Thermodynamics - Revision Session

$△_{f}$ G° for ( $N H_{4} C l$ ,s) at 310 K. Given $△_{f}$ H° ( $N H_{4} C l$ ,s) = -314.5 kJ/mol;
$△_{t} C_{p} = O$ is -
S° $N_{2}$ (g) = 192 $J K^{- 1} m o l^{- 1}$
S° $H_{2}$ (g) = 130.5 $J K^{- 1} m o l^{- 1}$
S° $C l_{2}$ (9) = 233 $J K^{- 1} m o l^{- 1}$
S° $N H_{4} C l$ (S) = 99.5 $J K^{- 1} m o l^{- 1}$
(All given data at 300 K)

1. -168.56 kJ/mol
2. -426.7 kJ/mol
3. -202.3 kJ/mol
4. None of the above

Using listed informations, calculate $△_{f}$ G° (in kJ/mol) at 27°C $C o_{3} O_{4} (s) + 4 C O (g) \to 3 C o (s) + 4 C O_{2} (g)$ Given: At 300 K
$△_{f}$ H° (kJ/mol) -891, -110.5, 0.0,-393.5 ;
$S^{0}$ (J/K-mol) 102.5, 197.7, 30.0, 213.7
1. -214.8
2. -195.0
3. -200.3
4. -256.45

Determine $△$ U° at 300 K for the following reaction using the listed enthalpies of
reactions
$4 C O (g) + 8 H_{2} (g) \to 3 C H_{4} (g) + C O_{2} (g) + 2 H_{2} O (l)$
$C (g r a p h i t e) + \frac{1}{2} O_{2} (g) \to C O (g); △ H_{1}^{0} = - 110.5 k J$
$C O (g) + \frac{1}{2} O_{2} (g) \to C O_{2} (g); △ H_{2}^{0} = - 282.9 k J$
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l); △ H_{3}^{0} = - 285.8 k J$
$C (g r a p h i t e) + 2 H_{2} (g) \to C H_{4} (g) △ H_{4}^{0} = - 74.8 k J$
1. -653.5 kJ
2. -686.2 kJ
3. -747.4kJ
4. None of these

$△_{f}$ H° (in kJ/mol) for $C r_{2} O_{3}$ from $△_{f}$ G° and the S° values provided at 27°C for the given reaction is -
4Cr(s)+ $3 O_{2}$ (g) $\to 2 C r_{2} O_{3}$ (s); $△ r$ G° = -2093.4 kJ/mol S° (J/k mol) : S (Cr,s) = 24; $S^{0}$ ( $O_{2}$ ,g) = 205; S° $(C r_{2} O_{3}, s)$ = 81

1. -2258.1 kJ/mol
2. -1129.05 kJ/mol
3. -964.35 kJ/mol
4. None of the above

The heat produced (in kJ) when 224 gm of CaO is completely converted to $C a C O_{3}$ by reaction with $C O_{2}$ at 27°C in a container of fixed volume is -
Given: $△_{f} H^{0}$ ( $C a C O_{3}$ ,s) = -1207 kJ/mol; $△_{f} H^{0}$ (CaO,s) = -635 kJ/mol $△_{f} H^{0}$ ( $C O_{2}$ ,g) = - 394 kJ/mol; [Use R = 8.3 $K^{- 1}$ $m o l^{- 1}$ ]

1. 302.04 kJ
2. 721.96 kJ
3. 712 J
4. 71 kJ

Enthalpy of neutralization of $H_{3} P O_{3}$ acid is - 106.68 kJ/mol using NaOH. If enthalpy of neutralization of HCl by NaOH is -55.84 kJ/mol. $△_{i o n i s a t i o n}$ of $H_{3} P O_{3}$ will be -

1. 50.84 kJ/mol
2. 5 kJ/mol
3. 2.5 kJ/mol
4. None of the above

The enthalpy of neutralization of weak monoprotic acid (HA) in 1 M solution with a
strong base is -55.95 kJ/mol. If the unionized acid is required 1.4 kJ/mol heat for it’s complete ionization and enthalpy of neutralization of the strong monobasic acid
with a strong monobasic base is -57.3 kJ/mol. What is the % ionization of the weak acid in molar solution
1. 1%
2. 3.57%
3. 35.7%
4. 10%

