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89701.

State the rate equation for a first order reaction. Derive the half-life period from the rate equation. A first order reaction takes 69.3 minutes for 50% completion. How much time will be needed for 80% completion?

Answer»

SOLUTION :Given that `t_1/2=69.3`MIN.
`K=0.693/(69.3min)=10^-2min^-1`Applying the relation:`t=2.303/K"LOG"100/100-80)`
(for`80%` completion ,when` a =100,K=80) t=2.303/(10^-2S^-1"log5"=(230.3xx0.6989)`minute=160.96minutes:
Discussion
89702.

A first order reaction takes 69.3 min for 50% completion. The time requiredfor 80% completion of this reaction is :

Answer»

104 MIN
161min
110.4 min
182 min

Answer :B
Discussion
89703.

A first order reaction takes 69.3 min for 50% completion. Set up on equation for determining the time needed for 80% completion.

Answer»

Solution :`K=0.693/t_(1//2)=0.693/69.3` min
`=10^(-2) "min"^(-1)`
`T=2.303/K "LOG"([R_0])/([R])`
`=2.303/10^(-2) log5`
=160.9 min
Discussion
89704.

A first order reaction takes 40 minutes for 30% decomposition. Calculate t_(1//2).

Answer»

SOLUTION :Applying the relation :
`k=(2.303)/(t)"log"([A]_(0))/([A])`
SUBSTITUTING the values, we have
`k=(2.303)/(40)"log"(100)/(70)=(2.303)/(40)xx0.1549=0.0089"min"^(-1)`
`t_(1//2)=(0.693)/(k)=(0.693)/(0.0089)=77.86` MINUTES
Discussion
89705.

A first order reaction takes 40 min for 30% decompositon.

Answer»

Solution :(i) To find K value `[R]_(0) = 100 [R] = 100 - 30 = 70, t = 40` MIN
`k = (2.303)/(t) log ""([R]_(0))/([R]), k = (2.303)/(40) log ""(100)/(70)`
`k = (2.303)/(40) xx 0.1549, k = 8.92 xx 10^(-3)` min.
(II) To find `t_(1//2)`
`t_(1//2) = (0.693)/(k) = (0.693)/(8.92xx10^(-3)) = 77.7` min.
Discussion
89706.

A first order reaction takes 40 min for 30% decomposition .Calculation t_((1)/(2))

Answer»

SOLUTION :The reaction is first ORDER,so it decomposes 30% decomposition so ,70% is remaining,
`therefore [R]_(t)70% of [R]_(0)`
`=(70)/(100)[R]_(0)=0.7 [R]_(0)`
Calculation of K:The reacton is first order ,so
`k=(2.303)/(t)` log `([R]_(0))/([R]_(t))`
`k=(2.303)/(t)` log `([R]_(0))/([R]_(t))`
`k=(2.303)/(t)` log `([R]_(0))/([R]_(t))` Where time t=40 MIN
`=(2.303)/(40 min `log `((10)/(7))` Calculation of `t_((1)/(2))`:Reaction is first order.
`therefore t_((t)/(2))=(0.693)/(k)=(0.693)/(8.9185xx10^(-3) min ^(-1))=77.7 min`
Discussion
89707.

A first order decomposition reaction takes 40min. For 30% decomposition. Calculate it's t_(1//2) value.

Answer»

17.69 min
7.77 min
77.69 min
27.69 min

Answer :C
Discussion
89708.

A first order reaction takes 40 min for 30% decomposition. Calculate t_(1//2).

Answer»

Solution :At `30%` DECOMPOSITION, `x=30%` of a i.e., 0.30 a.
Using FIRST order equation,
`k=(2.303)/(t) "log"(a)/(a-x)=(2.303)/(40" MIN")"log"(a)/(a-0.30a)=(2.303)/(40)XX"log"(10)/(7) "min"^(-1)`
`=(2.303)/(40)xx0.1549"min"^(-1) = 8.918xx10^(-3)"min"^(-1)`
For a 1ST order reaction, `t_(1//2)=(0.693)/(k)=(0.693)/(8.918xx10^(-3)"min"^(-1))=77.7` min.
Discussion
89709.

