This set of Class 11 Chemistry Chapter 6 Multiple Choice Questions & Answers (MCQs) focuses on “Gibbs Energy Change and Equilibrium”.
1. Find out whether the following reaction is spontaneous or not at 127 degrees centigrade? N2(g) + 3H2(g) \(\rightleftharpoons\) 2NH3(g); ΔH = 92.22 kJ/mol and ΔS = -198.75 J/K-mol.
a) it is spontaneous
b) it is not spontaneous
c) it may be spontaneous
d) cannot predict
Explanation: We have ΔG = ΔH – TΔS; by substituting ΔH = 92.22 kJ/mol and ΔS = -198.75 J/K-mol, we get ΔG = 92.22 kJ/mol – 400k(-198.75 J/K-mol) = 92.22 kJ/mol + 79.5 kJ/mole = 171.72 KJ/mol. As Gibbs free energy is positive the reaction is non spontaneous.
2. 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) -5.3 Kcal
c) 7.3 Kcal
d) -7.3 Kcal
Explanation: We have ΔG = ΔH – TΔS; by substituting ΔH = 19.07 kcal and ΔS = 90 cal/K, we get 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.
3. In a reaction, the change in entropy is given as 2.4 cal/K and the change in Gibbs free energy is given as 3.4 kcal, calculate the change in heat at the temperature of 20-degree centigrade?
a) 3.4 kcal
b) 3.4 cal
c) 3.4 kJ
d) 3.4 J
Explanation: Using the equation ΔG = ΔH – TΔS, where G is represented by the Gibbs free energy, H is represented by the change in enthalpy, T is represented by the temperature and S is represented by the entropy. ΔH = ΔG + TΔS = 3.4 kcal – 293 x 2.4 cal/K = 3.4 kcal.
4. The melting of ice into liquid water is an example of tube _______________ reaction.
Explanation: An unfavorable reaction or a nonspontaneous reaction that requires more energy than we get from that reaction is called endergonic reaction. It observes heat from surroundings and the change in entropy decreases it is also a type of endothermic reaction.
5. Which of the following is not a type of exergonic reaction?
a) formation of table salt from Sodium and chlorine
b) combustion reaction
Explanation: A reaction that is spontaneous and favorable is known as an exergonic reaction, it releases energy into the surroundings. The system’s free energy decreases here. The Gibbs free energy of an exergonic reaction is negative.
6. The Gibbs free energy is positive when a change in enthalpy and change in entropy are positive at ____________
a) high temperatures
b) low temperature
c) all temperatures
d) only at 0 Kelvin
Explanation: When a change in entropy and the change in enthalpy are positive the Gibbs free energy is positive at low temperatures and negative at high temperatures. We can obtain this through the equation ΔG = ΔH – TΔS.
7. Calculate the Gibbs free energy for the conversion of oxygen to Ozone at room temperature if KP is given as 2.47 x 10-29.
a) 163 kJ/mol
b) 163 J/mol
c) 163 kJ
d) 163 k/mol
Explanation: The chemical equation for the conversion of oxygen to Ozone is 3/2O2 → O3. We have the equation, ΔG = – 2.303 RT log Kp. So by substituting we get ΔG = – 2.303 x 8.314 J/K-mol x 293K x 2.47 x 10-29 = 163000 J/mol = 163 kJ/mol.
8. In the equation, ΔG = – 2.303 RT logK, what is K?
a) change in temperature
c) equilibrium constant
d) change in enthalpy
Explanation: In the above equation ΔG = – 2.303 RT logK, G is the change in standard Gibbs free energy, R is the universal gas constant, T is temperature, K is the equilibrium constant while log stands for the logarithm of base 10.
9. What is the relation between Gibbs free energy and the EMF of the cell? 10. Write 1 Faraday in terms of coulombs. Sanfoundry Global Education & Learning Series – Chemistry – Class 11. To practice all chapters and topics of class 11 Chemistry, here is complete set of 1000+ Multiple Choice Questions and Answers.
a) ΔG = -nFEcell
b) G = -nFEcell
c) ΔG = -nEcell
d) ΔG = -nFcell
Explanation: The relation between Gibbs free energy and EMF of the cell is ΔG = -nFEcell, where G is the caps free energy and as a number of electrons lost or gained while f depicts Faraday and E cell stand for standard electrode potential.
a) 96500 C
b) 95600 C
c) 9560 C
d) 9650 C
Explanation: Faraday is a unit which has no dimensions and it represents electric charge quantity. In the International System of units, it’s coulombs. 1 Faraday is equal to 96500 coulombs. F represents Faraday constant.
10. Write 1 Faraday in terms of coulombs.
Sanfoundry Global Education & Learning Series – Chemistry – Class 11.
To practice all chapters and topics of class 11 Chemistry, here is complete set of 1000+ Multiple Choice Questions and Answers.