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Gibbs Free Energy Questions

Gibbs free energy, also known as the Gibbs function, Gibbs energy, or free enthalpy, 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. Gibbs free energy is denoted by the symbol ‘G’. 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.

This property was determined by American scientist Josiah Willard Gibbs in the year 1876 when he was conducting experiments to predict the behaviour of systems when combined together or whether a process could occur simultaneously and spontaneously. Gibbs free energy was also previously known as “available energy.” It can be visualized as the amount of useful energy present in a thermodynamic system that can be utilized to perform some work.

Definition: Gibbs free energy is equal to the enthalpy of the system minus the product of the temperature and entropy. The equation is given as;

G = H – TS

Where,

G = Gibbs free energy

H = enthalpy

T = temperature

S = entropy

OR

G = U + PV – TS

Where,

U = internal energy (SI unit: Joule)

P = pressure (SI unit: Pascal)

V = volume (SI unit: m3 )

T = temperature (SI unit: Kelvin)

S = entropy (SI unit: joule/Kelvin)

Gibbs Free Energy Chemistry Questions with Solutions

Q1. If the reaction quotient (Q) is greater than the equilibrium constant (K), what is true about the Gibbs free energy?

a.) It is greater than zero.

b.) It is less than zero.

c.) More information is needed to determine the Gibbs free energy.

d.) It is equal to zero.

Correct Answer– (a.) It is greater than zero.

Explanation– If Q is greater than K, the reaction has exceeded the equilibrium state. It will proceed non-spontaneously, and this means that the ΔG must be positive, or greater than zero.

Q2. The standard reaction Gibbs energy for a chemical reaction at an absolute temperature T is given by: ΔrG° = A – BT

Where A and B are non-zero constants. Choose the correct statement about this reaction?

a.) The reaction is exothermic if B < 0.

b.) The reaction is exothermic if A > 0 and B < 0.

c.) The reaction is endothermic if A < 0 and B > 0.

d.) The reaction is endothermic if A > 0.

Correct Answer– (d.) The reaction is endothermic if A > 0.

Explanation– ΔG = ΔH – TΔS

ΔH = A

When ΔH = +ve, the reaction will be endothermic.

Q3. Calculate the Gibbs free energy for the reaction of conversion of ATP into ADP at 293 Kelvin the change in enthalpy is 19.07 Kcal and the change in entropy is 90 cal per Kelvin.

a.) 7.3 Cal

b.) –5.3Kcal

c.) 7.3 Kcal

d.) –7.3Kcal

Correct Answer– (d.) –7.3Kcal.

Explanation– ΔG = ΔH – TΔS; here ΔH = 19.07 kcal and ΔS = 90 cal/K,

ΔG = 19.07 Kcal – 293(90 cal/K)

= 19.07 Kcal – 26.37 Kcal

= –7300 cal = -7.3 Kcal.

The Gibbs free energy change is –7.3 Kcal.

Q4. Gibbs energy change ΔG is related to equilibrium constant K as:

a.) ΔG° = –RT lnk

b.) ΔG° = RT lnk

c.) lnk = – RT/ΔG°

d.) lnk = ΔG°/RT

Correct Answer– (a.) ΔG° = –RT lnk

Q5. Which is not the correct relationship between ΔGӨ and equilibrium constant KP?

a.) KP = –RTlogΔGӨ

b.) KP = [e/RT]ΔGӨ

c.) KP = –ΔGӨ/RT

d.) KP = e–ΔGӨ/RT

Correct Answer– (a.) KP = –RTlogΔGӨ, (b.) KP = [e/RT]ΔGӨ, (c.) KP = –ΔGӨ/RT

Q6. Fill in the blank.

The Gibbs free energy is positive when a change in enthalpy and change in entropy is positive at ____.

Answer. The Gibbs free energy is positive when a change in enthalpy and change in entropy is positive at low temperatures.

It is negative at high temperatures.

Q7. Is Gibbs free energy always negative?

Answer. Gibbs free energy is not always negative. When the reaction is spontaneous at a given temperature, it is negative, and for the non-spontaneous reaction, Gibbs free energy is positive.

Q8. State True or False.

ΔG = ΔGӨ + RT logK.

Answer. True.

ΔG = ΔH – TΔS

ΔG = ΔGӨ + RT logK

At equilibrium, ΔG = 0

Therefore, ΔGӨ = –RT logK.

Q9. When is the value of the equilibrium constant less than 1?

Answer. The value of the equilibrium constant is less than 1 when the reaction is non-spontaneous. It means that the equilibrium mixture contains more reactants than products. Therefore the value of ΔG° > 0.

Q10. What is Gibbs free energy for a reversible reaction at equilibrium?