A first order reaction takes 20 minutes for 25% decomposition. Calculate the time when 75% of the reaction will be completed.[Given : log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021]

Answer»

SOLUTION :The EQUATION for first ORDER reaction is :
`t=(2.303)/(k)"log"([R]_(0))/(R )`
For `25%` of reaction :
`20=(2.303)/(k)"log"([R]_(0))/(0.75[R]_(0))=(2.303)/(k) "log"(4)/(3)"" …(i)`
For 75% of the reaction:
`t=(2.303)/(k) "log"([R]_(0))/(0.25[R]_(0))= (2.303)/(k)log 4"" ...(II)`
Dividing (ii) by (i), we get
`(t)/(20)=(log 4)/("log"(4)/(3))=(log4)/(log 4-log3)=(0.6021)/(0.6021-0.4771)=(0.6021)/(0.1250)`
or ` t=(20xx0.6021)/(0.1250)=96.3` minutes
Discussion
89710.

A first order reaction takes 25 minutes for 25% decomposition. Calculate t_(1//2). ""["Give ":log 2=0.3010,log3=0.4771,log4=0.6021]

Answer»

SOLUTION :Apply the relation :
`K=(2.303)/(t)"log"([R]_(0))/([R])`
`25%` decomposition means `[R]=0.75[R]_(0)` and `t=25` minutes Substituting the VALUES in the EQUATION above,, we have
`k=(2.303)/(25)"log"([R]_(0))/(0.75[R]_(0))`
`=(2.303)/(25)"log"4/3`
`=(2.303)/(25)[log4-log3]`
`=(2.303)/(25)[0.6021-0.4771]`
`=(2.303)/(25)xx0.1250=0.011515" minutes"^(-1)`
`t_(1//2)=(0.693)/(k)`
`=(0.693)/(0.011515" minutes"^(-1))`
`=60.18" minutus"`
Discussion
89711.

A first order reaction takes 20 minutes for 25% decomposition. Calculate the time when 75% of the reaction will be completed. (Given : log2=0.3010, log3=0.4771,log4=0.6021)

Answer»


SOLUTION :PROCEED as in Solved Problem 3.
`K=(2.303)/(20)LOG""(a)/(a-0.25a)=(2.303)/(20)log""(4)/(3)=0.0144" min"^(-1)`
`t=(2.303)/(k)log""(a)/(a-0.75a)=(2.303)/(0.0144)log4=96.3" min"`
Discussion
89712.

A first order reaction occurs 40%in 120 min at 25^(@)C and in 15 min at 55^(@) C. The approximate value of (K_(35^(@)C))/(K_(25^(@) C))is :

Answer»


ANSWER :2
Discussion
89713.

A first order reaction takes 10 minutes for 25% decomposition. Calculate t_(1//2) for the reaction. [Given : log 2 = 0.3010, log 3 = 0.4771, log 4 = 0.6021]

Answer»

Solution :The equation for first ORDER reaction is
`t=(2.303)/(k)"log"([R]_(0))/([R])""…(1)`
For 25% decomposition, `[R]=(3[R]_(0))/(4)`
SUBSTITUTING the VALUES in equation (1), we have
`10=(2.303)/(k)"log"([R]_(0)xx4)/(3[R]_(0))`
or `10=(2.303)/(k) "log"(4)/(3)""...(2)`
For 50% decomposition, `[R]=([R]_(0))/(2)`
Substituting the values in equation (1), we have
`t_(1//2)=(2.303)/(k)"log"([R]_(0)xx2)/([R]_(0))`
or `t_(1//2)=(2.303)/(k)log2 ""...(3)`
Dividing (3) by (2), we have
`(t_(1//2))/(10)=(log2)/("log"(4)/(3)) or t_(1//2)=10xx(log2)/((log4-log3)) or t_(1//2) = 10xx (0.3010)/((0.6021-0.4771))`
or `t_(1//2)=(10xx0.3010)/(0.1250)=24` minutes
Discussion
89714.

