1
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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

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