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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-

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

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

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

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

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Explanation

$∆ H_{I} = C_{v} (T_{2} - T_{1}) + (P_{2} V_{1} - P_{1} V_{1})$

$∆ H_{I I} = ∆ U + P_{3} V_{2} - P_{2} V_{1} = P_{3} (V_{2} - V_{1}) + P_{3} V_{2} - P_{2} V_{1} = P_{3} V_{1} - P_{2} V_{1} ∆ H_{o v e r a l l} = C_{v} (T_{2} - T_{1}) + (P_{3} V_{1} - P_{1} V_{1})$

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

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

241.8 kJ

283 kJ

1312 kJ

1586 kJ

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Explanation

112 litre (at NTP) contains no. of moles = $\frac{112}{22.4}$ =5

$∴$ 2.5 moles of H₂ and 2.5 moles of CO present. Hence total amount of heat evolved

$= 2.5 \times 241.8 + 2.5 \times (283) = 1312 k J$

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

20 ml acid and 30 ml alkali

10 ml acid and 40 ml alkali

25 ml acid and 25 ml alkali

35 ml acid and 15 ml alkali

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Explanation

When 25 ml acid is reacting with 25 ml of alkali, maximum equivalent of acid is reacting with alkali since the volume of the resulting solution is the same in every cases, therefore, increase in temperature is the highest where maximum equivalents of acids and alkalies are reacting.

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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 :

A>B>C>D

A>D>C>B

D>C>B>A

D>B>C>A

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Explanation

Lower is heat of neutralisation, more is dissociation energy, weaker is acid

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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,

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

524.21

41.2

-262.5

-41.2

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Explanation

Given

$C + O_{2} \to C O_{2}; ∆ H^{\circ} = - 393.5 k J . . . . . . . (i) C + (1 / 2) O_{2} \to C O; ∆ H^{\circ} = - 110.5 k J . . . . . . . (i i) H_{2} + (1 / 2) O_{2} \to H_{2} O; ∆ H^{\circ} = - 241.8 k J . . . . . . . (i i i)$

By (ii) + (iii) - (i),

$C O_{2} + H_{2} \to C O + H_{2} O; ∆ H^{\circ} = + 41.2$

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

260 Kcal/mole

180 Kcal/mole

130 Kcal/mole

80 Kcal/mole

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Explanation

$C H_{4} \to C (s) + 2 H_{2} (g) ∆ H = 360 K c a l C_{2} H_{6} \to 2 C (s) + 3 H_{2} (g) ∆ H = 620$

The avg C-H bond energy = $\frac{360}{4}$ = 90 Kcal

$∴$ C-C bond energy = 620 - 90 $\times$ 6 = 620 - 540 = 80 Kcal

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For an endothermic reaction-

$\sum_{P r o d u c t s} H = \sum_{R e a c \tan t s} H$

$\sum_{P r o d u c t s} H < \sum_{R e a c \tan t s} H$

$\sum_{P r o d u c t s} H > \sum_{R e a c \tan t s} H$

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

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Explanation

For an endothermic reaction, $∆ H$ is positive

= $\sum H (p r o d u c t s) - \sum H (r e a c \tan t s) .$

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All the natural process in this universe produce

A decrease in entropy of the universe

An increase in entropy of the universe

No change in entropy

Sometimes increase or sometimes decrease in entropy

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Explanation

For any natural process there is always an increase in entropy.

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

166.5 kJ mol^-1

161.5 kJ mol^-1

164.0 kJ mol^-1

169 kJ mol^-1

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Explanation

$∆ H = ∆ E + ∆ n R T . ∆ n = - 1 164 \times 10^{3} = ∆ E - 1 \times 08.314 \times 298 = ∆ E - 2.478 \times 10^{3} ∆ E = 164 \times 10^{3} + 2.478 \times 10^{3} = 166.5 K J m o l^{- 1}$

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For $A \to B$ , $∆ H = 4 k c a l m o l^{- 1}, ∆ S = 10 c a l m o l^{- 1} K^{- 1} .$ Reaction is spontaneous when temperature can be :

400 K

300 K

500 K

none is correct

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Explanation

$∆ G = - v e$ for spontaneous charge

$∆ G = ∆ H - T ∆ S ∴ T ∆ S > ∆ H, T > \frac{∆ H}{∆ S} \Rightarrow T > \frac{4000}{10} \Rightarrow T > 400 K$

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