A firstorderreactiontakes100 minforcompletionof 60 %of reaction ,Thetimeeequiredfor completionof 90% of thereactionis

Answer»

`150 MIN `
`200`min
`220.9`min
`246.6` min

Solution :Forthe orderreaction ,`k=(2.303)/(t)log""(C_(0))/(C_(t))`
Case I:
`C_(0)=100 M,C_(t) =100 -60 =40, t= 100 min`
`k=(2.303)/(100)xxlog""(100)/(40)`
`k=(2.303xx0.3979)/(100)`
case-II :
`C_(0)=100 M,C_(t)=100-90=10M `
`t=(2.303)/(k) log""(100)/(10)`
`=(2.303)/(k)` .... substitutingK from I
`=(2.303xx100)/(2.303xx0.3979)=245.6 min`
Discussion
89715.

A first order reaction is half completed in 45 minutes.How long does it need 99.9% of the reaction to be completed.

Answer»

5 Hr
7.5 Hr
10 Hr.
20 Hr

Answer :B
Discussion
89716.

A first order reaction is half completed is 45 minutes. How long does it need for 99.9% of the reaction tobe completed?

Answer»

20 HOURS
10 hours
`"7 "1//2` hours
5 hours

Answer :C
Discussion
89717.

A first-order reaction is half completed in 45 minutes. How long does it need 99.9% of the reaction to be completed

Answer»

`5` HOURS
`7.5` hours
`10` hours
`20` hours

Solution :`k = (0.963)/(45) "min"^(-1) = (2.303)/(t_(99.9%))"log"(a)/(a-0.999)`
or `t_(99.9%) = (2.303 xx 45)/(0.693) log10^(3) = 448min ~~ 7.5 hr`
Discussion
89718.

A first order reaction is found to have a rate constant,k=5.5xx10^(-4)s^(-1).Find out the half-life of the reaction.

Answer»

SOLUTION :`t_(1/2)=0.693/k=0.693/(5.5xx10^(-4))`=1260sec
Discussion
89719.

A First order reaction is half completed in 45 minutes . How long does it need 99.9 % of the reaction to be completed

Answer»

5 hours
7.5 hours
10 hours
20 hours

SOLUTION :`k = (0.693)/(45) "min"^(-1) = (2.303)/(t_(99.9%)) "log" (a)/(a-0.999 a)` or
`t_(99.9%) = (2.303 XX 45)/(0.693) "log" 10^(3) = 448` min = 7.5 hrs.
Discussion
89720.

A first order reaction is found to have a rate constant,k=5.5xx10^-14S^-1.Find the half -life of the reaction.

Answer»

SOLUTION :`t_(1//2)=0.693/K,
t_(1//2)=0.693/(5.5xx10^(div-1).s^-1)=1.26s`
Discussion
89721.

A first order reaction is found to have a rate constant k=7.39xx10^(-5)sec^(-1). Find the half life of this reaction (log2=0.3010).

Answer»


SOLUTION :For a first order REACTION, `K=(2.303)/(t)LOG""(a)/(a-x)"":"""For "t=t_(1//2),x=a//2`
`t_(1//2)=(2.303)/(k)log""(a)/(a-a//2)=(2.303)/(k)log2=(2.303)/(7.39xx10^(-5)s^(-1))xx0.3010=9.38xx10^(3)s`
Discussion
89722.

A first order reaction is found to have a rate constant ,k=5.5xx10^(-14)s^(-1). Find the half-life of the reaction .

Answer»

Solution :The REACTION is of FIRST order
`K=5.5xx10^(-14)S^(-1),t_((t)/(2))=(?)`
`t_((t)/(2))=(0.693)/(k)`
`=(0.693)/(5.5xx10^(-14)s^(-1))`
Discussion
89723.

A first order reaction is found to have a rate constant k = 7.39xx10^(-5)s^(-1). Find the half life of this reaction.

