Worksheet on Gibb’s Free Energy





A thermodynamic potential that measures the process-initiating work obtained from a thermodynamic system at a constant pressure and temperature is called Gibbs Free Energy. Orginally Gibbs Free Energy is called Available energy.

Gibbs free energy was developed by Josiah Willard Gibbs in 1870s. Josiah Willard Gibbs was a American mathematician.

In this article you can find some questions related to “Gibbs Free Energy” and their answers:

Questions:

1. For a spontaneous chemical process, the free energy change is:

(a)    Negative

(b)   Positive

(c)    Either positive or negative

(d)   Zero

 

2. All the naturally occurring processes proceed spontaneously in a direction which leads to:

(a)    Increase of enthalpy

(b)   Decrease of free energy

(c)    Decrease of entropy

(d)   Increase of free energy

 

3. The free energy change for a reversible reaction at equilibrium is:

(a)    Slightly negative

(b)   Slightly positive ( ) (c)

(c)    Zero

(d)   Highly positive

 

4. In which one of the following cases AI-I and AE are not equal to each other?

(a)   The process is carried out in a closed vessel

(b)   The number of moles of gaseous reactants and gaseous products is equal to each other

(c)    The number of moles of gaseous reactants and gaseous products is not equal to each other

(d)   The reaction involves no gaseous reactant or product

 

5. Consider the following reaction occurring in an automobile:

2C_8H_{18} (g) + 25O_2(g) \to 16CO_2(g) + 18H_2O(g)

 

The sign of \Delta H, \Delta s and \Delta G would be

(a)    +, -, +

(b)   -, + , -

(c)    -, +, +

(d)   +, +, -

 

6. What is the free energy change, AG, when 1.0 mole of water at 100°C and 1 atm pressure is converted into steam at 100°C and 1 atm pressure?

(a)    0 cal

(b)   9800 cal

(c)    -9800 cal

(d)   540 cal

 

7. Gibb’s free energy G, enthalpy H and entropy S are related to one another by:

(a)    G = H + TS

(b)   S = H – G

(c)    G = H – TS

(d)   G-TS = H

 

 

8. The standard Gibb’s free energy change, G^0, is related to equilibrium constant, Kp as:

(a)    k_p = e^{-\Delta G^0 /RT}

(b)   K_P = -\dfrac{\Delta G}{RT}

(c)    K_P = -RT ln \Delta G^0

(d)   K_P = (e / RT)^{\Delta G^0}

Answers:

1.(a)         2. (b)        3. (c)       4. (c)        5. (b)      6. (a)       7. (c)       8. (a)



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