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What is Half-life of a reaction?

      The half-life of a reaction is stated as the time it takes for one half of a reactant to disappear.

 The half-life is given the symbol t1/2

Where, t denote that it is the time at which the concentration of reactant is one half its initial value. 

t(1/2) = 0.603/k

1/2

1) The half-life of a first-order reaction is 30 minutes. How will you calculate the rate of constant?

A first order reaction is given by :-

ln(Ro/R) = Kt

where,

Ro- Initial amount of substance

R - Amount of substance left

K - Rate constant

& t - is the time elapsed.

Now, talk about half life, Ro = 2R

i.e., 2.303 log(2) = kt

0.693/t = rate constant

0.693/30 =0.0231 min^-1

 = 0.00385 sec^-1 = rate constant

2) Half-life of a first-order reaction is 10 minutes. How much time will it take to complete 75% of the reaction?

Given that, t(1/2) = 10 min.

We know that-

 For 1st order reaction, t1/2 = 0.603/k, where k = rate constant.

Therefore, we get 10 = (0.693)/k

Or, k = 0.0693.

Now, t = (2.303/k) × log([A0]/ [A])

= (2.303/0.0693) × log([A0]/0.25[A0])

= 33.2323 × (log4) = 20 min.

So, 20 min is required to complete 75% of the reaction.

3) The rate constant (K) for a first order reaction was found to be 0.2 sec. What will be its half- life? (a) 10 seconds 5) 5 seconds (c) 2.5 seconds (d) 15 seconds​?

           A first order reaction, A --> B 

We know that- Integrated rate law-

 ln [A] = ln[Ao] - kt 

where k is the rate constant, t is time, [A] is molar concentration of reactant A at some time , t, during the reaction and [Ao] is the initial molar concentration of reactant A.

To derive the formula for half-life just plug in [A] = [Ao/2] and t = t1/2 so we get

ln [Ao/2] = ln[Ao] - kt1/2

ln [Ao] - ln 2 = ln[Ao] - kt1/2

( ln[Ao] drops out, so k is independent of [Ao] for 1st order reactions)

ln 2 = kt(1/2)

ln 2/k = t(1/2)

0.693/k = t(1/2)

so if you plug in k = 0.2 sec^-1 (1st order rate constant has units of inverse seconds), you get

0.693/0.2 = 3.5 sec.

4) The half-life of a reaction of the first order completes in 10 minutes. How much time will be needed for the 80% completion of this reaction?

For a First order Reaction,

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

where, t= time

a= Initial Concentration
(a-x)= Final Concentration
k= 0.693/ t(1/2), here t(1/2)= 10 min.
So k= 0.693/10= 0.0693.

Now t= 2.303/k x log[100-(100–80)] =2.303/0.0693 x (log5)
= 2.303/0.0693 x (0.699)
= 23.23 minutes.

5) The 1st order reaction completes 90% in 50 minutes. What is the half life period?

• The rate constant for a first order Reaction= k=2.303/t x log(a/(a-x)

t=50- min.

a=100
(a-x)= (100- 90)= 10
So k=2.303/50 x log(100/10)
=2.303/50 x 1= 4.606 x 100^-2- min^-1
Therefore half-life period t= 0.693/k

t½= 0.693/(4.606x 10^-2)

 

= 15.05  minutes.

6) The half-life period for a first-order reaction is 15 hour. How much reactant will remain unreacted at the end of 50 hours?

The expression for 1st order decay is:

At=A0e−kt${\mathrm{[A]}}_{\mathrm{t}}=[{\mathrm{A]}}_0{\mathrm{e^}}^{-\mathrm{k}\mathrm{t}}$

k=0.693t1/2$\mathrm{k}=\frac{0.693}{{\mathrm{t(}}_{1/\mathrm{2)}}}$

k=0.69315=0.0462hr−1$\mathrm{k}=\frac{0.693}{15}=0.0462\mathrm{h}{\mathrm{r}}^{-1}$

Taking natural logs of both sides ⇒$\Rightarrow$

lnAt=lnA0−kt$\mathrm{l}\mathrm{n[}{\mathrm{A]t\; = }}_{}\mathrm{l}\mathrm{n[}{\mathrm{A]}}_0-\mathrm{k}\mathrm{t}$

ln[AtA0]=−kt$\mathrm{l}\mathrm{n}[\frac{{\mathrm{A]}}_{\mathrm{t[}}}{{\mathrm{A]}}_0}=-\mathrm{k}\mathrm{t}$

In[AtA0]=−0.0462×50=−2.31$\mathrm{I}\mathrm{n}[\frac{{\mathrm{A]}}_{\mathrm{t[}}}{{\mathrm{A]}}_0}=(-\mathrm{0.0462)}\times 50=-2.31$

AtA0=0.1$\frac{{\mathrm{[A]}}_{\mathrm{t[}}}{{\mathrm{A]}}_0}=0.1$

This is the fraction of unreacted substance remaining which is about 10%.

7) The half-life of first order reactions is 60 minutes. How long will it take to consume 90% of the reactant?

t(1/2)= 60 min

We know tha: k= 0.693/t(1/2)

k= 0.693/60= 0.01155 /min
So, the time at which the Reaction will be 90% complete

=> t= 2.303/k(log 100)/10
=> 2.303/(0.01155)  log(10)
=> 2.303/(0.01155) (1)
=> 199.4  minutes.

8) Show that in a first order reaction time required for completion of 99.9% is 10 times of half life of the reaction?

We know that - Integrated rate law expression is :

K=2.303log(Ao/A)/t$K=2.303log(Ao/A)/t$ where,

K : rate constant,

Ao : Initial Concentration of the substrate,

A : the left over concentration,

t : the time of reaction.

So, when 99.9$99.9$% of reactants have been changed into products; left over concentration equals to (100$100$-99.9$\mathrm{99.9)\%}$=0.1$0.1$%.

Substituing the values in above equation & rearranging it we get:

t=2.303log(100/0.1)/k$t=2.303log(100/0.1)/k$ ............ (1)

Also given ..is that this time equals to 10 times half life of the reaction, so

t=10∗t1/2$\mathrm{t }=\mathrm{10\; (t1}/\mathrm{2)}$ .............. (2)

We know that t1/2=0.693/k$\mathrm{t(}1/\mathrm{2)}=0.693/k$

Equating (1) and (2) we get:

2.303log(100/0.1)/k=10∗0.693/k$2.303log(100/0.1)/k= \mathrm{10\; (}\mathrm{0.693)}/k$

=>6.90=6.93$= \mathrm{6.90\; => }6.93$ (approx. )

 9) A first order reaction is 20% complete in 10 minutes. What is time for the completion of 75% of reaction?

Given that 20% reaction completed in 10 min.

We know that- 

Time  period for 1 St order reactions is:

T = -2.303 log (final concentration/initial concentration)/k

For given reaction, we need to find k

Here, Given that:

Initial concentration A = 100

Final concentration B = 80

& time T = 10 min

Therefore,

k = -2.303 log (B/A) /T

=-2.303 log(80/100) / 10

=0.0223

Therefore k = 0.0223.

For new reaction

Final concentration C = 25 ( because 75 of reaction is finished).

Hence,

Time taken T = -2.303 log(C/A)/k

= -2.303 log(25/100) /0.0223

=62.17 min

Hence, time required to complete 75% of reaction is 62.17 min.