Answer»

SOLUTION :For first order REACTION
`k=(2.303)/(t)"log"(a)/(a-X)`
For `t=t_(1//2), "" x=(a)/(2)`
`t_(1//2)=(2.303)/(k)"log"(a)/(a-(a)/(2))`
`=(2.303)/(k)log 2 = (2.303xx0.3010)/(7.39xx10^(-5))`
`= 9.38xx10^(3)s`.
Discussion
89724.

A first order reaction is completed by 20% 2 minutes. How much further time is required for 64% of the initial concentration of the reactants to remain.

Answer»


SOLUTION :`(Kxx2)/(Kxxt)=(LOG(100/80))/(log(100/64)),2/t=(log1.25)/(log1.5625)=0.0969/0.19382=t=(2xx0.19382)/(0.0969~=4`
Discussion
89725.

A first order reaction is found to have a rate constant K=5.5xx10^(-14)S^(-1) .Find the half-life of the reaction.

Answer»

SOLUTION :`(0.693)/(K)=(0.693)/(5.5xx10^(-4))=1260" s"`
Discussion
89726.

A first order reaction is carried out with an initial concentration of 10ML^(-1) and 80% of the reactant changes into the product. Now if the same reaction is carried out with an initial concentration of 5ML^(-1), the percentage of reactant changing into the product. Now if the same reaction is carried out with an initial concentration of 5ML^(-1), the percentageof reactant changing to the product is

Answer»

`40`
`80`
`160`
can't be calculated

Answer :B
Discussion
89727.

A first order reaction is carried out with an initial concentration of 10 mole per litre and 80% of the reactant changes into the product. Now if the same - reaction is carried out with an initial concentration of5 mol per litre the percentage of the reactant changing to the product is:

Answer»

40
80
160
Cannot be calculated

Answer :B
Discussion
89728.

A first order reaction is carried out with an initial concentrationof 10 molper litre an 80% of the reactant changes into the product in 10 sec. Now if the same reaction is carried out with an initial concentration of 5 mol per litre. The percentage of the reactant changing to the product in 10 sec is:

Answer»

40
80
60
50

Solution :`10M(80%)/(t=10"SEC")2M 5M(80%)/(t=10"sec")1M`
Discussion
89729.

A first order reaction is carried out starting with 10 mol L^(-1) of the reactant . It is 40% complete in 50 min . If the same reaction is carried out with an initial concentration of 5 mol L^(-1) the percentage of reaction that is completed in 50min is

Answer»

40%
80%
20%
36.8%

SOLUTION :(A) For a first order reaction time TAKEN for a definite fraction to COMPLETE is independent of initial concentration .
Discussion
89730.

A first order reaction is 87.5% complete in an hour. The rate constant of the reaction is

Answer»

`0.0346min^(-1)`
`0.0693h^(-1)`
`0.0693min^(-1)`
`0.0346h^(-1)`

ANSWER :A
Discussion
89731.

A first order reaction is 60% complete in 20 minutes .How long will the reaction take to be 84% complete ?

Answer»

54 mins
68 mins
40 mins
76 mins

SOLUTION :( C) For 60% COMPLETION of reaction
`k= (2.303)/(t) "log"([R]_(0))/([R])`
`=(2.303)/(20)"log" (100)/(40)`
`=0.045 "MIN^(-1)`
For 84 % completion of reaction
`t = (2.303)/(k)"log"([R]_(0))/([R])`
`=(2.303)/(0.045)"log"(100)/(16)`
=40.7 min
Discussion
89732.

A first order reaction is 75% complete in 60 minutes. Find the half-life of this reaction.

Answer»


ANSWER :30 MINUTES
Discussion
89733.

A first order reaction is 60% complete in 20 minutes . How long will the reaction take to be 84% complete ?

