The difference in energy of activation for uncatalysed reaction and catalysed reaction is 20 kJ/mol. How many times the rate constant of catalysed reaction will increase over the uncatalysed reaction?

we have to count the difference between uncatalysed and catalysed reaction.
thus we know that,

k1 = Ae-Ea/RTk2 = Ae-Ea(c)/RT

logk2k1 = Ea-Ea(c)2.303 RTlog k2k1 = 20,00002.303×8.314×300=3.48k2k1 = 3020

Thus, the reaction rate has increased nearly by 3020.
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A first order reaction is 15% complete in 20 minutes. How long will it take to be 60% completed ?

For the first order reaction t= 2.303klog aa-xlet, a =100initally reaction is 15% completed, so x= 15, t=20 min so,  thus applying in the above equation20= 2.303klog 100100-1520 =2.303klog 1008520 =2.303klog 1.176420 =2.303k0.0706k= 0.00813

to completed 60% reaction 

For the first order reaction t= 2.303klog aa-xlet, a =100initally reaction is 15% completed, so x= 60, k= 0.00813so,  thus applying in the above equationt= 2.3030.00813log 100100-60t=2.3030.00813log 10040t =2.3030.00813log 2.5t=2.3030.00813×0.3979k= 283.27  x0.3979t =112.71 min

Ans. 112.7 min

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Show that the time required for the completion of 3/4th reaction of first order is twice the time required for completion of 1/2 of the reaction.


Since it is first order, thus

t= 2.303klog aa-xalso x =3a/4so the equation is t3/4= 2.303klog aa-3a/4.....1for x=1a/2t1/2= 2.303klog aa-a/2......2divide the 1 by 2, we gett3/4t1/2= 2.303klog aa-3a/42.303klog aa-a/2thus we gett3/4t1/2= log4log2t3/4t1/2= 2log2log2t3/4t1/2= 2 t3/4= 2t1/2

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75% of a reaction of the first order was completed in 32 minutes. When was its half-life completed?

The first order reaction can be 
t= 2.303klogaa-xwe have given t= 32let a= 100x=75thus putting the value in above equation 32= 2.303klog100100-7532= 2.303klog1002532= 2.303klog4k= 2.30332log4k =0.07196 log 4k= 0.07196  x 0.6020k= 0.0433half - time reaction of first order can be given by t1/2 =0.693kt1/2 =0.6930.0433 = 16

Ans. 16 minutes

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If half-life of a first order reaction involving reactant A is 5 min. How long will it take [A] to react 25% of its initial concentration?

Half -life of first order reaction can be given by 
t1/2 =0.693/k

for first order reaction :
 t= 2.303klog aa-x

let a=100
x= 25 (given)
time =5min

applying in above equation, we get

5 =2.303klog100100-255 =2.303klog10075k=2.3035log43 k=0.4606 log 1.33k=0.4606 x 0.1238k= 0.05704

thus for half life

t1/2 =0.693/0.05704

Ans. 12.14 min.

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