1

### AIPMT 2012 Prelims

Standard enthalpy of vaporisation $\Delta$vapHo for water at 100oC is 40.66 kJ mol$-$1. The internal energy of vaporisation of water at 100oC (in kJ mol$-$1) is
A
+37.56
B
$-$43.76
C
+ 43.76
D
+ 40.66

## Explanation

H2O(l) $\buildrel {100^\circ \,C} \over \longrightarrow$ H2O(g)

∆Ho = 40.66kJ mol–1

∆Ho = ∆uo + $\Delta$ng RT

$\Delta$ng = 1, R = 8.314 × 10–3 kJ mol–1 k–1

T = 100 + 273 = 373 K

$\Rightarrow$ 40.66 = ∆uo + (1) (8.314 × 10–3) × 373

$\Rightarrow$ ∆uo = 37.56 kJ mol–1
2

### AIPMT 2012 Prelims

In which of the following reactions, standard reaction entropy change ($\Delta$So) is positive and standard Gibb's energy change ($\Delta$Go) decreases sharply with increasing temperature ?
A
C(graphite) + ${1 \over 2}$O2(g) $\to$ CO(g)
B
CO(g) + ${1 \over 2}$O2(g) $\to$ CO2(g)
C
Mg(s) + ${1 \over 2}$O2(g) $\to$ MgO(g)
D
${1 \over 2}$C(graphite) + ${1 \over 2}$O2(g) $\to$ ${1 \over 2}$CO2(g)

## Explanation

$\Delta$So positive means entropy of products is more than that of reactants. Among the given reactions only in option (a) the $\Delta$So is positive cause products has 1 mole of gaseous products CO while reaction have half mole of gaseous O2 and Carbon (C) in solid state. Thus, reactions have less randomness than products. Also, the reaction is a combustion reaction which has a negative value of $\Delta$H. According to Gibbs energy change ($\Delta$So).

$\Delta$Go = $\Delta$Ho - T$\Delta$So

$\Rightarrow$ $\Delta$Go = -ve - T(+ve)

Thus, on increasing temperature value of $\Delta$Go decreases sharply.
3

### 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
4

### 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.