1

### AIPMT 2003

In Haber process 30 litres of dihydrogen and 30 litres of dinitrogen were taken for reaction which yielded only 50% of the expected product. What will be the composition of gaseous mixture underthe aforesaid condition in the end ?
A
20 litres ammonia, 20 litres nitrogen, 20 litres hydrogen
B
10 litres ammonia, 25 litres nitrogen, 15 litres hydrogen
C
20 litres ammonia, 10 litres nitrogen, 30 litres hydrogen
D
20 litres ammonia, 25 litres nitrogen, 15 litres hydrogen

## Explanation

Balanced chemical equation for Haber's process is as follows :

3H2 + N2 $\to$ 2NH3

It is given that only 50% of the expected product is formed hence only 10 litre of NH3 is formed.

Therefore, composition of gaseous mixture at the end is as follows :

N2 used = 5 litres

N2 left = 30 L – 5 L = 25 L

H2 used = 15 litres,

H2 legt = 30 L – 15 L = 15 L

NH3 = 10 L
2

### AIPMT 2003

Which one of the following statements is not true?
A
Among halide ions, iodide is the most powerful reducing agent.
B
Fluorine is the only halogen that does not show a variable oxidation state.
C
HOCl is a stronger acid than HOBr.
D
HF is a stronger acid than HCl.

## Explanation

F is more electronegative than Cl therefore HF bond is stronger than HCl and hence proton is not given off easily and hence HF is a weakest acid.
3

### AIPMT 2003

The equilibrium constants of the following are

N2 + 3H2 $\rightleftharpoons$ 2NH3;     K1

N2 + O2 $\rightleftharpoons$ 2NO;     K2

H2 + ${1 \over 2}$O2 $\rightleftharpoons$ H2O;     K3

The equilibrium constant (K) of the reaction :

2NH3 + ${5 \over 2}$ O2 $\rightleftharpoons$ 2NO + 3H2O will be
A
K2K33/K1
B
K2K3/K1
C
K23K3/K1
D
K1K33/K2

## Explanation

2NH3 $\rightleftharpoons$ N2 + 3H2;     ${1 \over {{K_1}}}$

N2 + O2 $\rightleftharpoons$ 2NO;     K2

3H2 + ${3 \over 2}$O2 $\rightleftharpoons$ 3H2O;     (K3)3

By adding all equations, we get

2NH3 + ${5 \over 2}$ O2 $\rightleftharpoons$ 2NO + 3H2O

$\therefore$ K = ${{{K_2} \times {{\left( {{K_3}} \right)}^3}} \over {{K_1}}}$
4

### AIPMT 2002

Solution of 0.1 N NH4OH and 0.1 N NH4Cl has pH 9.25. Then find out pKb of NH4OH
A
9.25
B
4.75
C
3.75
D
8.25

## Explanation

NH4OH and NH4Cl constitute to form a basic buffer.

pOH = pKb + log ${{\left[ {Salt} \right]} \over {\left[ {Base} \right]}}$

We know, pOH+ pH = 14 or pOH = 14 – pH

$\therefore$ 14 - pH - $\log {{\left[ {Salt} \right]} \over {\left[ {Base} \right]}}$ = pKb

$\Rightarrow$ 14 - 9.25 - $\log {{0.1} \over {0.1}}$ = pKb

$\Rightarrow$ 14 – 9.25 – 0 = pKb

$\Rightarrow$ pKb = 4.75