1
MCQ (Single Correct Answer)

AIPMT 2003

What is the entropy change (in J K$$-$$1 mol$$-$$1) when one mole of ice is converted into water at 0oC? (The enthalpy change for the conversion of ice to liquid water is 6.0 kJ mol$$-$$1 at 0oC).
A
20.13
B
2.013
C
2.198
D
21.98

Explanation

H2O(s) → H2O(l)

$$\Delta $$H = 6.0 kJ mol–1

T = 0 + 273 K = 273 K; $$\Delta $$S = ?

$$\Delta $$S = $${{\Delta H} \over T}$$ = $${{6000} \over {273}}$$ = 21.98 JK–1 mol–1
2
MCQ (Single Correct Answer)

AIPMT 2003

Formation of solution from two components can be considered as
(i) Pure solvent $$ \to $$ separated solvent molecules, $$\Delta $$H1
(ii) Pure solute $$ \to $$ separated solute molecules, $$\Delta $$H2
(iii) Separated solvent and solute molecules $$ \to $$ solution, $$\Delta $$H3
Solution so formed will be ideal if
A
$$\Delta $$Hsoln = $$\Delta $$H1 + $$\Delta $$H2 + $$\Delta $$H3
B
$$\Delta $$Hsoln = $$\Delta $$H1 + $$\Delta $$H2 $$-$$ $$\Delta $$H3
C
$$\Delta $$Hsoln = $$\Delta $$H1 $$-$$ $$\Delta $$H2 $$-$$ $$\Delta $$H3
D
$$\Delta $$Hsoln = $$\Delta $$H3 $$-$$ $$\Delta $$H1 $$-$$ $$\Delta $$H2

Explanation

For ideal solution,

$$\Delta $$Hsoln = $$\Delta $$H1 + $$\Delta $$H2 + $$\Delta $$H3
3
MCQ (Single Correct Answer)

AIPMT 2003

The densities of graphite and diamond at 298 K are 2.25 and 3.31 g cm$$-$$3, respectively. If the standard free energy difference $$\left( {\Delta {G^o}} \right)$$ is equal to 1895 J mol$$-$$1, the pressure at which graphite will be transformed into diamond at 298 K is
A
11.14 $$ \times $$ 108 Pa
B
11.14 $$ \times $$ 107 Pa
C
11.14 $$ \times $$ 106 Pa
D
11.14 $$ \times $$ 105 Pa

Explanation

C(graphite) $$ \to $$ C(diamond)

Volume of graphite = $${{12} \over {2.25}} = 5.33$$ cm3 mol–1

Volume of diamond = $${{12} \over {3.31}} = 3.63$$ cm3 mol–1

$$\Delta $$V = Vgraphite – Vdiamond

= 1.70 cm3 mol–1

= 1.70 × 10–3 L mol–1

So, $$\Delta $$Go = P$$\Delta $$V

$$ \Rightarrow $$ 1895 J mol–1 = P(1.70 × 10–3 L mol–1)

$$ \Rightarrow $$ 1114.70 × 103 J/L = P

$$ \Rightarrow $$ $${{1114.7 \times {{10}^3}} \over {101.33}}$$ = P

$$ \Rightarrow $$ P = 11000.69 atm × 1.013 × 105 Pa

= 11143.69 × 105 Pa

= 11.14 × 108 Pa
4
MCQ (Single Correct Answer)

AIPMT 2003

For the reaction,
C3H8(g) + 5O2(g) $$ \to $$ 3CO2(g) + 4H2O(l)
at constant temperature, $$\Delta $$H $$-$$ $$\Delta $$E is
A
+ RT
B
$$-$$3RT
C
+ 3RT
D
$$-$$ RT

Explanation

$$\Delta $$H = $$\Delta $$E + $$\Delta $$ng RT

$$\Delta $$ng = 3 – (1 + 5) = – 3

$$ \Rightarrow $$ $$\Delta $$H = $$\Delta $$E + (– 3RT)

$$ \Rightarrow $$ $$\Delta $$H – $$\Delta $$E = – 3RT

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