Thermodynamics Practice Questions - DPP for NEET

 Page 1


1. (a) Internal energy is dependent upon temperature and according to
first law of thermodynamics total energy of an isolated system
remains same, i.e., in a system of constant mass, energy can
neither be created nor destroyed by any physical or chemical
change but can be transformed from one form to another
For closed insulated container, q = 0, so, ?E = + W, as work is done by
the system
2. (a)  =  J K
–1
 mol
–1
 = 9.77 J
K
–1
 mol
–1
3. (b) Desired equation is 
(Equation II - Equation I)
?H = –245.5 kJ – (–286 kJ) = 40.5 kJ
4. (b)  kJ mol
–1
Let x be the bond enthalpy of N – H bond then
[Note : Enthalpy of formation or bond formation enthalpy is given
which is negative but the given reaction involves bond breaking
hence values should be taken as positive.]
?H = S Bond energies of reactants – S Bond energies of products
2 × – 46 = 712 + 3 × (436) – 6x
– 92 = 2020 – 6x
6x = 2020 + 92
6x = 2112
x = + 352 kJ/mol
5. (b)
Hence, 
Then, ?H for  is 8 × (–64) = –512k cal
6. (a) Given, for reaction
(i)H
2
O ( l) 
Page 2


1. (a) Internal energy is dependent upon temperature and according to
first law of thermodynamics total energy of an isolated system
remains same, i.e., in a system of constant mass, energy can
neither be created nor destroyed by any physical or chemical
change but can be transformed from one form to another
For closed insulated container, q = 0, so, ?E = + W, as work is done by
the system
2. (a)  =  J K
–1
 mol
–1
 = 9.77 J
K
–1
 mol
–1
3. (b) Desired equation is 
(Equation II - Equation I)
?H = –245.5 kJ – (–286 kJ) = 40.5 kJ
4. (b)  kJ mol
–1
Let x be the bond enthalpy of N – H bond then
[Note : Enthalpy of formation or bond formation enthalpy is given
which is negative but the given reaction involves bond breaking
hence values should be taken as positive.]
?H = S Bond energies of reactants – S Bond energies of products
2 × – 46 = 712 + 3 × (436) – 6x
– 92 = 2020 – 6x
6x = 2020 + 92
6x = 2112
x = + 352 kJ/mol
5. (b)
Hence, 
Then, ?H for  is 8 × (–64) = –512k cal
6. (a) Given, for reaction
(i)H
2
O ( l) 
(ii)
For  reaction (i)
      …(iii)
For reaction (ii)
 
On replacing this value in eq. (iii) we have
7. (b)
The given reaction is a combustion reaction. Hence this will be an
exothermic reaction.
i.e. ?H = –ve
Further ?n = + ve          i.e.     ?S = +ve
?G = ?H – T?S = –?H – T?S = –ve
8. (b) We have to calculate the enthalpy of the reaction
OH (g) ? O(g) + H(g)
From the given reactions, this can be obtained as follows.
– ; ?H = – 42.09 kJ mol
–1
+  [H
2
(g) ? 2H(g)] ; ?H =  × 435.89 kJ mol
–1
+  [O
2
(g) ? 2O(g)];  ?H =  × 495.05 kJ mol
–1
Add
;  ?H = 
Page 3


1. (a) Internal energy is dependent upon temperature and according to
first law of thermodynamics total energy of an isolated system
remains same, i.e., in a system of constant mass, energy can
neither be created nor destroyed by any physical or chemical
change but can be transformed from one form to another
For closed insulated container, q = 0, so, ?E = + W, as work is done by
the system
2. (a)  =  J K
–1
 mol
–1
 = 9.77 J
K
–1
 mol
–1
3. (b) Desired equation is 
(Equation II - Equation I)
?H = –245.5 kJ – (–286 kJ) = 40.5 kJ
4. (b)  kJ mol
–1
Let x be the bond enthalpy of N – H bond then
[Note : Enthalpy of formation or bond formation enthalpy is given
which is negative but the given reaction involves bond breaking
hence values should be taken as positive.]
?H = S Bond energies of reactants – S Bond energies of products
2 × – 46 = 712 + 3 × (436) – 6x
– 92 = 2020 – 6x
6x = 2020 + 92
6x = 2112
x = + 352 kJ/mol
5. (b)
Hence, 
Then, ?H for  is 8 × (–64) = –512k cal
6. (a) Given, for reaction
(i)H
2
O ( l) 
(ii)
For  reaction (i)
      …(iii)
For reaction (ii)
 
