1

### JEE Main 2019 (Online) 11th January Morning Slot

Consider the reaction
N2(g) + 3H2(g) $\rightleftharpoons$ 2NH3(g)

The equilibrium constant of the above reaction is Kp. If pure ammonia is left to dissociate, the partial pressure of ammonia at equilibrium is given by (Assume that PNH3 << Ptotal at equilibrium)
A
${{{3^{{3 \over 2}}}{K_P^{{1 \over 2}}}{P^2}} \over 4}$
B
${{K_P^{{1 \over 2}}{P^2}} \over 4}$
C
${{{3^{{3 \over 2}}}{K_P^{{1 \over 2}}}{P^2}} \over 16}$
D
${{K_P^{{1 \over 2}}{P^2}} \over 16}$

## Explanation

N2(g) + 3H2(g) $\rightleftharpoons$ 2NH3(g) ; Keq = Kp

Write this equation reverse way,

2NH3(g)  $\rightleftharpoons$ N2(g) + 3H2(g) ; Keq = ${1 \over {{K_p}}}$

2NH3(g) N2(g) + 3H2(g)
At t = 0 Po 0 0
At t = teq PNH3 p 3p

At equillibrium

PTotal = PNH3 + PN2 + PH2

= PNH3 + p + 3p

(As PNH3 << Ptotal so we can ignore PNH3)

$\therefore$ PTotal = 4p

$\Rightarrow$ p = ${{{{P_{total}}} \over 4}}$

Formula of
Keq = ${{{p_{{N_2}}} \times {{\left( {{p_{{H_2}}}} \right)}^3}} \over {{{\left( {{p_{N{H_3}}}} \right)}^2}}}$ = ${1 \over {{K_p}}}$

$\Rightarrow$ ${1 \over {{K_p}}}$ = ${{p \times 27{p^3}} \over {{{\left( {{p_{N{H_3}}}} \right)}^2}}}$

$\Rightarrow$ ${{{\left( {{p_{N{H_3}}}} \right)}^2}}$ = Kp $\times$ 27 $\times$ ${{{\left( {{{{P_{total}}} \over 4}} \right)}^4}}$

$\Rightarrow$ PNH3 = $\sqrt {{K_p}} \times {\left( {27} \right)^{{1 \over 2}}} \times {\left( {{{{P_{Total}}} \over 4}} \right)^{{4 \over 2}}}$

= ${{{3^{{3 \over 2}}}K_p^{{1 \over 2}}P_{Total}^2} \over {16}}$
2

### JEE Main 2019 (Online) 11th January Evening Slot

For the equilibrium,
2H2O $\rightleftharpoons$ H3O+ + OH$-$, the value of $\Delta$Go at 298 K is approximately :
A
$-$ 80 kJ mol–1
B
100 kJ mol$-$1
C
$-$ 100 kJ mol$-$1
D
80 kJ mol–1

## Explanation

2H2O $\rightleftharpoons$ H3O+ + OH$-$

Here Keq = Kw(H2O)

At 298 K, Kw(H2O) = 10-14

$\therefore$ Keq = 10-14

$\Delta$Go = -RTln(Keq)

= -RTln(Kw)

= -2.303RT log (10-14)

= -2.303 $\times$ 8.314 $\times$ 298 $\times$ $\times$ (-14)

= 80 kJ/mol
3

### JEE Main 2019 (Online) 11th January Evening Slot

Given the equilibrium constant:

KC of the reaction :

Cu(s) + 2Ag+ (aq) $\to$ Cu2+ (aq) + 2Ag(s) is

10 $\times$ 1015, calculate the E$_{cell}^0$ of this reaciton at 298 K

[2.303 ${{RT} \over F}$ at 298 K = 0.059V]
A
0.4736 mV
B
0.04736 V
C
0.4736 V
D
0.04736 mV

## Explanation

We know,

$\Delta$Go = -RTln(KC) ....(1)

Also $\Delta$Go = -nF$E_{cell}^o$ ....(2)

$\therefore$ -nF$E_{cell}^o$ = -RTln(KC)

$\Rightarrow$ $E_{cell}^o$ = ${{RT} \over {nF}}\ln \left( {{K_C}} \right)$

= $2.303{{RT} \over {nF}}\log \left( {{K_C}} \right)$

= ${{0.059} \over 2}\log \left( {10 \times {{10}^{15}}} \right)$

( n = no of electron transferred = 2 )

= 0.059 $\times$ ${{16} \over 2}$

= 0.059 $\times$ 8

= 0.472 V
4

### JEE Main 2019 (Online) 12th January Morning Slot

In a chemical reaction, the initial concentration of B was 1.5 times of the concentration of A, but the equilibrium concentrations of A and B were found to be equal. The equilibrium constant (K) for the aforesaid chemical reaction is -
A
16
B
1
C
1/4
D
4

## Explanation At equilibrium [A] = [B]

$\Rightarrow$ a - x = 1.5a - 2x

$\Rightarrow$ x = 0.5a

Kc = ${{{{\left[ C \right]}^2}\left[ D \right]} \over {\left[ A \right]{{\left[ B \right]}^2}}}$

= ${{{{\left( a \right)}^2}\left( {0.5a} \right)} \over {\left( {0.5a} \right){{\left( {0.5a} \right)}^2}}}$ = 4