1

### 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

### 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

### 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

### 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