One mole of a real gas is subjected to heating at constant volume from P 1 , V 1 , T 1 state to P 2 , V 2 , T 2 state. Then it is subjected to irreversible adiabatic compression against constant external pressure of P 3 atm till syatem reaches the final state P 3 , V 3 , T 3 . If the constant volume molar heat capacity of real gas is C V . Find out correct expression for ∆ H from state 1 to state 3-

C v T 2 - T 1 + P 3 V 1 - P 1 V 1

C v T 3 - T 1 + P 3 V 1 - P 1 V 1

C v T 2 - T 1 + P 3 V 2 - P 1 V 1

C p T 2 - T 1 + P 3 V 1 - P 1 V 1

The standard heat of combustion of Al is -837.8 kJ mol -1 at 25 ∘ C which of the following releases 250 kcal of heat ?

The formation of 0.624 mol of Al 2 O 3

The reaction of 0.624 mol of Al

The reaction of 0.312 mol of Al

The formation of 0.150 mol of Al 2 O 3

Chemistry - Thermodynamics - Quiz-7

One mole of a real gas is subjected to heating at constant volume from $(P_{1}, V_{1}, T_{1})$ state to $(P_{2}, V_{2}, T_{2})$ state. Then it is subjected to irreversible adiabatic compression against constant external pressure of P₃ atm till syatem reaches the final state $(P_{3}, V_{3}, T_{3})$ . If the constant volume molar heat capacity of real gas is C_V. Find out correct expression for $∆ H$ from state 1 to state 3-

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$C_{v} (T_{3} - T_{1}) + (P_{3} V_{1} - P_{1} V_{1})$
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$C_{v} (T_{2} - T_{1}) + (P_{3} V_{2} - P_{1} V_{1})$
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$C_{v} (T_{2} - T_{1}) + (P_{3} V_{1} - P_{1} V_{1})$
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$C_{p} (T_{2} - T_{1}) + (P_{3} V_{1} - P_{1} V_{1})$

$[\begin{matrix} H_{2} (g) + \frac{1}{2} O_{2} (g) = H_{2} O (g); ∆ H = 241.8 k J \\ C O (g) + \frac{1}{2} O_{2} (g) = C O_{2} (g); ∆ H = 283 k J \end{matrix}]$

The heat evolved in the combustion of 112 litres of water gas(mix of equal volumes of H₂ and CO)

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241.8 kJ
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3. 1312 kJ
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2. 283 kJ
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1586 kJ

In which case of mixing of a strong acid and a base, each of 1(N) concentration, temperature-increase is the highest ?

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20 ml acid and 30 ml alkali
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10 ml acid and 40 ml alkali
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25 ml acid and 25 ml alkali
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35 ml acid and 15 ml alkali

The heats of neutralisation of four acids A, B, C and D are -13.7, -9.4, -11.2 and -12.4 kcal respectively when they are neutralised by a common base. The acidic character obeys the order :

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A>B>C>D
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A>D>C>B
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D>C>B>A
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D>B>C>A

The $∆ H_{f}^{\circ}$ for CO₂(g), CO(g) and H₂O(g) are -393.5, -110.5 and -241.8 kJ mol^-1 respectively. The standard enthalpy change (in kJ) for the reaction,

$C O_{2} (g) + H_{2} (g) \to H_{2} O (g) + C O (g)$ is :

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524.21
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41.2
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-262.5
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-41.2

The dissociation energy of CH₄ and C₂H₆ are respectively 360 and 620 Kcal/mole. The bond energy of C-C is-

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260 Kcal/mole
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180 Kcal/mole
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130 Kcal/mole
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80 Kcal/mole

For an endothermic reaction-

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$\sum_{P r o d u c t s} H = \sum_{R e a c \tan t s} H$
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$\sum_{P r o d u c t s} H < \sum_{R e a c \tan t s} H$
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$\sum_{P r o d u c t s} H > \sum_{R e a c \tan t s} H$
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$\sum_{P r o d u c t s} H = 0 b u t \sum_{R e a c \tan t s} H i s p o s i t i v e .$

