1

### JEE Main 2017 (Online) 9th April Morning Slot

The rate of a reaction quadruples when the temperature changes from 300 to 310 K. The activation energy of this reaction is :

(Assume activation energy and preexponential factor are independent of temperature; ln 2 = 0.693; R = 8.314 J mol−1 K−1)
A
107.2 kJ mol$-$1
B
53.6 kJ mol$-$1
C
26.8 kJ mol$-$1
D
214.4 kJ mol$-$1
2

### JEE Main 2018 (Offline)

At 518oC the rate of decomposition of a sample of gaseous acetaldehyde initially at a pressure of 363 Torr, was 1.00 Torr s–1 when 5% had reacted and 0.5 Torr s–1 when 33% had reacted. The order of the reaction is
A
0
B
2
C
3
D
1

## Explanation

For a nth order reaction, the rate of reaction at time t ,

Rate = K [Pt] n

Here Pt = pressure at time t, k = constant.

Note :

Here instead of concentration of product, pressure of product is given.

When 5% is reacted at a rate 1 Toss S$-$1

Then un-reacted is 95%..

As initial pressure is 363 Torr

then after 5% reaction completed the pressure will be

= 363 $\times$ ${{95} \over {100}}$ Torr.

$\therefore\,\,\,$ 1 = K ${\left[ {363 \times {{95} \over {100}}} \right]^n}$ . . . . . . . .(1)

When 33% is reacted at a rate 0.5 , Torr S$-$1 then un-reacted is 67%

So, after 33% reaction completion, the pressure is = $363 \times {{67} \over {100}}$ Torr.

$\therefore\,\,\,$ 0.5 = K ${\left[ {363 \times {{67} \over {100}}} \right]^n}.......$ (2)

Dividing (1) by (2), we get

${1 \over {0.5}} = {{{{\left[ {363 \times {{95} \over {100}}} \right]}^n}} \over {{{\left[ {363 \times {{67} \over {100}}} \right]}^n}}}$

$\Rightarrow \,\,2 = {\left[ {{{95} \over {67}}} \right]^n}$

$\Rightarrow \,\,$ 2 = ${\left[ {1.41} \right]^n}$

$\Rightarrow \,\,$ 2 = ${\left[ {\sqrt 2 } \right]^n}$

$\Rightarrow \,\,\,\,2 = {2^{{n \over 2}}}$

$\therefore\,\,\,$ ${n \over 2}$ = 1

$\Rightarrow \,\,\,\,$ n = 2

This is a 2nd order reaction.
3

### JEE Main 2018 (Online) 15th April Morning Slot

N2O5 decomposes to NO2 and O2 and follows first order kinetics. After 50 minutes, the pressure inside the vessel increases from 50 mmHg to 87.5 mmHg. The pressure of the gaseous mixture after 100 minute at constant temperature will be :
A
175.0 mmHg
B
116.25 mmHg
C
136.25 mmHg
D
106.25 mmHg

## Explanation

N2O5 $\rightleftharpoons$ 2NO2 + ${1 \over 2}$O2
At t = 0 50 0 0
At t = 50 min 50 $-$ P1 2P1 ${{{P_1}} \over 2}$

Total pressure after 50 min,

= 50 $-$ P1 + 2P1 + ${{{P_1}} \over 2}$ = 87.5

$\therefore\,\,\,$ 50 + ${{3{P_1}} \over 2}$ = 87.5

$\Rightarrow$ $\,\,\,$ P1 = 25

So, 50 min is the Half $-$ life period of the reaction.

Then 100 min is two half life

N2O5 $\rightleftharpoons$ 2NO2 + ${1 \over 2}$O2
At t = 100 min 50 $-$ P2 2P2 ${{{P_2}} \over 2}$

$\therefore\,\,\,$ 50 $-$ P2 = ${{25} \over 2}$

P2 = 37.5

$\therefore\,\,\,$ Total pressure at t = 100 min

= 50 $-$ P2 + 2P2 + ${{{P_2}} \over 2}$

= 50 + ${3 \over 2}$ $\times$ 37.5

= 106.25 mm of Hg
4

### JEE Main 2018 (Online) 15th April Evening Slot

For a first order reaction, A $\to$ P, t1/2 (half-life) is 10 days The time required for ${1 \over 4}$th conversion of A (in days) is : (ln 2 = 0.693,    ln 3 = 1.1)
A
5
B
3.2
C
4.1
D
2.5

## Explanation

For first order reaction,

The half life, t${1 \over 2}$ = ${{0.693} \over k}$

Here given t${1 \over 2}$ = 10 days

$\therefore\,\,\,$ k = ${{0.693} \over {10}}$ = 0.0693 days$-$1

Now, the time required for ${1 \over 4}$th conversion of A is ,

t = ${{2.303} \over k}$ log10 $\left( {{a \over {a - x}}} \right)$

= ${{2.303} \over {0.0693}}$ log $\left( {{1 \over {1 - {1 \over 4}}}} \right)$

= ${{2.303} \over {0.0693}}$ log $\left( {{4 \over 3}} \right)$

= 4.1 days.