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Question-1

Calculate the entropy change in surroundings when 1.00 mol of H2O(l) is formed under standard conditions. Δ rH0 = -286 kJ mol-1.

Solution:
1 mol of H2O(l) is formed under standard conditions.Δ fH0 (surroundings) = 286 kJ mol-1
Δ S = ? 


Δ S =

      = = 1048 J K-1= mol -1.

Question-2

Comment on the thermodynamic stability of NO(g), given
N2g) + O2(g) NO(g); Δ rH0 = 90 kJ mol-1
NO(g) + O2(g) NO2(g); Δ rH0 = -74 kJ mol-1.

Solution:
As Δ fH0 of NO is positive, NO is thermodynamically unstable with reference to its decomposition into constituent elements. Δ fH0 of oxidation of NO is negative which means that NO readily oxidizes to NO2.

Question-3

The equilibrium constant for a reaction is 10. What will be the value of Δ G0? R = 8.314 LK mol-1, T = 300 K.

Solution:
Given: Keq = 10 ; R = 8.314 J K-1 mol-1
Δ G0 = ? T = 300 KΔ G0 = -2.303 RT log Keq

        
= -2.303 × 8.314 JK-1 mol-1 × 300 K × log 10

        = -5744.1426 × 1

        = -5744.14 J mol-1.

Question-4

For the reaction 2A (g) + B (g) 2D (g)     Δ U0 = -10.5 kJ and Δ S0 = -44.1 JK-1. Calculate Δ GO for the reaction, and predict whether the reaction may occur spontaneously.

Solution:
Δ H0 = (Δ U0 + Δ noRT) = - 10.5 J + (-1) × 298 × 8.314 = -12.98 kJ
ΔG0 = ΔH0 - TΔ S0 = 161.8 Jmol−1. SinceΔGο + the reactionis non-spontaneous.  

Question-5

For the reaction, 2Cl (g) Cl2 (g), what are the signs of Δ H and Δ S?

Solution:
Δ H is –ve as the process involves formation of bond between Cl atoms. Δ S is also –ve.

Question-6

For the reaction at 298 K, 2A + B C Δ H = 400 kJ mol-1 and Δ S = 0.2 kJ mol-1 At what temperature will the reaction become spontaneous considering Δ H and Δ S to be constant over the temperature range?

Solution:
T = 298 K

2A + B C Δ H = 400 kJ mol-1
Δ S = 0.2 kJ K-1 mol-1

We have,Δ S = T = = 2000 K T = 2000 K.

Question-7

In a process, 701 J of heat is absorbed by a system and 394 J of work is done by the system. What is the change in internal energy for the process?

Solution:
Here,

q = +701 J

w= -394 J (since work isw done by the system)

According to first law of thermodynamics,Δ U = q + w = 701 - 394 = 307 J.

Question-8

The reaction of cyanamide, NH2CN(s) with dioxygen was carried out in a bomb calorimeter, and Δ U was found to be –742.7 kJ mol-1 at 298 K. Calculate enthalpy change for the reaction at 298 K.NH2CN(g) + O2(g) N2(g) + CO2(g) + H2O(l)

Solution:
The combustion equation of cyanamide is given as

NH2CN(g) + O2(g) N2(g) + CO2(g) + H2O(l)Δ U = -742.7 mol-1
Δ n = 2 - = 0.5 mol

Now, applying the relation, Δ H = Δ U + (Δ n)RT

= -742.7 (kJ) + (-0.5) mol × (8.314 × 10-3 kJ K-1 mol-1) (298 K)

= -742.7 + 1.239 = -741.5 kJ mol-1.

Question-9

Calculate the quantity of kJ of heat necessary to raise the temperature of 60.0 g of aluminium from 35° C to 55° C. Molar heat capacity of Al is 24 J mol-1 K-1.

Solution:
q = c × Δ T     c = n × Cm

Here, Cm = 24.0 J mol-1 K-1; n = = 2.22 mol

c = 2.22 mol × 24.0 J mol-1 K-1 = 53.28 JK-1

Now, q = 53.28 JK-1 × Δ T = 53.28 JK-1 × 20 K = 1065.6 J or 1.065 kJ.

