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1

### 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
2

### 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
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) 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

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