Answer»

54 MINS
68 mins
40 mins
76 mins

SOLUTION :If `[A]_(0) = 100` then `[A] = 100 - 60 = 40 ` in 20 minutes
Rate constant (k) = `(2.303)/(t) "log" ([A]_(0))/([A])`
`k = (2.303)/(20) "log" (100)/(40)`
`k = (2.303)/(20) (1- 0.6020) =(2.303)/(20) xx 0.397 "min"^(-1) … (1)`
The time for 84 % completion of the REACTION is `t_(84 %) = (2.303)/(k) "log" (100)/(100 - 84)`
substituting the VALUE of k from eq. (1) we get `t_(84 %) = (2.303 xx 20)/(2.303 xx 0.397) "log" (100)/(16)`
`t_(84 %) = (20)/(0.397) ( 2 - 1.20)`
`t_(84 %) = 50.377 xx 0.795 = 40.09` min
Discussion
89734.

A first order reaction is 50%completed in 30 minutes at 27^(@) and in 10 minutes at 47^(@)C. The energy of activation of the reaction is

Answer»

42.84kJ `//`MOL
34.84 KJ` //` mol
84.00 KJ `//` mol
30.00 KJ` //`mol

ANSWER :A
Discussion
89735.

A first order reaction is 50% completed in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction. [Given : log 2=0.3010, log 4=0.6021, R=8.314 JK^(-1)"mol"^(-1)]

Answer»

Solution :For a first ORDER reaction
`k=(2.303)/(t)"log"([R]_(0))/([R])`
At 300K
`k_(1)=(2.303)/(40)"log"([R]_(0))/([R]_(0)//2)=(2.303)/(40)"log 2"=(2.303)/(40)xx0.3010`
AT 320 K
`k_(2)=(2.303)/(20)"log"([R]_(0))/([R]_(0)//2)=(2.303)/(20)LOG2=(2.303)/(20) xx 0.3010`
`(k_(2))/(k_(1))=(2.303xx0.3010xx40)/(20xx2.303xx0.3010) =2`
Applying Arrhenius equation, we have
`"log"(k_(2))/(k_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]`
Substituting the values, we have
`log 2=(E_(a))/(2.303xx8.314JK^(-1)"mol"^(-1))[(320K-300K)/(320K xx 300K)]`
or `0.3010=(E_(a))/(19.1471)[(20)/(96000)]`
or `E_(a)=(0.3010xx19.1471xx96000)/(20)=27663.7J`
Discussion
89736.

A first order reaction is 50% completed in 20 minutes at 27^(@)C. The energy of activation of the reaction is-

Answer»

43.58 KJ
55.14 KJ
11.97 KJ
6.65 KJ

Solution :`"Log"((K_(2))/(K_(1)))=(Ea)/(2.303R)((1)/(T_(1))-(1)/(T_(2)))`
`Log4=(Ea)/(2.303xx8)((1)/(300)-(1)/(320))`
`{K_(1)=(1n2)/(20),K_(2)=(n2)/(5)}`
Discussion
89737.

A first order reaction is 50% completed in 20 minutes at 27^(@) C and in 5 minutes at 47^(@) . The energy of activation of the reaction is :

Answer»

43.85 KJ/mol
55.14 KJ/mol
11.97 KJ/mol
6.65 KJ/mol

Answer :B
Discussion
89738.

A first order reaction is 50% completed in 1.26xx10^(14) s. How much time would it take for 100% completion ?

Answer»

`1.26xx10^(15)s`
`2.52xx10^(14)s`
`2.52xx10^(28)s`
infinite

Solution :Whole of the substance never REACTS because in every half life, `50%` of the substance reacts. Hence, TIME taken for `100%` completion of a reaction is infinite.
Discussion
89739.

A first order reaction is 50% completed in 1.26xx10^(14)S.How much time would it take for 100% completion?

Answer»

`1.26xx10^(15)s`
`2.52xx10^(14)s`
`2.52xx10^(28)s`
infinite

Solution :Whole of the substance never reacts because in every HALF life.505 of the substance reacts.Hence ,TIME taken for 100% completion of a reaction is infinite.
`t_((1)/(2))=1.26xx10^(4)s`
`therefore k=(0.693)/(t^((1)/(v)))=(0.693)/(1.26xx10^(4))`
The FINAL concentration when the reaction is completed 100% =`[R]_(t)`=zero
So `t_(100%)=(2.303)/(k)` log `([R]_(0))/([R]_(t))` `(2.303)/(k)` log (infinite)
`~~(2.303)/(k)xx10^(prop)=prop`
If any reaction `t_((1)/(2))` =exact value ,then it will take infinite time for 100% completion .So option (D) is correct.
Discussion
89740.