Question-10

Enthalpy of combustion of carbon of CO2 is –393.5 kJ mol-1. Calculate the heat released upon formation of 35.2 g of CO2 from carbon and di-oxygen gas.

Solution:
C + O2 CO2 Δ CH = -393.5 kJ mol-1
Molecular mass of CO2 = 44g
44g of CO2 is formed by releasing –393.5 kJ mol-1 of energy 35.2 g of CO2 will be formed by releasing = -314.8 kJ mol-1

Question-11

Enthalpies of formation of CO(g), CO2(g), N2O(g) and N2O4(g) are –110, -393, 81 and 9.7 kJ mol-1 respectively. Find the value of Δ rH for the reaction:
N2O4(g) + 3CO(g) N2O(g) + 3CO2(g)

Solution:
Δ rH0 for the reaction is given as   Δ rH0 = Δ fH0 (Products) - Δ fH0 (Reactants)
        = (3 Δ fH(CO2) + Δ fH(N2O) )
3 Δ fH(CO) + Δ fH(N2O4))

Substituting the given values, we getΔ rH0 = 3(-393) + 81.0 - 3(-110) - 9.7
        = -1179 + 81.0 + 330 - 9.7 = -777.7 kJ

Question-12

N2(g) + 3H2(g) 2NH3(g); Δ rH0 = -92.4 kJ mol-1
What is the standard enthalpy of formation of NH3 gas?

Solution:

ΔH =ΔU + ΔnRT or ΔU = ΔH - ΔnRT

ΔH = -92.4 KJ T = 298K

Δn = Total number of products - Total number of reactants
= N2 + 3H2
2 NH3

Δn = 2 - 4 = -2

Δu = -92.4 - (-2 RT)
= -92.3 + (2
× 8.314 × 10-3 KJ mol-1 k-1 × 298)
= -92.3 + 4.955


Δu = 87.345 KJ mol-1.
________________________________________
 

Question-13

Calculate the standard enthalpy of formation of CH3OH(I) from the following data:
CH3OH(I) + O2(g) CO2(g) +2H2O(I); ΔrH0 = -726 kJ mol-1
 
.................... (i)
 
C(g) + O2(g) CO2(g); ΔcH0 = -393 kJ mol-1
 
.................... (ii)
 
H2(g) + O2(g) H2O(l); ΔfH0 = -286 kJ mol-1
 
.................... (iii)
 

Solution:
The equation we want is:

C(s) + 2H2(g) + O2(g) CH3OH(l) Δ H = ?

Multiply equation (iii) by 2 and add it to equation (ii)

(iv) C(s) + 2H2(g) + 2O2(g) CO2(g) + 2H2O(l) Δ H =
965.0 kJ

Subtract equation (i) from equation (iv)

C(s) + 2H2(g) + 2O2(g) - CH3OH(l) - O2(g)
CO2(g) + 2H2O(l) - CO2(g) - 2H2O(l); Δ H =965 –(726) = 239.0 KJ

Or

C(s) + 2H2(g) + O2(g) CH3OH(l); Δ H = -2 39.0 kJ.

Question-14

Calculate the enthalpy change for the process

CCl4(g) C(g) + 4 Cl(g) and calculate bond enthalpy of C - Cl in CCl4(g).Δ vapH0(CCl4) = 30.5 kJ mol-1
Δ rH0(CCl4) = -135.5 kJ mol-1
Δ aH0(C) = 715.0 kJ mol-1, where Δ aH0 is enthalpy of atomizationΔ aH0(Cl2) = 242 kJ mol-1.

Solution:
Δ rH0 = Δ aH(C) + 2Δ aH(Cl2) - Δ vapH(CCl4) - Δ fH(CCl4)

         =715.0 + 2(242) - 30.5 - (-135.5) kJ

         = 1304 kJ mol-1; Δ HC - Cl = = 326 kJ mol-1.
Enthalpy change of the process = 1304 kJ mol-1
Bond enthalpy of C-Cl in CCl4 = 326 kJ mol-1

 

Question-15

For an isolated system, Δ U = 0, what will be Δ S?

Solution:
Δ U = 0 does not necessarily means that Δ S is also zero. Δ S can be = 0 or > 0.




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