# a first order reaction is 20% complete in 10 minutes. calculate the time required for the completion of 75% of the reaction.

k = log

So, when 20% reaction is completed, then 80% reactants are left unreacted.

And, t = 10 min = 600s. Let, the initial concentration of reactants be 100 = [Ro]

So, k = log-4

When the reaction is 75% completed, then 25% reactants are left unreacted.

So, k = log. Thus, t = log

t = log

t = 0.372 x 104 s

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???? = [Ro] and ?? = [R]

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i didnt get the formula u used

plss explain ur answer again

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k = 2.303/t log [Ro] / [R].

So, when 20% of the reaction is completed, then 80% reactants are left unreacted.

And, t = 10min = 600s. Let, the initial concentration of reactants be 100 = [Ro].

So, k = 2.303 / 600 log 100 / 80. Thus k = 3.719 x 10-4.

Now, when the reaction is 75% completed then 25 % reactants are left unreacted.

Also, t = 2.303 / k log [Ro] / [R].

Thus, t = 2.303 / 3.719 x 10-4 log 100 / 25

t = 0.372 x 104 s.

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thank uuuu soo much sarthak mahajan for ur kind help :)

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Welcome Jeevani...... :)

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its my pleasure sarthak

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A second order reaction in which initial concentration for both reactants are same is 25% complete in 600 see. How long Will it take for the reaction to be 75%completed?
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For first order reaction
a= 100 at t=0
X1= 20% at t =10

K= 2.303/10 *log(a/a-x1)
K= 2.303/10*log (100/100-20)

K=0.2303 log (1.25)

K=0.2303*0.0969

K=0.0223

Now, for 60% t=?
a =100
X= 60

K=2.303/t*log(a/a-x)

t=2.303/0.0223*log(100/25)

t=103.273*log4

t=103.273*0.6021

t=62.2minutes
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Vvxh
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hello
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Because ln is perfect than log
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Reproduction
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Final concentration C = 25 ( because?75?of?reaction?is finished). Therefore?time taken?to?complete 75% of?reaction?is 62.17?min.
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