For which of the following equations, $$\Delta H > \Delta E$$ ?

$$1.~~\mathrm{N}_{2}(g)+3 \mathrm{H}_{2}(\mathrm{~g}) \longrightarrow 2 \mathrm{NH}_{3}(\mathrm{g})\\ \\ 2.~~\mathrm{H}_{2}(g)+\frac{1}{2} \mathrm{O}_{2}(\mathrm{~g}) \longrightarrow \mathrm{H}_{2} \mathrm{O}(\mathrm{l})\\ \\ 3.~~\mathrm{PCl}_{5}(g) \rightarrow \mathrm{PCl}_{3}(g)+\mathrm{Cl}_{2}(g)\\ 4.~~\mathrm{H}_{2}(\mathrm{~g})+\mathrm{I}_{2}(\mathrm{~g}) \rightarrow 2 \mathrm{HI}(\mathrm{g})$$
Subtopic:  Enthalpy & Internal energy |
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If water vapour is assumed to be a perfect gas, molar enthalpy change for vapourisation of 1 mol of water at 1 bar and 100°C is 41kJ mol–1. The internal energy change, when 1 mol of water is vapourised at 1 bar pressure and 100°C is:

1. 35.5 kJ mol–1

2. 37.9 kJ mol–1

3.  41 kJ mol–1

4. 44.2 kJ mol–1

Subtopic:  Enthalpy & Internal energy |
58%
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A swimmer coming out from a pool is covered with a film of water weighing about 18g. The internal energy of vaporization at 298K. is-

vap H for water at 298K= 44.01kJ mol–1

1. 38.63 kJ
2. 43.82 J
3. 41.53 kJ
4. 40.33 J



Subtopic:  Enthalpy & Internal energy |
67%
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 Assertion (A): The increase in internal energy $$(\Delta E )$$ for the vapourisation of one mole of water at 1 atm and 373 K is zero. Reason (R): For all isothermal processes, $$\Delta E = 0$$.

 1 Both (A) and (R) are True and (R) is the correct explanation of (A). 2 Both (A) and (R) are True but (R) is not the correct explanation of (A). 3 (A) is True and (R) is False. 4 (A) and (R) both are False.
Subtopic:  Enthalpy & Internal energy |
75%
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The enthalpy of fusion of water is 1.435 kcal/mol. The molar entropy change for the melting of ice at 0 °C is:

1. 10.52 cal/(mol K)

2. 21.04 cal/(mol K)

3. 5.260 cal/(mol K)

4. 0.526 cal/(mol K)

Subtopic:  Enthalpy & Internal energy |
80%
From NCERT
NEET - 2012
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