A first order reaction is 50% completed in 1.26xx10^(14)s. How much time would it take for 100% completion?

Answer»

`1.26xx10^(15)s`
`2.52xx10^(14)s`
`2.52xx10^(28)s`
INFINITE

ANSWER :D
Discussion
89741.

A first order reaction is 50% completed in 1.26xx10^(14). How much time would is take for 100 % completion ?

Answer»

`1.26xx10^(15)s`
`2.52xx10^(14)s`
`2.52xx10^(28)s`
INFINITE

ANSWER :D
Discussion
89742.

A first order reaction is 50% complete in 50 minutres at 300K and the same reaction is again 50% complete in 25 minutes at 350K . Calculate activation energy of the reaction .

Answer»

Solution : The RELATION between HALF- life period and rate CONSTANT for a first order reaction is GIVEN by
`t_(1//2)= (0.693)/(K)`
Substitutiong the value , we have
`K_(2)(350K)= (0.693)/(25)` and `K_(1)(300K)= (0.693)/(50)`
`therefore (K_(2))/(K_(1)= 2)`
Applying Arrhenius equation , we have
log `(K_(2))/(K_(1)) = (e_(a))/(2.303R)[(1)/(T_(1))-(1)/(T_(2))]`
Or log `2= (E_(a))/(2.303xx8.314)[(350-300)/(350xx300)]`
Or `e_(a)=12.104Kj//mol`
Discussion
89743.

A first order reaction is 50% complete in 40 minutes at 300 K and in 20 minutes at 320 K. Calculate the activation energy of the reaction (Given: log 2 = 0.3010. log 4 = 0.6021, R = 8.314 JK^–1 mol^–1)

Answer»

Solution :`K_2=0.693//20`,
`k_1=0.693//40`
`LOG k_2/k_1=E_2/2.303 R [1/T_1-1/T_2]`
`k_2//k_1=2`
`log 2 =E_8/(2.303 times 1.314) [(320-300)/(320 times 300)],Ea=27.66 KJ Mol^-1`
Discussion
89744.

A firstorderreaction is 50% completed in1.26 xx 10^(14)s. Howmuchtimewouldittakefor100%completion ?

Answer»

`1.26 xx 10^(15)s`
`2.52 xx10^(14) s`
`2.52 xx 10^(28)s`
Infinite

Solution :The TIME takenfor halfthe reaction to complete i.e., the time in whichtheconcentrationof a reactantis reduced to halfof its origianal value is CALLED half- lineperiod of the reaction.
Butit isimpossibleto perform 100% of the reaction. wholeof the substance new reacts because in very half - life, 50% of the substance reacts. HENCE, time taken 100% COMPLETION of a reaction is infinite.
Discussion
89745.

A first order reaction is 50% complete in 30 minutes at 27^(@)C and in 10 minutes at 47^(@)C. The rate constant at 47^(@)C and energy of activation of the reaction in kJ//mole will be

Answer»

`0.0693, 43.848 kJ MOL^(-1)`
`0.0560,45.621 kJ mol^(-1)`
`0.0625, 42.926 kJ mol^(-1)`
`0.0660, 46.189 kJ mol^(-1)`

Solution :We know that `k = (2.303)/(t) log((a)/(a-x))`
Subtituting the VALUES at the two given conditions
`k_(27^(@)) = (0.0693)/(30) = 0.0231 , k_(47^(@)) = (0.693)/(10)= 0.0693`
We ALSO know that
`"log" (k_(2))/(k_(1)) = (2.303)/(t)"log"((a)/(a-x))`
or `E_(a) = (2.303 R xx T_(1)T_(2))/(T_(2) - T_(1)) "log"(k_(2))/(k_(1))`
`= (2.303 xx 8.314 xx 10^(-3) xx 300 xx 320)/(320 - 300) xx "log" (0.0693)/(0.231)`
`= 43.848 kJ mol^(-1)`
Discussion
89746.

