1

### 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

### 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

### 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

### 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.