On replacing this value in eq. (iii) we have
7. (b)
The given reaction is a combustion reaction. Hence this will be an
exothermic reaction.
i.e. ?H = –ve
Further ?n = + ve          i.e.     ?S = +ve
?G = ?H – T?S = –?H – T?S = –ve
8. (b) We have to calculate the enthalpy of the reaction
OH (g) ? O(g) + H(g)
From the given reactions, this can be obtained as follows.
– ; ?H = – 42.09 kJ mol
–1
+  [H
2
(g) ? 2H(g)] ; ?H =  × 435.89 kJ mol
–1
+  [O
2
(g) ? 2O(g)];  ?H =  × 495.05 kJ mol
–1
Add
;  ?H = 
9. (a) C + O
2
 ? CO
2
 + 393.5 kJ/mol
12g 44g
44g is formed from 12g of carbon
35.2g is formed from 
= 9.6 g of C = 9.6/12 = 0.8 mole
1 mole release heat 393.5 kJ
0.8 mole release heat = 393.5 × 0.8
= 314.8 kJ ˜ 315 kJ
10. (d) (i) 2C(s) + H
2
(g)  H – C C – H(g) 
?H = 225 kJ mol
–1
(ii) 2C(s) 2C(g) ?H = 1410 kJ mol
–1
(iii) H
2
(g)   2H(g) ?H = 330 kJ mol
–1
From equation (i) :
225 =  
225 = [1410 + 1 × 330] – [2 × 350 + 1 × BE
C C
]
225 = [1410 + 330] – [700 + BE
C C
]
225 = 1740 – 700 – BE
C C
BE
C C
 = 815 kJ mol
–1
11. (a) Given ?H  35.5 kJ mol
–1
?S = 83.6 JK
–1
 mol
–1
Q ?G = ?H – T?S
For a reaction to be spontaneous, ?G = –ve
i.e., ?H < T?S
?   T > 
So, the given reaction will be spontaneous at T > 425 K
12. (b) ?G = ?H – T?S
Page 4


1. (a) Internal energy is dependent upon temperature and according to
first law of thermodynamics total energy of an isolated system
remains same, i.e., in a system of constant mass, energy can
neither be created nor destroyed by any physical or chemical
change but can be transformed from one form to another
For closed insulated container, q = 0, so, ?E = + W, as work is done by
the system
2. (a)  =  J K
–1
 mol
–1
 = 9.77 J
K
–1
 mol
–1
3. (b) Desired equation is 
(Equation II - Equation I)
?H = –245.5 kJ – (–286 kJ) = 40.5 kJ
4. (b)  kJ mol
–1
Let x be the bond enthalpy of N – H bond then
[Note : Enthalpy of formation or bond formation enthalpy is given
which is negative but the given reaction involves bond breaking
hence values should be taken as positive.]
?H = S Bond energies of reactants – S Bond energies of products
2 × – 46 = 712 + 3 × (436) – 6x
– 92 = 2020 – 6x
6x = 2020 + 92
6x = 2112
x = + 352 kJ/mol
5. (b)
Hence, 
Then, ?H for  is 8 × (–64) = –512k cal
6. (a) Given, for reaction
(i)H
2
O ( l) 
(ii)
For  reaction (i)
      …(iii)
For reaction (ii)
 
