NEET Questions Solved


The plot of log k vs 1/T helps to calculate 

1. Energy of activation

2. Rate constant of reaction 

3. Order of reaction

4. Energy of activation as well as the frequency factor

(4) logK= logA - Ea/2.303RT

comparing with y = mx+c

log A = intercept

-Ea/2.303RT = slope

Ea = slope x (-2.303R)

So, Ea (energy of activation ) and A (Frequancy factor) can be calculated from above plot.

Difficulty Level:

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Two reactions of the same order have equal Pre-exponential factors but their activation energies differ by 24.9 kJ/mol. Calculate the ratio between the rate constants (K2/K1) these reactions at 27°C :

(1) 3 x 104

(2) 1/3 x 10-4

(3) 1/3 x 104

(4) 3 x 10-4

(1) log10K1 = log10A - Ea1/2.303RT   ----(1)

     log10K2 = log10A - Ea2/2.303RT   ----(2)

     log10(K2/K1) = (Ea1 - Ea2)/2.303RT

     = log10(K2/K1) = (24.9x1000) / (2.303x8.314x300)

     = K2/K1 = 2.99x104 = 3 x 104

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NEET - 2017

A first order reaction has a specific reaction rate of 10-2s-1. How much time will it take for 20 g of the reactant to reduce to 5 g?

(a) 238.6 s      (b) 138.6 s        (c) 346.5  s        (d) 693.0 s

(b) For a first order reaction,

Rate constant (k) = 2.303/t . log(a/a-x)

where, a = initial concentration

a-x = concentration  after  time 't'

t= time in 'sec'

Given, a= 20 g, a-x = 5g, k=10-2

... t =2.303/10-2 . log(20/5) = 138.6 s

Alternatively,

Half-life for the first order reaction,

    t1/2/2 = 0.693/k = 0.693/10-2 = 69.3s

Two half-lives  are required for the reduction of 20 g of reactant into 5g

    20g t1/210gt1/2

 

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NEET - 2016

The decomposition of phosphine (PH3) on tungsten at low pressure is a first-order reaction. It is because the

(a) rate is proportional to the surface coverage

(b) rate is inversely proportional to the surface coverage

(c) rate is independent of the surface coverage

(d) rate of decomposition is very slow

(a) PH3 wP +3/2. H2

This is an example of surface catalysed unimolecular decomposition.

For the above reaction, rate is given as

 Rate = kαρ/1+αρ

where, ρ = partial pressure of absorbing substrate.

At low pressure, αρ

 

 

Difficulty Level:

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NEET - 2016

The rate of a first-order reaction is 0.04 mol L-1 s-1 at 10 sec and 0.03 mol L-1 s-1 at 20 sec after initiation of the reaction. The half-life period of the reaction is

(a) 34.1 s (b) 44.1 s (c) 54.1 s (d) 24.1 s

(d) Given, order of reaction = 1
Rate of reaction at 10 s = 0.04 mol L-1 s-1
Rate of reaction at 20 s = 0.03 mol L-1s-1

Half-lite period (t1/2) = ?
We have the equation for rate-constant 'k' in first
order reaction.

                      k=2.303tlogAtA0  =2.303105log0.040.03  = 2.303105×0.124k= 0.028 s-1 

Difficulty Level:

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NEET - 2013

What is the activation energy for a reaction if its rate doubles when the temperature is raised from 20°C to 35°C?  (R=8.314 J mol-1 K-1)

(a) 342 kJ mol-1

(b) 269 kJ mol-1

(c) 34.7 kJ mol-1

(d) 15.1 kJ mol-1

(c) Given, initial temperature,
T1=20+273=293K
Final temperature
T2=35+273=308K

R=8.314 J mol-1 K-1

Since, rate becomes double on raising temperature,
∴ r2=2r1 or r2/r1=2

As rate constant, k∝r
∴ k2/k1=2

From Arrnhenius equation, we know that

log k2/k1= -Ea______ [T1-T2/T1T2]
                 2.303R

log 2= -Ea______         [293-308/293x308]
            2.303x8.314

0.3010=   -Ea______         [-15/293x308]
            2.303x8.314

∴ Ea=0.3010x2.303x8.314x293x308
                           15

=34673.48 J mol-1=34.7 kJ mol-1



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NEET - 2007

If 60% of a first order reaction was completed in 60 min, 50% of the same reaction would be completed in approximately.

(a) 50 min
(b) 45 min
(c) 60 min
(d) 40 min

(log 4=0.60, log 5=0.69)

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BOARD

The rate of  a reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (Ea) of the reaction assuming that it does not change with temperature. [R=8.314 J K-1 mol -1, log =0.6021]

We know, log K2K1=Ea2.303RT2-T1T1T2

      Given : T1 = 293 K,     T2 = 313 K,

      Let K1 = K(consider)  K2 = 4K,

          R = 8.314,     E= ?

As,    log4KK=Ea2.303×8.314313-293293×313 

  or  log 4 =Ea19.1471×2091709

  or  0.6020 =Ea19.1471×2091709

 Ea0.6020×19.1471×9170920

        = 52854.43 = 52.8 K J mol-1

Difficulty Level:

BOARD

The following data were obtained during the first order thermal decomposition of SO2Cl2 at a constant volume :

SO2Cl2(g) SO2 (g) + Cl2(g)

Experiment           time/s-1              Total pressure/atm

   1                        0                          0.4

   2                     100                          0.7

Calculate the rate constant.

(Given :log 4 = 0.6021, log 2 = 0.3010)

a) 3.386× 10-2 s-1

b) 2.386× 10-2 s-1

c) 1.986× 10-2 s-1

d)1.386× 10-2 s-1

SO2Cl2(g) SO2 (g) + Cl2(g)

Using formula, K = 2.303tlog P2P- Pt

When t=100 s

     K = 2.303100log 0.42×0.4- 0.7

         = 2.303100log 0.40.8- 0.7  = 2.303100log 0.40.1

          = 2.303100log 41 = 2.303100log 4

           = 2.303100×0.6021  = 1.386× 10-2 s-1

Difficulty Level:

BOARD

Hydrogen peroxide, H2O2 (aq) decomposes to H2O(l) and O2(g) in a reaction that is first order in H2O2 and has a rate constant k = 1.06×10-3 min-1.

(i) How long will it take for 15% of a sample of H2O2  to decompose?

(ii) How long will it take for 85% of a sample  to decompose?

Given : k = 1.06×10-3min-1

For first order reaction

  k =2.303tlog aa-x where [ x= 15100a=0.15a ]

or    t=2.303klog aa-x 

or    t=2.3031.06×10-3log aa-0.15a

       t= 2172.64 log 1.1764

 Time (t) =2172.64 ×0.07055

               = 153.29 min

 

(ii)   t=2.303klog aa-x            [for 85% completion]

       t=2.3031.06×10-3log aa-0.85a

   t= 2172.64 log 6.66

 Time (t) =2172.64 ×0.8234

               = 1789.11 min

    

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