1

### AIPMT 2001

When 1 mol of gas is heated at constant volume temperature is raised from 298 to 308 K. Heat supplied to the gas is 500 J. Then which statement is correct ?
A
q = w = 500 J, $\Delta$E = 0
B
q = $\Delta$E = 500 J, w = 0
C
q = w = 500 J, $\Delta$E = 0
D
$\Delta$E = 0, q = w = $-$ 500 J

## Explanation

As, $\Delta$E = q + W

Also, W = – p$\Delta$V

As $\Delta$V = 0

So, W = 0

$\Rightarrow$ $\Delta$E = q

Now, q = 500 J

Thus, $\Delta$E = q = 500 J
2

### AIPMT 2001

PbO2  $\to$ PbO; $\Delta$G298 < 0
SnO2 $\to$ SnO;  $\Delta$G298 > 0

Most probable oxidation state of Pb and Sn will be
A
Pb4+, Sn4+
B
Pb4+, Sn2+
C
Pb2+, Sn2+
D
Pb2+, Sn4+

## Explanation

PbO2 → PbO

Pb has +4 oxidation state In PbO2.

Pb has +2 oxidation state In PbO.

Here $\Delta$G is negative that is why reaction is spontaneous and Pb4+ reduces Pb2+ thus, Pb2+ is most probable oxidation state of Pb.

SnO2 $\to$ SnO;

Sn has +4 oxidation state In SnO2.

Sn has +2 oxidation state In SnO.

$\Delta$G is positive that is why reaction is not spontaneous thus, +4 oxidation state of Sn is more stable as it does not change to +2 oxidation state spontaneously.
3

### AIPMT 2001

Change in enthalpy for reaction,

2H2O2(l) $\to$ 2H2O(l) + O2(g)

if heat of formation of H2O2(l) and H2O(l) are $-$188 and - 286 kJ/mol respectively, is
A
$-$ 196 kJ/mol
B
+ 196 kJ/mol
C
+948 kJ/mol
D
$-$ 948 kJ/mol

## Explanation

2H2O2(l) $\to$ 2H2O(l) + O2(g), $\Delta$Hr = ?

H2(g) + O2(g) → H2O2(l), $\Delta$H = – 188 kJ mol ...(1)

H2(g) + O2(g) → H2O(l), $\Delta$H = –286 kJ/mol ....(2)

(1) - (2)

H2O2(l) → H2O(l) + ${1 \over 2}$O2(g)

$\Delta$H = –286 – (– 188) = –98 kJ mol–1

Multiplying this equation by 2 we get the required equation

2H2O2(l) → 2H2O(l) + O2(g)

$\Delta$H = 2 (– 98) kJ/mol = – 196 kJ/mol
4

### AIPMT 2000

For the reaction,
C2H5OH(l) + 3O2(g) $\to$ 2CO2(g) + 3H2O(l)
which one is true
A
$\Delta$H = $\Delta$E $-$RT
B
$\Delta$H = $\Delta$E + RT
C
$\Delta$H = $\Delta$E + 2RT
D
$\Delta$H = $\Delta$E $-$ 2RT

## Explanation

As we know, $\Delta$H = $\Delta$E + $\Delta$ng RT

Now, $\Delta$ng = Number of gaseous moles of products – number of gaseous moles of reactions

= 2 – 3 = – 1

$\Rightarrow$ $\Delta$H = $\Delta$E + (–1) RT

$\Delta$H = $\Delta$E – RT