All the natural process in this universe produce

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A decrease in entropy of the universe
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An increase in entropy of the universe
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No change in entropy
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Sometimes increase or sometimes decrease in entropy

For the reaction, $2 N_{2} (g) + O_{2} (g) \to 2 N_{2} O (g)$ , at 298K $∆ H$ is 164 kJ mol^-1. The $∆ E$ of the reaction is-

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166.5 kJ mol^-1
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141.5 kJ mol^-1
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104.0 kJ mol^-1
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-169 kJ mol^-1

$F o r A \to B$ , $∆ H = 4 kcal {mol}^{- 1}, ∆ S = 10 cal {mol}^{- 1} K^{- 1, the r}$ eaction is spontaneous when the temperature is:

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400 K
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300 K
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500 K
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None of the above

44.0 kJ of heat is required to evaporate one mole of water at 298 K. If $∆ H_{f}$ of $H_{2} O (l)$ is -286 kJ mol^-1, $∆ {H^{\circ}}_{f}$ of $H_{2} O (g)$ is

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-330 kJ mol^-1
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+242 kJ mol^-1
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-242 kJ mol^-1
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-198 kJ mol^-1

The correct calculate value for $△ H$ for:

$M (s) \to M^{2 +} (aq)$

From following arbitrary values :

$M (s) \to M (g); △ H = 1000 KJ {mol}^{- 1} M (g) \to M^{+} (g); △ H = 750 KJ {mol}^{- 1} M^{+} (g) \to M^{2 +} (g); △ H = 1200 KJ {mol}^{- 1} M^{2 +} (g) + aq \to M^{2 +} (aq .); △ H = - 1800 KJ {mol}^{- 1}$

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1950 KJ mol^-1
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1150 KJ mol^-1
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2300 KJ mol^-1
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None of the above.

The molar heat capacity, $C_{v}$ of an ideal gas whose energy is that of translational motion only is

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$2.98 J d e g^{- 1} m o l^{- 1}$
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$12.47 J d e g^{- 1} m o l^{- 1}$
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$6.43 J d e g^{- 1} m o l^{- 1}$
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$9.41 J d e g^{- 1} m o l^{- 1}$

The lattice energy of NaCl is 780 kJ mol^-1. The enthalpy of hydration of Na⁺(g) and Cl^-(g) ions are -406 kJ mol^-1 and -364 kJ mol^-1. The enthalpy of the solution of NaCl(s) is-

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23 kJ mol^-1
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10 kJ mol^-1
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-10 kJ mol^-1
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-82 kJ mol^-1

Enthalpy of fusion of a liquid is 1.435 kcal mol^-1 and molar entropy change is 5.26 cal mol^-1K^-1. Hence melting point of liquid is :

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$0^{\circ} C$
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373 K
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$100^{\circ} C$
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$- 273^{\circ} C$

Following reaction occurs at $25^{\circ} C$ :

$2 N O (g, 1 \times 10^{- 5} a t m) + C l_{2} (g, 1 \times 10^{- 2} a t m) ⇌ 2 N O C l (g, 1 \times 10^{- 2} a t m) ∆ G^{\circ}$ is-

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-45.65 kJ
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3. -282 kJ
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2. -28.53 kJ
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-57.06 kJ

When 1 mole of an ideal gas to 20 atm pressure and 15 L volume expands such that the final pressure becomes 10 atm and the final volume become 60 L. Calculate entropy change for the reaction (C_p.m = 30.96)

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$80.2 J k^{- 1} m o l^{- 1}$
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$62.42 J k^{- 1} m o l^{- 1}$
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$120 \times 10^{2} J k^{- 1} m o l^{- 1}$
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$27.22 J k^{- 1} m o l^{- 1}$

If a process is both endothermic and spontaneous, then :

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$∆ S > 0$
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$∆ S < 0$
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$∆ H < 0$
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$∆ G > 0$

The bond energies of C=C and C-C at 298 K are 590 and 331 kJ mol^-1 respectively. The enthalpy of polymerization per mole of ethylene is

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-70 kJ
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-72 kJ
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72 kJ
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-68 kJ

1 mole of NH₃ gas at $27^{\circ} C$ is expanded adaibatic condition to make volume 8 times ( $γ$ =1.33). Final temperature and work done respectively are-

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150 K, 900 cal
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150 K, 400 cal
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250 K, 1000 cal
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200 K, 800 cal

Which of the following statements is correct with regard to $∆$ G of a cell reaction and EMF of the cell (E) in which the reaction occurs ?

