1

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

$\underline A \,\,\buildrel {4KOH,{O_2}} \over \longrightarrow \,\,\mathop {2\underline B }\limits_{\left( {Green} \right)} \,\, + \,\,2{H_2}O$

$3\underline B \,\,\buildrel {4HCl} \over \longrightarrow \,\,\mathop {2\underline C }\limits_{\left( {Purple} \right)} \,\, + \,\,Mn{O_2} + 2{H_2}O$

$2\underline C \,\,\buildrel {{H_2}O.KI} \over \longrightarrow \,\,\mathop {2\underline A }\limits_{\left( {Purple} \right)} \,\, + \,\,2KOH\,\, + \,\,\underline D$

In the above sequence of reactions, ${\underline A }$ and ${\underline D }$, respectively, are :
A
Kl and KMnO4
B
Kl and K2MnO4
C
KlO3 and MnO2
D
MnO2 and KlO3

## Explanation

$Mn{O_2}(A) \,\,\buildrel {4KOH,{O_2}} \over \longrightarrow \,\,\mathop {2K_2MnO_4(B) }\limits_{\left( {Green} \right)} \,\, + \,\,2{H_2}O$

$3K_2MnO_4(B) \,\,\buildrel {4HCl} \over \longrightarrow \,\,\mathop {2K_2MnO_4(C) }\limits_{\left( {Purple} \right)} \,\, + \,\,Mn{O_2} + 2{H_2}O$

$2K_2MnO_4(C) \,\,\buildrel {{H_2}O.KI} \over \longrightarrow \,\,\mathop {2 Mn{O_2}(A) }\limits_{\left( {Purple} \right)} \,\, + \,\,2KOH\,\, + \,\,KIO_3(D)$

$\therefore$ A $\to$ MnO2

D $\to$ KIO3
2

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

The reaction 2X $\to$ B is a zeroth order reaction. If the initial concentration of X is 0.2 M, the half-life is 6 h. When the initial concentration of X is 0.5 M, the time required to reach its final concentration of 0.2 M will be:
A
18.0 h
B
9.0 h
C
7.2 h
D
12.0 h

## Explanation

For zero order reaction,

t1/2 = ${{{a_0}} \over {2k}}$

$\Rightarrow$ k = ${{{a_0}} \over {2{t_{1/2}}}}$ = ${{0.2} \over {2 \times 6}}$ = 1.67 $\times$ 10-2 mol L–1h–1

For zero order reaction,

A0 - At = kt

$\Rightarrow$ 0.5 - 0.2 = 1.67 $\times$ 10-2 t

$\Rightarrow$ t = ${{0.3} \over {1.67 \times {{10}^{ - 2}}}}$ = 18 h
3

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

Decomposition of X exhibits a rate constant of 0.05 $\mu$g/year. How many year are required for the decomposition of 5$\mu$g of X into 2.5 $\mu$g?
A
50
B
20
C
25
D
40

## Explanation

According to unit of rate constant it is a zero order reaction, then half life of zero order reaction.

t1/2 = ${{{a_0}} \over {2k}} = {5 \over {2 \times 0.05}}$ = 50 years
4

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

For a reaction consider the plot of $\ell$n k versus 1/T given in the figure. If the rate constant of this reaction at 400 K is 10–5 s–1, then the rate constant at 500 K is –

A
10$-$4 s$-$1
B
4 $\times$ 10$-$4 s$-$1
C
10$-$6 s$-$1
D
2 $\times$ 10$-$4 s$-$1

## Explanation

From Arhenius equation,

K = Ae$^{ - {{Ea} \over {RT}}}$

taking log both sides,

lnk = lnA $-$ ${{{Ea} \over {RT}}}$

So in lnK vs ${1 \over T}$ graph

slope = $-$ ${{{Ea} \over R}}$

Rate constant at 400 K is = 10$-$5 s$-$1

Let rate constant at 500 K is = K

$\therefore$   ln(10$-$5) = ln A $-$ ${{Ea} \over {R\left( {400} \right)}}$ . . . (1)

ln K = ln A $-$ ${{Ea} \over {R\left( {500} \right)}}$ . . . .(2)

Subtracting (2) from (1) we get,

ln ${K \over {{{10}^{ - 5}}}}$ = ${{{Ea} \over R}}$ $\left[ {{1 \over {400}} - {1 \over {500}}} \right]$

$\Rightarrow$   2.303 log ${K \over {{{10}^{ - 5}}}}$ = 4606 $\left[ {{1 \over {400}} - {1 \over {500}}} \right]$

$\Rightarrow$   K = 10$-$4 s$-$1

NEET