On replacing this value in eq. (iii) we have
7. (b)
The given reaction is a combustion reaction. Hence this will be an
exothermic reaction.
i.e. ?H = –ve
Further ?n = + ve          i.e.     ?S = +ve
?G = ?H – T?S = –?H – T?S = –ve
8. (b) We have to calculate the enthalpy of the reaction
OH (g) ? O(g) + H(g)
From the given reactions, this can be obtained as follows.
– ; ?H = – 42.09 kJ mol
–1
+  [H
2
(g) ? 2H(g)] ; ?H =  × 435.89 kJ mol
–1
+  [O
2
(g) ? 2O(g)];  ?H =  × 495.05 kJ mol
–1
Add
;  ?H = 
9. (a) C + O
2
 ? CO
2
 + 393.5 kJ/mol
12g 44g
44g is formed from 12g of carbon
35.2g is formed from 
= 9.6 g of C = 9.6/12 = 0.8 mole
1 mole release heat 393.5 kJ
0.8 mole release heat = 393.5 × 0.8
= 314.8 kJ ˜ 315 kJ
10. (d) (i) 2C(s) + H
2
(g)  H – C C – H(g) 
?H = 225 kJ mol
–1
(ii) 2C(s) 2C(g) ?H = 1410 kJ mol
–1
(iii) H
2
(g)   2H(g) ?H = 330 kJ mol
–1
From equation (i) :
225 =  
225 = [1410 + 1 × 330] – [2 × 350 + 1 × BE
C C
]
225 = [1410 + 330] – [700 + BE
C C
]
225 = 1740 – 700 – BE
C C
BE
C C
 = 815 kJ mol
–1
11. (a) Given ?H  35.5 kJ mol
–1
?S = 83.6 JK
–1
 mol
–1
Q ?G = ?H – T?S
For a reaction to be spontaneous, ?G = –ve
i.e., ?H < T?S
?   T > 
So, the given reaction will be spontaneous at T > 425 K
12. (b) ?G = ?H – T?S
Since ?G = ?H – T?S for an endothermic reaction,
?H = +ve and at low temperature ? S = + ve
Hence ?G = (+) ?H – T ( + )?S
and if T ? S < ?H (at low temp)
?G  = +ve (non spontaneous)
But at high temperature, reaction becomes
spontaneous i.e. ?G = –ve.
because at higher temperature T?S > ?H. 
13. (c) The standard enthalpy of the combustion of glucose can be
calculated by the eqn.
C
6
H
12
O
6
(s) + 6O
2
(g) ? 6CO
2
(g) + 6H
2
O(l)
?H
C
 = 6 × ?H
f
(CO
2
) + 6 × ?H
f 
(H
2
O) – ?H
f 
[C
6
H
12
O
6
]
?H° = 6 (–400) + 6(–300) – (–1300)
?H° = –2900 kJ/mol
For one gram of glucose, enthalpy of combustion  ?H° = – 
14. (c) CH
2
 = CH
2
 (g) + H
2
 (g)  CH
3
 - CH
3
Enthalpy change = Bond energy of reactants – Bond energy of products.
H = 1(C = C) + 4 (C – H) + 1 (H - H) - 1 (C - C) - 6 (C - H)
= 1 (C = C) + 1 ( H – H) – 1 (C – C) – 2 (C– H)
= 615 + 435 – 347 – 2 × 414  = 1050 – 1175 = –125 kJ.
15. (c) For a reaction to be at equilibrium ?G = 0. Since  
so at equilibrium or  
For the reaction
;    (given)
Calculating ?S for the above reaction, we get
=50 – (30 + 60) JK
–1
 = – 40 JK
–1
At equilibrium,      
? 
Page 5


1. (a) Internal energy is dependent upon temperature and according to
first law of thermodynamics total energy of an isolated system
remains same, i.e., in a system of constant mass, energy can
neither be created nor destroyed by any physical or chemical
change but can be transformed from one form to another
For closed insulated container, q = 0, so, ?E = + W, as work is done by
the system
2. (a)  =  J K
–1
 mol
–1
 = 9.77 J
K
–1
 mol
–1
3. (b) Desired equation is 
(Equation II - Equation I)
?H = –245.5 kJ – (–286 kJ) = 40.5 kJ
4. (b)  kJ mol
–1
Let x be the bond enthalpy of N – H bond then
[Note : Enthalpy of formation or bond formation enthalpy is given
which is negative but the given reaction involves bond breaking
hence values should be taken as positive.]
?H = S Bond energies of reactants – S Bond energies of products
2 × – 46 = 712 + 3 × (436) – 6x
– 92 = 2020 – 6x
6x = 2020 + 92
6x = 2112
x = + 352 kJ/mol
5. (b)
Hence, 
Then, ?H for  is 8 × (–64) = –512k cal
6. (a) Given, for reaction
(i)H
2
O ( l) 
(ii)
For  reaction (i)
      …(iii)
For reaction (ii)
 