Bond enthalpy of C-C and C-H bond enthalpy (in kJ/mol) is -
Given:
$△_{f} H^{0} (C_{2} H_{6}, g)$ = -85 kJ/mol, $△_{f} H^{0} (C_{3} H_{8}, g)$ = - 104 kJ/mol; $△_{s u b} H^{0} (C, s)$ = 718 kJ/mol, B.E. (H-H) = 436 kJ/mol

1. 414, 345
2. 345,414
3. 287,404.5
4. None of the above.

Consider the following data : $△_{f}$ H°( $(N_{2} H_{4}, l)$ = 50 kJ/mol, $△_{f}$ H° $(N H_{3}, g)$ = -46 kJ/mol B.E. (N-H) = 393 kJ/mol and B.E. (H-H) = 436 kJ/mol $△_{v a p}$ $H (N_{2} H_{4}, l)$ = 18 kJ/mol The N-N bond energy in $N_{2} H_{4}$ is-
1. 236 kJ/mol
2. 154 kJ/mol
3. 190 kJ/mol
4. None of the above

10.

From given following equations and $△ H^{0}$ values, determine the enthalpy of reaction at 298 K for the reaction:
$C_{2} H_{4} (g) + 6 F_{2} (g) \to 2 F_{4} (g) + 4 H F (g)$
$H_{2} (g) + F_{2} (g) \to 2 H F (g); △ H_{1}^{0} = - 537 k J$
$C (s) + 2 F_{2} (g) \to C F_{4} (g); △ H_{2}^{0} = - 680 k J$
$2 C (s) + 2 H_{2} (g) \to C_{2} H_{4} (g); △ H_{3}^{0} = 52 k J$
1. -1165
2. -2486
3. +1165
4. +2486

11.

Consider the following reactions:
C(s) + $O_{2}$ (g) $\to {CO}_{2} (g)$ + x kJ
CO(g) + $\frac{1}{2} O_{2} (g)$ ----> ${CO}_{2}$ + y kJ
The heat of formation of CO(g) is-
1. -(x+y) kJ/mol

2. (x-y) kJ/mol

3. (y-x) kJ/mol

4. None of the above

12.

The fat, $C_{57} H_{104} O_{6} (s)$ C57H1O6(s) is metabolized via the following reaction. Given the enthalpies of formation, calculate the energy (kJ) liberated when 1.0 g of this fat reacts
$C_{57} H_{104} (s) + 80 O_{2} \to C O_{2} (g) + 57 H_{2} O (l)$
$△_{f} H^{0} (C_{57} H_{104} O_{6}, s) = - 70870 k J / m o l;$
$△_{f} H^{0} (H_{2} O, l) = - 285.8 k J / m o l;$
$△_{f} H^{0} (C_{2}, g) = - 393.5 k J / m o l;$
1. -37.98
2. -40.4
3. -33.4
4. -30.2

13.

Determine the enthalpy of formation for $H_{2} O_{2} (l)$ , using the following enthalpies of reaction :
$N_{2} H_{4} (l) + 2 H_{2} O_{2} (l) \to N_{2} (g) + 4 H_{2} O (l);$ $△ H_{1}^{0} = - 818$ kJ/mol
$N_{2} H_{4} (l) + O_{2} (g) \to N_{2} (g) + 2 H_{2} O (l);$ $△ H_{2}^{0} = - 622$ kJ/mol
$H_{2} (g) + \frac{1}{2} O_{2} (g) \to H_{2} O (l);$ $△ H_{3}^{0} = - 285$ $k J / m o l$

1.	-383 kJ/mol	2.	-187 kJ/mol
3.	-49 kJ/mol	4.	None of the above

14.

Enthaply of neutralization of HCl by NaOH is 55.84 kJ/mol and by $N H_{4} O H$ is -51.34 kJ/mol. The enthalpy of ionization of $N H_{4} O H$ is :
1. 10.18 kJ/mol
2. 4.5 kJ/mol
3. -4.5 kJ/mol
4. None of the above

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