# Question-1

**Calculate the entropy change in surroundings when 1.00 mol of H**

_{2}O(l) is formed under standard conditions. Î”_{r}H^{0}= -286 kJ mol^{-1}.**Solution:**

1 mol of H

_{2}O(l) is formed under standard conditions.Î”

_{f}H

^{0}(surroundings) = 286 kJ mol

^{-1}

Î” S = ?

Î” S =

= = 1048 J K^{-1}= mol ^{-1}.

# Question-2

**Comment on the thermodynamic stability of NO(g), given**

N

NO(g) + O

N

_{2}g) + O_{2}(g) â†’ NO(g); Î”_{r}H^{0}= 90 kJ mol^{-1}NO(g) + O

_{2}(g) â†’ NO_{2}(g); Î”_{r}H^{0}= -74 kJ mol^{-1}.**Solution:**

As Î”

_{f}H

^{0}of NO is positive, NO is thermodynamically unstable with reference to its decomposition into constituent elements. Î”

_{f}H

^{0}of oxidation of NO is negative which means that NO readily oxidizes to NO

_{2}.

# Question-3

**The equilibrium constant for a reaction is 10. What will be the value of Î” G**

^{0}? R = 8.314 LK mol^{-1}, T = 300 K.**Solution:**

Given: K

_{eq}= 10 ; R = 8.314 J K

^{-1}mol

^{-1}

Î” G

^{0}= ? T = 300 KÎ” G

^{0}= -2.303 RT log K

_{eq }= -2.303 Ã— 8.314 JK

^{-1}mol

^{-1}Ã— 300 K Ã— log 10

= -5744.1426 Ã— 1

= -5744.14 J mol

^{-1}.

# Question-4

**For the reaction**

**2A (g) + B (g) â†’ 2D (g)**

**Î” U**

^{0}= -10.5 kJ and Î” S^{0}= -44.1 JK^{-1}.**Calculate Î” G**

^{O}for the reaction, and predict whether the reaction may occur spontaneously.**Solution:**

Î” H

^{0}= (Î” U

^{0}+ Î” n

_{o}RT) = - 10.5 J + (-1) Ã— 298 Ã— 8.314 = -12.98 kJ

Î”

**G**= Î”H

^{0}^{0}- TÎ” S

^{0 }= 161.8 Jmolâˆ’1. Since

**Î”G**Î¿ + the reactionis non-spontaneous.

# Question-5

**For the reaction, 2Cl (g) â†’ Cl**

_{2}(g), what are the signs of Î” H and Î” S?**Solution:**

Î” H is â€“ve as the process involves formation of bond between Cl atoms. Î” S is also â€“ve.

# Question-6

**For the reaction at 298 K,**

**2A + B â†’ C Î” H = 400 kJ mol**

^{-1}and Î” S = 0.2 kJ mol^{-1 }At what temperature will the reaction become spontaneous considering Î” H and Î” S to be constant over the temperature range?**Solution:**

T = 298 K

2A + B â†’ C Î” H = 400 kJ mol

^{-1}

Î” S = 0.2 kJ K

^{-1}mol

^{-1}

We have,Î” S = â‡’ T = = 2000 Kâˆ´ 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,Î” U = q + w = 701 - 394 = 307 J.

# Question-8

**The reaction of cyanamide, NH**

_{2}CN(s) with dioxygen was carried out in a bomb calorimeter, and Î” U was found to be â€“742.7 kJ mol^{-1}at 298 K. Calculate enthalpy change for the reaction at 298 K.NH_{2}CN(g) + O_{2}(g) â†’ N_{2}(g) + CO_{2}(g) + H_{2}O(l)**Solution:**

The combustion equation of cyanamide is given as

NH

_{2}CN(g) + O

_{2}(g) â†’ N

_{2}(g) + CO

_{2}(g) + H

_{2}O(l)Î” U = -742.7 mol

^{-1}

Î” n = 2 - = 0.5 mol

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

= -742.7 (kJ) + (-0.5) mol Ã— (8.314 Ã— 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 35Â° C to 55Â° C. Molar heat capacity of Al is 24 J mol**

^{-1}K^{-1}.**Solution:**

q = c Ã— Î” T c = n Ã— C

_{m}

Here, C

_{m}= 24.0 J mol

^{-1}K

^{-1}; n = = 2.22 mol

c = 2.22 mol Ã— 24.0 J mol

^{-1}K

^{-1}= 53.28 JK

^{-1}

Now, q = 53.28 JK

^{-1}Ã— Î” T = 53.28 JK

^{-1}Ã— 20 K = 1065.6 J or 1.065 kJ.