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Both $∆$ G and E are extensive properties
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$∆$ G is an intensive property but E is an extensive property
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$∆$ G is an extensive property and E is an intensive property
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Both $∆$ G and E are intensive properties.

Temperature of 1 mol of a gas is increased by $1^{\circ}$ at constant pressure. Work done is-

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-R
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2R
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R/2
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3R

When 0.16 g of glucose was burnt in a bomb calorimeter, the temperature rose by 4 deg. Calculate the calorimeter constant (water equivalent of the calorimeter) given that $∆ H^{\circ} = - 2.8 \times 10^{6} J m o l^{- 1} .$ [molar enthalpy of combustion]. Molar mass of glucose = 180 mol^-1.

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$5.73 \times 10^{2} J / d e g$
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$7.53 \times 10^{2} J / d e g$
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$6.22 \times 10^{2} J / d e g$
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$3.57 \times 10^{2} J / d e g$

The C-Cl bond energy can be calculated from :

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$∆ {H^{\circ}}_{f} (C C l_{4}, l) o n l y$
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${∆^{\circ}}_{f} (C C l_{4}, l) a n d ∆ (C l_{2})$
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$∆ {H^{\circ}}_{f} (C C l_{4}, l) ∆ (C l_{2})$
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$∆ {H^{\circ}}_{f} (C C l_{4}, l) ∆ (C l_{2}), ∆ {H^{\circ}}_{f} (C, g) a n d ∆ {H^{\circ}}_{v a p} (C C l_{4})$

Given $∆ {H^{\circ}}_{f}$ of DyCl₃ (s) = -994.30 kJ mol^-1

$\frac{1}{2} H_{2} (g) + \frac{1}{2} C l_{2} (g) \overset{+ a q}{\to} H C l (a q . 4 M); ∆ H^{\circ} = - 158.31 k J m o l^{- 1} D y C l_{3} (s) \to_{H C l}^{a q} D y C l_{3} (a q . i n 4 M H C l); ∆ H^{\circ} = - 180.06 k J m o l^{- 1} D y (s) \to_{a q . 4 M}^{+ 3 H C l} D y C l_{3} (a q . 4 M H C l) + \frac{3}{2} H_{2} (g); ∆ H^{\circ} = x, c a l c u l a t e x .$

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-966.5 kJ/mol
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-699.43 kJ/mol
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-596.6 kJ/mol
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-569.6 kJ/mol

When 1 g H₂ gas at S.T.P is expanded to twice its initial volume, then the work done is -

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22.4 L atm
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5.6 L atm
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11.2 L atm
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44.8 L atm

The gas absorbs 100 J heat and is simultaneously compressed by a constant external pressure of 1.50 atm from 8 lit. to 2 lit. in volume. Hence $∆ E$ will be-

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-812 J
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812 J
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1011 J
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911 J

If $∆ G = ∆ H - T ∆ S$ and $∆ G = ∆ H + T {(\frac{d (∆ G)}{d T})}_{P}$ then variation of EMF of a cell E, with temperature T, is given by

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$\frac{∆ H}{n F}$
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$\frac{∆ G}{n F}$
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$\frac{∆ S}{n F}$
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$- \frac{∆ S}{n F}$

The standard heat of combustion of Al is -837.8 kJ mol^-1 at $25^{\circ} C$ which of the following releases 250 kcal of heat ?

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The reaction of 0.624 mol of Al
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The formation of 0.624 mol of Al₂O₃
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The reaction of 0.312 mol of Al
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The formation of 0.150 mol of Al₂O₃

$C_{P} - C_{V} = R$ . This R is :

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Change in K.E.
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Change in rotational energy
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work done which system can do on expanding the gas per mol per degree increase in temperature
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All correct

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Practice Chemistry Quiz Questions and Answers

0:0:1

Practice Chemistry Quiz Questions and Answers

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