1

### JEE Main 2019 (Online) 10th January Morning Slot

Consider the given plots for a reaction obeying Arrhenius equation (0oC < T < 300oC) : (K and Ea are rate constant and activation energy, respectively)

Choose the correct option :
A
I is right but II is wrong
B
Both I and II are correct
C
Both I and II are wrong
D
I is wrong but II is right

## Explanation

Arrhenius Equation -

Arrhenius gave the quantitative dependence of rate constant on temperature by the Arrhenius equation.

- wherein

$k = A{e^{ - {E_a}/RT}}$

${\mathop{\rm lnk}\nolimits} = lnA - {{{E_a}} \over {RT}}$

k = Rate constant

as we have learned in Arrhenius equation when we increase Ea, K should decrease As T increases, power of exponential increases so k also increase.

Hence, both the plots are correct.
2

### JEE Main 2019 (Online) 10th January Evening Slot

For an elementary chemical reaction,

the expression for ${{d\left[ A \right]} \over {dt}}$ is
A
2K1[A2] – K –1 [A]2
B
K1[A2] – K –1 [A]2
C
K1[A2] + K –1 [A]2
D
2K1[A2] – 2K –1 [A]2

## Explanation

3

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

If a reaction follows the Arrhenius equation, the plot ln k vs ${1 \over {\left( {RT} \right)}}$ gives straight line with a gradient ($-$ y) unit.

The energy required to active the reactant is :
A
y unit
B
y/R unit
C
yR unit
D
$-$y unit

## Explanation

According to Arrhenius equation,
k = A${e^{ - {{{E_a}} \over {RT}}}}$

or ln k = ln A - ${{{{E_a}} \over {RT}}}$

Comparing the above equation with straight line equation,

y = mx + c,

we get, slope or gradient (m) = –Ea
and Intercept (c) = ln A

Also given that slope or gradient (m) = -y

$\therefore$ -y = –Ea

$\Rightarrow$ Ea = y

So the activation energy of the reactant, Ea = y unit
4

### 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