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Question: Calculate the equilibrium constant at \[25degrees{\text{ }}Celsius\] given the Standard Free Energy ...

Calculate the equilibrium constant at 25degrees Celsius25degrees{\text{ }}Celsius given the Standard Free Energy value of − 107.2 kJ. - {\text{ }}107.2{\text{ }}kJ.
A) − 43.2 - {\text{ }}43.2
B) 43.243.2
C) 6.18×1086.18 \times {10^8}
D) 1.041.04
E) 6.18×1096.18 \times {10^9}

Explanation

Solution

Standard free energy change of the reaction = ΔG∘= {\text{ }}\Delta G^\circ (which is equal to the difference in the free energies of formation of the products and reactants both in their standard states) according to the equation.
At equilibrium,
ΔG∘=− RT In K(eq)\Delta G^\circ = -{\text{ }}RT{\text{ }}In{\text{ }}K\left( {eq} \right)
R=8.314 Jmol−1 K−1 or 0.008314 kJ mol−1 K−1.R = 8.314{\text{ }}Jmo{l^{ - 1{\text{ }}}}{K^{ - 1}}{\text{ }}or{\text{ }}0.008314{\text{ }}kJ{\text{ }}mo{l^{ - 1}}{\text{ }}{K^{ - 1.}}
TT =temperature on the Kelvin scale
  K\;K =equilibrium constant

Complete Step by step answer: Gibbs free energy is a quantity that is used to measure the maximum amount of work done in a thermodynamic system when the temperature and pressure are kept constant. Its value is usually expressed in Joules or Kilojoules. Gibbs free energy can be defined as the maximum amount of work that can be extracted from a closed system.

Gibbs free energy is a state function, So change in Gibbs free energy is ΔG=ΔH−Δ(TS)              \Delta G = \Delta H - \Delta \left( {TS} \right)\;\;\;\;\;\;\;
  ΔH\;\Delta H= enthalpy change, T = temperature ΔS = entropy change.
Under standard conditions, the Gibbs free energy is expressed as a - ΔG∘=ΔH∘−TΔS∘\Delta G^\circ = \Delta H^\circ - T\Delta S^\circ
ΔG∘=\Delta G^\circ = Standard free energy change of the reaction,

The standard free energy of a substance represents the free energy change associated with the formation of the substance from the elements in their most stable forms as they exist under standard conditions.
The standard Gibbs free energy of all elements in their standard state is zero.
ΔG=ΔG∘+RT InQ                                                    \Delta G = \Delta G^\circ + RT{\text{ }}InQ\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;\;
Where QQ is the reaction quotient.
At equilibrium,
ΔG=0 and Q\Delta G = 0{\text{ }}and{\text{ }}Q Become equal to the equilibrium constant. Hence the equation becomes,
⇒ΔG∘=− RT In K(eq)\Rightarrow \Delta G^\circ = -{\text{ }}RT{\text{ }}In{\text{ }}K\left( {eq} \right)
As per question to calculate value of KK(equilibrium constant) by putting value in above equation,
Given,
T=25  degrees Celsius =298 KT = 25\;degrees{\text{ }}Celsius{\text{ }} = 298{\text{ }}K
ΔG∘\Delta G^\circ = - 107.2 kJ = =−107200J
⇒ΔG∘=−RT ln K\Rightarrow \Delta G^\circ = - RT{\text{ }}ln{\text{ }}K
⇒ln K(eq)=−ΔG∘RT  \Rightarrow ln{\text{ }}K\left( {eq} \right) = - \dfrac{{\Delta G^\circ }}{{RT\;}}
⇒K(eq)=e−ΔGRT\Rightarrow K\left( {eq} \right) = {e^{\dfrac{{ - \Delta G}}{{RT}}}} =e−(−107200J)8.314  J/K⋅mol(298  K){e^{\dfrac{{ - ( - 107200J)}}{{8.314\;J/K \cdot mol\left( {298\;K} \right)}}}} = e43.268{e^{43.268}}
⇒K=6.18×108.\Rightarrow K = 6.18 \times {10^8}.

So the option (C)\left( C \right) is correct.

Note: In a reversible reaction, the free energy of the reaction mixture is lower than the free energy of reactants as well as products. Hence, free energy decreases whether we start from reactants or products i.e., ΔG\Delta G is −ve - ve in backward as well as forward reactions.