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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

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