1

### AIPMT 2001

Enthalpy of CH4 + ${1 \over 2}$ O2 $\to$ CH3OH is negative. If enthalpy of combustion of CH4 and CH3OH are x and y respectively. Then which relation is correct?
A
x > y
B
x < y
C
x = y
D
x3 y

## Explanation

CH4 + ${1 \over 2}$ O2 $\to$ CH3OH

Given that for this reaction, $\Delta$Hr = -ve

CH4 + 2O2 → CO2 + 2H2O,    $\Delta$H = x ....(1)

CH3OH + ${3 \over 2}$ O2 → CO2 + 2H2O,    $\Delta$H = y ....(2)

Eqn. (1) – Eqn. (2)

CH4 + ${1 \over 2}$ O2 $\to$ CH3OH

$\Delta$Hr = x - y = -ve

$\therefore$ x < y
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

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
4

### 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