In the Arrhenius equation for certain reaction, the values of the frequency factor and energy of activation are 4 x 1013 sec–1 and 98.6 kJ/mol respectively. If the reaction is of first order, at what temperature will its half-life be 10 minutes.

We have given that

A = 4 × 1013 s-1Ea = 98.6 kJ/mol = 98600 J/molt1/2 = 10 min.


                        k=0.69310=0.0693 min-1 = 0.069360k = 0.001155 s-1

We  know that     log k = log A-Ea2.303 R+1
             Ea2.303 R.1T = logA-logB                         = log 4 × 1013-log 0.00155                        = 13.602 + 2.937   Ea2.303 R.1T = 16.539

or
                        T = 986002.303 × 8.314 × 16.539 = 311.4 K
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Consider the following data for the reaction A + B → Products.

Run

Initial

Initial

Initial rate (mol s–1)

 

concentration

concentration

 

1

0.10 M

0.1 M

2.1 x 10–3

2

0.20 M

1.0 M

8.4 x 10–3

3

0.20 M

2.0 M

8.4 x 10–3

Determine the order of reaction with respect to A and with respect to B and the overall order of reaction.


The rate law may be expressed as

                     Rate = kAp Bq

Comparing experiments 2 and 3
                    (Rate)2 = k0.2p [1.0]q = 8.4×10-3                        ...(i)(Rate)3 = k[0.2]p 2.0q = 8.4 × 10-3                      ...(ii)

Dividing eqn. by (ii) by (i),
                         (Rate)3(Rate)2 = k0.2p 2.0qk0.2p 1.0q = 8.4 × 10-38.4 × 10-3
                             
  2q = 20    or  q = 0.

Comparing experiments (i) and (ii)
                    (Rate)2 = k0.20p 1.0q =8.4 × 10-3                   ...(iii)(Rate)1 = k0.10p 1.0q = 2.1 × 10-3                 ...(iv)


Dividing eqn. (iii) by (iv),
 

(Rate)2(Rate)1 = k0.20p 1.0qk[0.10]p [1.0]q = 8.4 × 10-32.7 × 10-3 = 4     2q = 22    or   q = 2.

Order with respect to A = 2.
Order with respet to B = 0.
Overall order of reaction = 2.

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The reaction:
CH3COF + H2O  CH3COOH+HF

has been studied under the following initial conditions.

CASE I:                                                 CASE II:CH2O = 1.00 M                                        CH2O = 0.02 M.CCH3COF = 0.01 M                                 CCH3COF = 0.80 M                    
Concentrations were monitored as a function of time and are given below:

Determine the order of the reaction and the rate constant for the reaction.


Let rate k(CCH3COF)(CH2O)b
CH2O >> CCH3COF and in the second case, CH2O << CCH3COF. In the first case we determine the order of the reaction with respect to CH3 COF. We note that the reaction is not of zero order as rate of reaction changes with time.

Let rate k(CCH3COF)a (CH2O)bCH2O >> CCH3COF and in the secon

Therefore k(CH2O) = 0.0154 min–1 and we note the order of reaction with respect to CH3COF is 1.
Now we determine the order of raction with respect to water.
Again,

Let rate k(CCH3COF)a (CH2O)bCH2O >> CCH3COF and in the secon
The reaction is first order in H2O and we have

Let rate k(CCH3COF)a (CH2O)bCH2O >> CCH3COF and in the secon

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The rate constant of a reaction is 1.2 x 10–3 sec–3 at 30°C and 2.1 x 10–3 sec–1 at 40°C. Calculate the energy of activation of the reaction.

We have given that

   k1 = 1.2 × 10-3 sec-1T1 = 30+273 = 303 Kk2 = 2.1 × 10-3 sec-1T2 = 40+273 = 313

Substituting these values in the equation

   
                      logk2k1 = Ea2.303T2-T1T1T2

we get
         log 2.1 × 10-31.2 × 10-3 = Ea2.303 × 8.314313-303303 × 313
 
           or

log 2.11.2 = Ea2.303 × 8.314×10303×313


                     Ea = 44.13 kJ mol-1.

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The rate of chemical reaction becomes double for every 10° rise in temperature because of
  • decreases in activation energy
  • increase in activation energy
  • increase in number of activated molecules
  • increase in number of molecular collisions

D.

increase in number of molecular collisions
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