# Question-10

**Enthalpy of combustion of carbon of CO**

_{2}is â€“393.5 kJ mol^{-1}. Calculate the heat released upon formation of 35.2 g of CO_{2}from carbon and di-oxygen gas.**Solution:**

C + O

_{2}â†’ CO

_{2}Î”

_{C}H = -393.5 kJ mol

^{-1}

Molecular mass of CO

_{2}= 44g

44g of CO

_{2}is formed by releasing â€“393.5 kJ mol

^{-1}of energyâˆ´ 35.2 g of CO

_{2}will be formed by releasing = -314.8 kJ mol

^{-1}

# Question-11

**Enthalpies of formation of CO(g), CO**

N

_{2}(g), N_{2}O(g) and N_{2}O_{4}(g) are â€“110, -393, 81 and 9.7 kJ mol^{-1}respectively. Find the value of Î”_{r}H for the reaction:N

_{2}O_{4}(g) + 3CO(g) â†’ N_{2}O(g) + 3CO_{2}(g)**Solution:**

Î”

_{r}H

^{0}for the reaction is given as Î”

_{r}H

^{0}= âˆ‘ Î”

_{f}H

^{0}(Products) - âˆ‘ Î”

_{f}H

^{0}(Reactants)

= (3 Î”

_{f}H(CO

_{2}) + Î”

_{f}H(N

_{2}O) ) âˆ’ 3 Î”

_{f}H(CO) + Î”

_{f}H(N

_{2}O

_{4}))

Substituting the given values, we getÎ”

_{r}H

^{0}= 3(-393) + 81.0 - 3(-110) - 9.7

= -1179 + 81.0 + 330 - 9.7 = -777.7 kJ

# Question-12

**N**

What is the standard enthalpy of formation of NH

_{2}(g) + 3H_{2}(g) â†’ 2NH_{3}(g); Î”_{r}H^{0}= -92.4 kJ mol^{-1}What is the standard enthalpy of formation of NH

_{3}gas?**Solution:**

Î”H =Î”U + Î”nRT or Î”U = Î”H - Î”nRT

Î”H = -92.4 KJ T = 298K

Î”n = Total number of products - Total number of reactants

= N2 + 3H2 â†’ 2 NH3

Î”n = 2 - 4 = -2

Î”u = -92.4 - (-2 RT)

= -92.3 + (2 Ã— 8.314 Ã— 10-3 KJ mol-1 k-1 Ã— 298)

= -92.3 + 4.955

Î”u = 87.345 KJ mol-1.

________________________________________

# Question-13

**Calculate the standard enthalpy of formation of CH**_{3}OH(I) from the following data:CH_{3}OH(I) + O_{2}(g) â†’ CO_{2}(g) +2H_{2}O(I); Î”_{r}H^{0} = -726 kJ mol^{-1} |
.................... (i) |

C(g) + O_{2}(g) â†’ CO_{2}(g); Î”_{c}H^{0} = -393 kJ mol^{-1} |
.................... (ii) |

H_{2}(g) + O_{2}(g) â†’ H_{2}O(l); Î”_{f}H^{0} = -286 kJ mol^{-1} |
.................... (iii) |

**Solution:**

The equation we want is:

C(s) + 2H

_{2}(g) + O

_{2}(g) â†’ CH

_{3}OH(l) Î” H = ?

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

(iv) C(s) + 2H

_{2}(g) + 2O

_{2}(g) â†’ CO

_{2}(g) + 2H

_{2}O(l) Î” H = âˆ’965.0 kJ

Subtract equation (i) from equation (iv)

C(s) + 2H

_{2}(g) + 2O

_{2}(g)

**-**CH

_{3}OH(l) - O

_{2}(g)â†’ CO

_{2}(g) + 2H

_{2}O(l) - CO

_{2}(g) - 2H

_{2}O(l); Î” H =âˆ’965 â€“(âˆ’726) = âˆ’ 239.0 KJ

Or

C(s) + 2H

_{2}(g) + O

_{2}(g) â†’ CH

_{3}OH(l); Î” H = -2 39.0 kJ.

# Question-14

**Calculate the enthalpy change for the process**

CCl

Î”

Î”

CCl

_{4}(g) â†’ C(g) + 4 Cl(g) and calculate bond enthalpy of C - Cl in CCl_{4}(g).Î”_{vap}H^{0}(CCl_{4}) = 30.5 kJ mol^{-1}Î”

_{r}H^{0}(CCl_{4}) = -135.5 kJ mol^{-1}Î”

_{a}H^{0}(C) = 715.0 kJ mol^{-1}, where Î”_{a}H^{0}is enthalpy of atomizationÎ”_{a}H^{0}(Cl_{2}) = 242 kJ mol^{-1}.**Solution:**

Î”

_{r}H

^{0}= Î”

_{a}H(C) + 2Î”

_{a}H(Cl

_{2}) - Î”

_{vap}H(CCl

_{4}) - Î”

_{f}H(CCl

_{4})

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

= 1304 kJ mol^{-1}; Î” H_{C - Cl} = = 326 kJ mol^{-1}.

Enthalpy change of the process = 1304 kJ mol^{-1}

Bond enthalpy of C-Cl in CCl_{4} = 326 kJ mol^{-1}

# Question-15

**For an isolated system, Î” U = 0, what will be Î” S?**

**Solution:**

Î” U = 0 does not necessarily means that Î” S is also zero. Î” S can be = 0 or > 0.