1

### AIPMT 2005

The absolute enthalpy of neutralisation of the reaction :

Mg(O)(s) + 2HCl(aq) $\to$ MgCl2(aq) + H2O(l) will be
A
$-$57.33 kJ mol$-$1
B
greater than $-$ 57.33 kJ mol$-$1
C
less than $-$ 57.33 kJ mol$-$1
D
57.33 kJ mol$-$1

## Explanation

We know that enthalpy of neutralization of a strong acid and a strong base is –57.33 kJ mol–1.

Here, MgO is the weak base and HCl is a strong acid thus, a small amount of energy is used in the ionization of MgO thus, the heat of neutralization decreases. Therefore, enthalpy of neutralization is less than – 57.33 kJ mol–1
2

### AIPMT 2005

A reaction occurs spontaneously if
A
T$\Delta$S < $\Delta$H and both $\Delta$H and $\Delta$S are +ve
B
T$\Delta$S > $\Delta$H and $\Delta$H is +ve and $\Delta$S are $-$ve
C
T$\Delta$S > $\Delta$H and both $\Delta$H and $\Delta$S are +ve
D
T$\Delta$S = $\Delta$H and both $\Delta$H and $\Delta$S are +ve

## Explanation

$\Delta$G = $\Delta$H – T$\Delta$S

For spontaneous reaction, $\Delta$G has to be negative.

Among the given options, it is positive only when T$\Delta$S > $\Delta$H and both $\Delta$H and $\Delta$S are +ve .
3

### AIPMT 2005

Which of the following pairs of a chemical reaction is certain to result in a spontaneous reaction ?
A
Exothermic and increasing disorder
B
Exothermic and decreasing disorder
C
Endothermic and increasing disorder
D
Endothermic and decreasing disorder

## Explanation

$\Delta$G = $\Delta$H – T$\Delta$S

For a reaction to become spontaneous, $\Delta$G must be negative.

For case (a) exothermic and increasing disorder

For exothermic reaction,

$\Delta$H = –ve and increasing disorder,

$\Delta$S = +ve

$\Delta$G = –ve –T(+ve)

Thus, $\Delta$G is negative for all temperature range

For case (b) exothermic and decreasing disorder For exothermic reaction, $\Delta$H = –ve and for decreasing disorder,

$\Delta$S = –ve

$\Rightarrow$ $\Delta$G = –ve – T(–ve)

Thus, $\Delta$G is not negative for all temperature range.

For case (c) endothermic and increasing disorder. For endothermic reaction, $\Delta$H = +ve and increasing disorder,

$\Delta$S = +ve

Thus, $\Delta$G is not negative for all temperature range.

For case (d) endothermic and decreasing disorder

For endothermic reaction,

$\Delta$H = +ve and decreasing disorder,

$\Delta$S = –ve

$\Delta$G = +ve – T (–ve)

Thus, $\Delta$G is positive for all temperature range.
4

### AIPMT 2004

The work done during the expansion of a gas from a volume of 4 dm3 to 6 dm3 against a constant external pressure of 3 atm is (1 L atm = 101.32 J)
A
$-$ 6 J
B
$-$ 608 J
C
+ 304 J
D
$-$ 304 J

## Explanation

Work done during the expansion, W = – pdV

W = –3 atm (6 dm3 – 4 dm3)

= – 3 atm ( 2 dm3 ) (1 dm3 = 1 L)

= – 3 atm × 2 L

= – 6 L atm

As, 1 L atm = 101.32 J

$\therefore$ W = – 6 × 101.32 J = – 607.92 J ≈ – 608 J