(\mathrm{aq})+\mathrm{H}_{2}\mathrm{O}(\mathrm{l}) \rightleftharpoons\) \(\mathrm{CH}_{3} \mathrm{CO}_{2}^{-}(\mathrm{aq})+\mathrm{H}_{3}\mathrm{O}^{+}(\mathrm{aq})\) is spontaneous in the reverse direction.

Step-by-step solution

Determine the Gibbs Energy change under standard conditions

The standard Gibbs energy change(\(\Delta G^o\)) given in the problem is \(27.07 \, kJmol^{-1}.\) We will need to convert this energy into \(Jmol^{-1}\) in order to put it in the correct units for later calculations. So, \(\Delta G^o = 27.07 \times 10^3 \, Jmol^{-1}\).

Calculate the Reaction Quotient (Q)

The concentrations given in the problem allow us to determine the reaction quotient Q. For a reaction of the form \(aA + bB \rightleftharpoons cC + dD\), Q is equal to \((C^c \times D^d)/(A^a \times B^b)\). Substituting the provided concentrations into the Q expression yields \(Q = ([CH_{3}CO_{2}^{-}] \times [H_{3}O^{+}])/([CH_{3}CO_{2}H]) = (1.0 \times 10^{-3} \times 1.0 \times 10^{-3})/0.10 = 0.01.\)

Calculate the Gibbs Energy change under non-standard conditions

The Gibbs energy change based on the concentrations can be calculated using the relationship \(\Delta G = \Delta G^o + RT \ln Q\), where R is the gas constant (8.314 J/Kmol) and T is the temperature in Kelvin (298 K in this case). Substituting in these values gives us \(\Delta G = 27.07 \times 10^3 + 8.314 \times 298 \times \ln 0.01 = 27.07 \times 10^3 - 4795.21 = 22274.79 \, Jmol^{-1} = 22.27 \, kJmol^{-1}\).

Determine the Direction of the Reaction

As the Gibbs energy change calculated is positive, it indicates that the reaction is not spontaneous in the forward direction, but it is in the reverse direction. So, under the given concentrations, the reaction will proceed towards the left, from products to reactants.

Key concepts

Reaction Quotient

The Reaction Quotient, denoted as \( Q \), is an essential concept in chemical thermodynamics. It helps predict the direction in which a reaction will proceed at any given set of concentrations. To calculate \( Q \), you use the formula based on the balanced chemical equation:

• For a reaction \( aA + bB \rightleftharpoons cC + dD \), \( Q = \frac{[C]^c[D]^d}{[A]^a[B]^b} \).

This formula reflects the ratio of the concentrations of the products over the reactants, each raised to the power of their respective coefficients in the balanced chemical equation.
In the exercise provided, the reaction quotient \( Q \) was calculated to be 0.01 using the given concentrations. This small value of \( Q \) compared to any equilibrium constant \( K \) suggests that there are more reactant molecules present, indicating that the reaction might need to proceed to the right to reach equilibrium. This helps in eventually determining the spontaneity of the reaction under non-standard conditions.

Spontaneity of Reactions

The spontaneity of a reaction can be intuitively understood as the natural tendency of a process to occur without the need of external energy input. The concept of Gibbs free energy \( \Delta G \) is crucial here.
If \( \Delta G < 0 \), the reaction is spontaneous in the forward direction. Conversely, if \( \Delta G > 0 \), it is non-spontaneous in that direction and will favor the reverse. Lastly, if \( \Delta G = 0 \), the system is at equilibrium, and no net change occurs.
In practical terms, calculating \( \Delta G \) under non-standard conditions involves the reaction quotient:

• Formula: \( \Delta G = \Delta G^o + RT \ln Q \).

Here, \( R \) is the gas constant, and \( T \) is the temperature in Kelvin. In the exercise, a positive \( \Delta G \) value of 22.27 kJ/mol indicates that the reaction is not spontaneous in the forward direction at the provided conditions, meaning it is more likely to proceed in the reverse direction (from right to left).

Standard Conditions

Standard conditions refer to a specific set of circumstances for reactions that allow chemists to compare results uniformly. These conditions generally include:

• Pressure of 1 bar (or sometimes 1 atm)
• Temperature of 298 K (25°C)
• Concentrations of 1 M for any aqueous solutions involved

Under these conditions, the standard Gibbs energy change \( \Delta G^o \) is measured and can be used as a baseline to calculate \( \Delta G \) under more varied conditions.
In the exercise, the \( \Delta G^o \) was initially provided at standard conditions as 27.07 kJ/mol. When analyzing a reaction beyond standard conditions, like in a lab or real-world scenario, you need this benchmark to understand how changes in concentration or temperature affect the reaction's spontaneity. This helps predict whether a process will proceed naturally, considering varying initial conditions.