1

### JEE Main 2018 (Online) 16th April Morning Slot

The gas phase reaction 2NO2(g) $\to$ N2O4(g) is an exothermic reaction. The decomposition of N2O4, in equilibrium mixture of NO2(g) and N2O4(g), can be increased by :
A
lowering the temperature.
B
increasing the pressure.
C
addition of an inert gas at constant volume.
D
addition of an inert gas at constant pressure.

## Explanation

2NO2 (g) $\buildrel \, \over \longrightarrow$ N2O4 (g); $\Delta$H = $-$ Ve

At equilibrium, N2O4 $\rightleftharpoons$ 2NO2 ; $\Delta$H = + Ve

Decomposition N2O4 is endothermic

Increasing the pressure on a gas reaction shifts the position of equilibrium towards the side with fewer molecules.

So, it will move towards backward direction which lead to formation of N2O4 from NO2.
2

### JEE Main 2019 (Online) 9th January Morning Slot

20 mL of 0.1 M H2SO4 solution is added to 30 mL of of 0.2 M NH4OH solution. The pH of the resultant mixture is : [pkb of NH4OH = 4.7].
A
5.2
B
9.0
C
5.0
D
9.4

## Explanation

H2SO4 + 2NH4OH $\to$ (NH4)2SO4 + H2O

Initially,

H2SO4 present = 20 $\times$ 0.1 $\times$ 2 = 4 miliequivalent

NH4OH present = 30 $\times$ 0.2 = 6 miliequivalent

Here H2SO4 is the limiting reagent,

So, finally. H2SO4 present = 0

and NH4OH present = (6 $-$ 4) = 2

and (NH4)2SO4 produced = 4 miliequivalent.

As in the solution there is (NH4)2 SO4 present so it a basic buffer.

$\therefore$   POH = PKb + log ${{\left[ {Salt} \right]} \over {\left[ {base} \right]}}$

= 4.7 + log ${4 \over 2}$

= 4.7 + log2

= 4.7 + 0.3

= 5

$\therefore$   PH = 14 $-$ POH

= 14 $-$ 5

= 9
3

### JEE Main 2019 (Online) 9th January Evening Slot

The pH of rain water, is approximately :
A
5.6
B
7.5
C
7.0
D
6.5

## Explanation

pH of rain water is approximate 5.6
4

### JEE Main 2019 (Online) 9th January Evening Slot

If the standard electrode potential for a cell is 2 V at 300 K, the equilibrium constant (K) for the reaction
Zn(s) + Cu2+ (aq) $\rightleftharpoons$ Zn2+(aq) + Cu(s)
at 300 K is approximately
(R = 8 JK$-$1mol$-$1, F = 96000 C mol$-$1)
A
e$-$80
B
e$-$160
C
e320
D
e160

## Explanation

$\Delta$Go = $-$ RT lnk = $-$nFEocell

lnk = ${{n \times F \times {E^o}} \over {R \times T}} = {{2 \times 96000 \times 2} \over {8 \times 300}}$

lnk = 160

k = e160