A first order reaction is 50% complete in 30 minutes at 27^@C and in 10 minutes at 47^@C. The energy of activation of the reaction is

Answer»

`43.83 kJ mol^(-1)`
`34.84kJmol^(-1)`
`84.00kJ mol^(-1)`
`30.00kJmol^(-1)`

Solution :`k_(27^@C)= (0.693)/30 MI n^(-1)`,
`k_(47^@C)=0.693/10 mi n^(-1)`
`:. K_(47^@C)/k_(27^@C)=3`
`log. (k_(47^@C))/(k_(27^@C))= log 3 = 0.4771`
`:. 0.4771= E_a/(2.303 xx 8.314)(20/(300xx 320))`
`E_a=(0.4771 xx 2.303 xx 8.314xx 300xx320)/20`
`= 43.84 kJ`
Discussion
89747.

A first order reaction is 40% complete in 80 minutes. Calculate the value of rate constant (k). In what time will the reaction be 90% completed ? [Given : log2=0.3010,log3=0.4771,log4=0.6021,log5=0.6771,log6=0.7782]

Answer»

Solution :Equation for first order reaction is
`K=(2.303)/(t)log([R]_(0))/([R])`
For 40% complete reaction, [R] = 0.6 `[R]_(0)`, t = 80 minutes SUBSTITUTING the values in the above equation, we have
`k=(2.303)/(80)log([R]_(0))/(0.6[R]_(0))`
or `""k=(2.303)/(80)[log5-log3]=(2.303)/(80)[0.6771-0.4771]`
or `""k=(0.4606)/(80)=0.005758`
Now, `""0.005758=(2.303)/(t)log([R]_(0))/(0.1[R]_(0))=(2.303)/(t)XX1`
or `""t=(2.303)/(0.005758)=400` minutes
Discussion
89748.

A first order reaction is 40% complete in 50 minutes. Calculate the value of the rate constant. In what time will the reaction be 80% complete ?

Answer»

SOLUTION :For the first order reaction
`k=((2.303)/(t))log((a)/(a-x))`
When `x=((40)/(100))a=0.4 a`
`t = 50 m`
`THEREFORE k = ((2.303)/(50))log((a)/(a-0.4a))`
`k=((2.303)/(50))log((1)/(0.6))`
`=0.010216 "min"^(-1)`
t = ? When x = 0.8 a
From above `K = 0.010216 "min"^(-1)`
`therefore t ((2.303)/(0.010216))log((a)/(a-0.8a))`
`((2.303)/(0.010216))log((1)/(0.2))=157.58` min
The time at which the reaction will be 80% COMPLETE is 157.58 min.
Discussion
89749.

A first order reaction is 40% complete in 50 minutes . Calculate the value of the rate constant. In wha time the reaction be 80% complete ?

Answer»

SOLUTION :(i) For the first order reaction `k=(2.303)/tloga/((a-x))`
Assume , a = 100% x = 40% t = 50 MINUTES
Therefore , a-x = 100-40 = 60
k = (2.303/50)log (100/60)
`k = 0.010216 "min"^(-1)` .
Hence the value of the rate constant is `0.010216"min"^(-1)`
(II) t = ? , wheb x = 80%
Therefore , a - x = 100-80 = 20
From above , k = `0.010216"min"^(-1)`
`t=(2.303//0.010216)log (100//20)`
t = 157.58 min
The time at which the reaction will be 80% COMPLETE is 157.58 min.
Discussion
89750.

A first order reaction is 30% completed in 30 minutes. Calculate the half-life.

Answer»

SOLUTION :`K=2.303/309"LOG"[A]_0/(75/100[A]_0)=2.303/30"log"4/3=(2.303xx0.1250)/30=0.0096 min^-1=9.6xx10^-3min^-1`
`t_(1/2)=0.693/9.6xx10^3=6930/96=72.18min`.`
Discussion