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.

Step by step solution

01

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, \(\Delta G^o = 27.07 \times 10^3 \, Jmol^{-1}\).

02

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 \(Q = ([CH_{3}CO_{2}^{-}] \times [H_{3}O^{+}])/([CH_{3}CO_{2}H]) = (1.0 \times 10^{-3} \times 1.0 \times 10^{-3})/0.10 = 0.01.\)

03

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 \(\Delta G = 27.07 \times 10^3 + 8.314 \times 298 \times \ln 0.01 = 27.07 \times 10^3 - 4795.21 = 22274.79 \, Jmol^{-1} = 22.27 \, kJmol^{-1}\).

04

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.

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