1
MCQ (Single Correct Answer)

### 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 = ${{{K_2} \times {{\left( {{K_3}} \right)}^3}} \over {{K_1}}}$
2
MCQ (Single Correct Answer)

### 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
3
MCQ (Single Correct Answer)

### AIPMT 2002

Which has highest pH?
A
CH3COOK
B
Na2CO3
C
NH4Cl
D
NaNO3

## Explanation

Na2CO3 is a salt of weak acid H2CO3 and strong base NaOH, therefore, its aqueous solution will be basic hence has pH more than 7.
4
MCQ (Single Correct Answer)

### AIPMT 2002

Reaction BaO2(g) $\rightleftharpoons$ BaO(s) + O2(g); $\Delta$H = +ve. In equilibrium condition, pressure of O2 depends on
A
increase mass of BaO2
B
increase mass of BaO
C
increase temperature on equilibrium
D
increase mass of BaO2 and BaO both.

## Explanation

For the reaction

BaO2(g) $\rightleftharpoons$ BaO(s) + O2(g); H = +ve.

At equilibrium Kp = PO2

[For solid and liquids concentration term is taken as unity]

Hence, the value of equilibrium constant depends only upon partial pressure of O2. Further on increasing temperature formation of O2 increases as this is an endothermic reaction.

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