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1

### NEET 2022

The given graph is a representation of kinetics of a reaction The y and x axes for zero and first order reactions, respectively are

A
zero order (y = concentration and x = time), first order (y = t1/2 and x = concentration)
B
zero order (y = concentration and x = time), first order (y = rate constant and x = concentration)
C
zero order (y = rate and x = concentration), first order (y = t1/2 and x = concentration)
D
zero order (y = rate and x = concentration), first order (y = rate and x = t1/2)

## Explanation

$$\bullet$$ For zero order reaction

r = k[A]0

r = k (constant)

hence, 'y' as 'rate' and 'x' as concentration will give desired graph.

$$\bullet$$ For first order reaction

$${t_{1/2}} = {{0.693} \over k}$$ (constant)

hence, 'y' as 't1/2' and 'x' as concentration will given desired graph.

2

### NEET 2021

The slope of Arrhenius plot $$\left( {\ln \,k\,v/s{1 \over T}} \right)$$ of first order reactino is $$-$$5 $$\times$$ 103K. The value of Ea of the reaction is. Choose the correct option for your answer.

[Given R = 8.314 JK$$-$$1mol$$-$$1]
A
$$-$$83 kJ mol$$-$$1
B
41.5 kJ mol$$-$$1
C
83.0 kJ mol$$-$$1
D
166 kJ mol$$-$$1

## Explanation

Arrhenius equation

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

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

$$\ln k = \ln A - {{{E_a}} \over R}\left( {{1 \over T}} \right)$$ .... (1)

Slope of $$\ln k\,$$ vs $${1 \over T}$$ curve,

$$m = - {{{E_a}} \over R}$$

$$- 5 \times {10^3} = - {{{E_a}} \over R}$$

Ea = 5 $$\times$$ 103 $$\times$$ 8.314 J/mol

= 41.57 $$\times$$ 103 J/mol

$$\simeq$$ 41.5 kJ/mol
3

### NEET 2021

For a reaction, A $$\to$$ B, enthalpy of reaction is $$-$$4.2 kJ mol$$-$$1 and enthalpy of activation is 9.6 kJ mol$$-$$1. The correct potential energy profile for the reaction is shown in option.
A B C D ## Explanation

The enthalpy of reaction is negative, – 4.2 kJ mol−1 . The reaction is an exothermic reaction i.e. the energy of product B, is less than the energy of reactant A. So, the potential energy profile for the reaction is 4

### NEET 2020 Phase 1

The rate constant for a first order reaction is 4.606 $$\times$$ 10-3 s-1. The time required to reduce 2.0 g of the reactant to 0.2 g is :
A
200 s
B
500 s
C
1000 s
D
100 s

## Explanation

$$k = \left( {{{2.303} \over t}} \right)$$ log$$\left( {{{{A_0}} \over A}} \right)$$ for first order reaction

$$4.606 \times {10^{ - 3}} = \left( {{{2.303} \over t}} \right)$$ log$$\left( {{2 \over {0.2}}} \right)$$

$$\Rightarrow$$ t = 500 s

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Class 12