Answer. Gibbs free energy is given by ΔG = ΔH – TΔS

For a reaction to be reversible, ΔG must be equal to 0.

Q11. Give the correct relation between equilibrium constant (K), standard free energy (ΔG°) and temperature (T).

Answer. The correct relation between equilibrium constant (K), standard free energy (ΔG°) and temperature (T) can be given as-

Consider a reaction, A +B ⇄ C + D

ΔG = ΔGӨ + RT lnK

For equilibrium, ΔG = 0

0 = ΔGӨ + RT lnK

ΔGӨ = –RT lnK

ΔGӨ = –2.303RT logK

\(\begin{array}{l}K=10^{-\Delta G^{\circ }/2.303RT}\end{array} \)

Q12. Calculate the Standard Free Energy Change at 25℃ given the Equilibrium constant of 1.3 × 104.

Answer. ΔGӨ = –2.303 RTlogK

ΔGӨ = –2.303 RTlogK = –8.314 × 298 × log 1.3 × 104 = –23469 J = –23.4 KJ

Q13. Using the Gibbs change, ΔGӨ = +63.3kJ for the reaction Ag2CO3 ⇄ 2Ag+ + CO32–. Calculate the Ksp for Ag2CO3 in water at 25°C.

Answer. The relationship between solubility product and Gibbs free energy is-

ΔGӨ = –2.303 RTlogKsp

On substituting the values-

63.3 × 103 = –2.303 × 8.31 × 298 logKsp.

–11.09 = logKsp.

8 × 10–12 = Ksp.

Hence, the Ksp for AgCO3 in water at 25°C is 8 × 10–12.

Q14. Calculate the standard free energy at 1 atm 60℃ for N2O4. It is 50% dissociated at this temperature.

Answer. N2O4 ⇄ 2NO2

If 50% is dissociated, then the mole fraction of both substances will be-

\(\begin{array}{l}X_{N_{2}O_{4}}=\frac{1.05}{1+1.05}\end{array} \)
\(\begin{array}{l}X_{NO_{2}}=\frac{2\times .05}{1+1.05}\end{array} \)

Pressure–

\(\begin{array}{l}P_{N_{2}O_{4}}=\frac{0.5}{1.5}\times 1atm\end{array} \)
\(\begin{array}{l}P_{NO_{2}}=\frac{1}{1.5}\times 1atm\end{array} \)

KP = PNO2/PN2O4 = 1.5/(1.5)2(0.5) = 1.33

ΔG° = –2.303 RT log KP,

T=333 K

ΔG° = –2.303 × 8.314 ×333 log (1.33)

ΔG° = – 789.34 KJ/mol.

Q15. Is the reaction rate affected by Gibbs free energy?

Answer. Concentration, pressure, temperature, and surface area are all factors that influence reaction rate.

There are two variables in the Gibbs Free Energy equation–

ΔH, which is a system’s enthalpy, and ΔS, which is its entropy.

Gibbs Free Energy indicates whether a chemical change is thermodynamically possible.

The Gibbs Free energy ΔG must be –ve for a reaction to be spontaneous. ΔH < 0 and ΔS >0 to attain this.

As a result, the temperature and entropy that affect Gibbs Free Energy also affect the reaction rate indirectly.

Hence, we can say that the reaction rate is affected by Gibbs energy.

Practise Questions on Gibbs Free Energy

Q1. If the reaction quotient (Q) is less than the equilibrium constant (K). Choose the correct statement about Gibbs free energy?

a.) We must know the reaction enthalpy to determine it.

b.) It is greater than zero.

c.) It is equal to zero.

d.) It is less than zero

Q2. What is the relation between Gibbs free energy and the EMF of the cell?

a.) ΔG = –nFEcell

b.) G = –nFEcell

c.) ΔG = –nEcell

d.) ΔG = –nF

Q3. Calculate the Gibbs free energy for the conversion of oxygen to Ozone at room temperature if Kp is given as 2.47 × 10–29.

Q4. Use the given the standard Gibbs energy changes for these equations:

2Fe2O3 (s) → 4Fe (s) + 3O2 (g), ΔG° = −742.2 kJ/mol

Fe(OH)3 (s) → 3Fe (s) + OH (g), ΔG° = −696.5 kJ/mol

Fe3O4 (s) → 3Fe (s) + 2O2 (g), ΔG° = −1015 kJ/mol

To identify the ΔG° for the following reaction

6Fe2O3 (s) → O2 (g) + 4Fe3O4 (s)

Q5. CalculateΔG° for the reaction 2CO + O2 → 2CO2 at 298.15 K.

Given ΔH = –128.3 kJ, ΔS = –159.5 JK–1

Click the PDF to check the answers for Practice Questions.
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