1
MCQ (Single Correct Answer)

AIPMT 2011 Mains

Consider the following processes :

H (kJ/mol)
1/2A $$ \to $$ B +150
3B  $$ \to $$ 2C + D -125
E + A  $$ \to $$ 2D +350

For B + D $$ \to $$ E + 2C, $$\Delta $$H will be
A
525 kJ/mol
B
$$-$$175 kJ/mol
C
$$-$$ 325 kJ/mol
D
325 kJ/mol

Explanation

H (kJ/mol)
1/2A $$ \to $$ B +150
3B  $$ \to $$ 2C + D -125
E + A  $$ \to $$ 2D +350
___________________________________
   B + D $$ \to $$ E + 2C;

$$\Delta $$H = (300 - 125 - 350) = - 175 kJ/mol
2
MCQ (Single Correct Answer)

AIPMT 2011 Prelims

Which of the following is correct option for free expansion of an ideal gas under adiabatic condition ?
A
q = 0, $$\Delta $$T $$ \ne $$ 0, w = 0
B
q $$ \ne $$ 0,   $$\Delta $$T = 0,   w = 0
C
q = 0,   $$\Delta $$T = 0,   w = 0
D
q = 0,  $$\Delta $$T < 0, w $$ \ne $$ 0

Explanation

For adiabatic condition, change in heat does not take place thus, q = 0.

If q = 0 then change in temperature does not take place thus, $$\Delta $$T = 0.

Also for a free expansion of ideal gas work done, W = 0 as there is no external pressure on it.
3
MCQ (Single Correct Answer)

AIPMT 2011 Prelims

Enthalpy change for the reaction,
4H(g)  $$ \to $$  2H2(g) is $$-$$869.6 kJ
The dissociation energy of H $$-$$ H bond is
A
434.8 kJ
B
$$-$$ 869.6 kJ
C
+ 434.8 kJ
D
+ 217.4 kJ

Explanation

4H(g) → 2H2(g),    ∆H = – 869.6 kJ

Reverse the above equation

2H2(g) → 4H(g),    ∆H = + 869.6 kJ

Divide the above equation by 2,

H2(g) → 2H(g), $$\Delta H = {{869.6} \over 2}$$ kJ = 434.8 kJ
4
MCQ (Single Correct Answer)

AIPMT 2011 Prelims

If the enthalpy change for the transition of liquid water to steam is 30 kJ mol$$-$$1 at 27oC, the entropy change for the process would be
A
10 J mol$$-$$1 K$$-$$1
B
1.0 J mol$$-$$1 K$$-$$1
C
0.1 J mol$$-$$1 K$$-$$1
D
100 J mol$$-$$1 K$$-$$1

Explanation

H2O(l) $$\buildrel {300K} \over \longrightarrow $$ H2O(g)

∆H = 30 kJ mol–1

$$\Delta S = {{\Delta H} \over T}$$ = $${{30 \times {{10}^3}} \over {300}}$$

= 100 J mol–1 K–1

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