On replacing this value in eq. (iii) we have
7. (b)
The given reaction is a combustion reaction. Hence this will be an
exothermic reaction.
i.e. ?H = –ve
Further ?n = + ve          i.e.     ?S = +ve
?G = ?H – T?S = –?H – T?S = –ve
8. (b) We have to calculate the enthalpy of the reaction
OH (g) ? O(g) + H(g)
From the given reactions, this can be obtained as follows.
– ; ?H = – 42.09 kJ mol
–1
+  [H
2
(g) ? 2H(g)] ; ?H =  × 435.89 kJ mol
–1
+  [O
2
(g) ? 2O(g)];  ?H =  × 495.05 kJ mol
–1
Add
;  ?H = 
9. (a) C + O
2
 ? CO
2
 + 393.5 kJ/mol
12g 44g
44g is formed from 12g of carbon
35.2g is formed from 
= 9.6 g of C = 9.6/12 = 0.8 mole
1 mole release heat 393.5 kJ
0.8 mole release heat = 393.5 × 0.8
= 314.8 kJ ˜ 315 kJ
10. (d) (i) 2C(s) + H
2
(g)  H – C C – H(g) 
?H = 225 kJ mol
–1
(ii) 2C(s) 2C(g) ?H = 1410 kJ mol
–1
(iii) H
2
(g)   2H(g) ?H = 330 kJ mol
–1
From equation (i) :
225 =  
225 = [1410 + 1 × 330] – [2 × 350 + 1 × BE
C C
]
225 = [1410 + 330] – [700 + BE
C C
]
225 = 1740 – 700 – BE
C C
BE
C C
 = 815 kJ mol
–1
11. (a) Given ?H  35.5 kJ mol
–1
?S = 83.6 JK
–1
 mol
–1
Q ?G = ?H – T?S
For a reaction to be spontaneous, ?G = –ve
i.e., ?H < T?S
?   T > 
So, the given reaction will be spontaneous at T > 425 K
12. (b) ?G = ?H – T?S
Since ?G = ?H – T?S for an endothermic reaction,
?H = +ve and at low temperature ? S = + ve
Hence ?G = (+) ?H – T ( + )?S
and if T ? S < ?H (at low temp)
?G  = +ve (non spontaneous)
But at high temperature, reaction becomes
spontaneous i.e. ?G = –ve.
because at higher temperature T?S > ?H. 
13. (c) The standard enthalpy of the combustion of glucose can be
calculated by the eqn.
C
6
H
12
O
6
(s) + 6O
2
(g) ? 6CO
2
(g) + 6H
2
O(l)
?H
C
 = 6 × ?H
f
(CO
2
) + 6 × ?H
f 
(H
2
O) – ?H
f 
[C
6
H
12
O
6
]
?H° = 6 (–400) + 6(–300) – (–1300)
?H° = –2900 kJ/mol
For one gram of glucose, enthalpy of combustion  ?H° = – 
14. (c) CH
2
 = CH
2
 (g) + H
2
 (g)  CH
3
 - CH
3
Enthalpy change = Bond energy of reactants – Bond energy of products.
H = 1(C = C) + 4 (C – H) + 1 (H - H) - 1 (C - C) - 6 (C - H)
= 1 (C = C) + 1 ( H – H) – 1 (C – C) – 2 (C– H)
= 615 + 435 – 347 – 2 × 414  = 1050 – 1175 = –125 kJ.
15. (c) For a reaction to be at equilibrium ?G = 0. Since  
so at equilibrium or  
For the reaction
;    (given)
Calculating ?S for the above reaction, we get
=50 – (30 + 60) JK
–1
 = – 40 JK
–1
At equilibrium,      
? 
or or 750 K
16. (a) +   HX
Let the bond enthalpy of X – X  bond be x.
 = – 50 = 
H– H
 +   – 
= 2x + x – 2x = 
x = 50 × 2 = 100 
17. (b)
= (213.6 + 2 × 69.9) – (186.2 + 2 × 205.2)
= – 242. 8 J K
–1
 mol
–1
.
18. (b)
?  B.E. of HCl = 215 + 120 + 90
= 425 kJ mol
–1
19. (b) For 5 moles of gas at temperature T,
 = 5RT
For 5 moles of gas at temperature T – 2,
 = 5R(T – 2)
P = 5R(T – 2 – T);
= – 10R,
–  = 10R
When  is negative, W is + ve.
20. (a) 
Bomb calorimeter gives ?U of the reaction