1
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
2
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
3
MCQ (Single Correct Answer)

AIPMT 2002

2 mole of ideal gas at 27oC temperature is expanded reversibly from 2 lit. to 20 lit. Find entropy change. (R = 2 cal/mol K)
A
92.1
B
0
C
4
D
9.2

Explanation

For isothermal reversible expansion

w = q = nRT $$ \times $$ 2.303 $$\log {{{v_2}} \over {{v_1}}}$$

= 2RT $$ \times $$ 2.303 $$\log {{{20}} \over {{2}}}$$

= 2 × 2 × T × 2.303 × 1 = 9.2 T

Entropy change, $$\Delta $$S = $${q \over T}$$ = $${{9.2T} \over T}$$ = 9.2 cal/mol K
4
MCQ (Single Correct Answer)

AIPMT 2002

Which reaction is not feasible?
A
2KI + Br2 $$ \to $$ 2KBr + I2
B
2KBr + I2 $$ \to $$ 2KI + Br2
C
2KBr + Cl2 $$ \to $$ 2KCl + Br2
D
2H2O + 2F2 $$ \to $$ 4HF + O2

Explanation

2KBr + I2 $$ \to $$ 2KI + Br2

reaction is not possible because Br ion is not oxidised in Br2 with I2 due to higher electrode (oxidation) potential of I2 than bromine.

EXAM MAP

Joint Entrance Examination

JEE Advanced JEE Main

Medical

NEET

Graduate Aptitude Test in Engineering

GATE CE GATE ECE GATE ME GATE IN GATE EE GATE CSE GATE PI