For the first order reaction, calculate the ratio of the time taken to complete `99%` of the reaction to the time taken to complete `90%` of the reaction.
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The ratio of the times of 99.9 % of the reaction to complete and half of the reaction to complete is
Correct Answer - D `t_(1//2)=(2.303)/(k) log ""(100)/(0.1)=(2.303)/(k)x3` `t_(1//2)=(2.303)/(k)log2""=(2.303)/(k)xx0.301` `(t)/(t_(1//2))=(3)/(0.301)~~10`
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If initial concentration is reduced to `1//4^(th)` in a zero order reaction , the time taken for half the reaction to complete
Correct Answer - C `t_(1//2)prop [conc.]"for" 0^(th) ` order reaction .
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A first order reaction takes 100 min for completion of 60 % of reaction ,The time eequired for completion of 90% of the reaction is
Correct Answer - D For the order reaction ,`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)` .... substituting...
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The half-life period or a first order reaction is 1 hrs. What is the time in hour taken for `87.5%` completion of the reaction?
Correct Answer - C 87.5 % completion ` therefore 12.5 % ` remaining i.e., 1/8 reaming `therefore` 3 half life period .
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Show that for a first order reaction, time required for `99.99%` of the reaction to take place is 10 times the time required for the completion of hal
For first order reaction: `t=2.303/k log a/(a-x)` Ist case, `a=100%, x=50%, (a-x)=100-50 = 50%`. `t_(1//2)=2.303/k log2(100%)/(50%)` `=(2.303)/k log 2= (2.303 xx 0.3010)/k=0.693/k` ...........(i) IInd case, `a=100%, x=99.9%, (a-x) = (100-99.9)=0.1%`...
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A first order reaction is `50%` complete in 30 minutes at `27^(@)`C and in 10 minutes at `47^(@)`C. Calculate the reaction rate constants at these tem
The half-life period for first order reaction is given by: `t_(1//2) = 0.693/k` or `k=0.693/t_(1//2)` Therefore, `k_(1) = 0.693/(30 min) = 0.231 min^(-1)` and `k_(2) = 0.693/(10 min) = 0.0693...
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In the first order reaction, half of the reaction is completed in 100 seconds. The time for `99%` of the reaction to occur will be
Correct Answer - A a) `k=0.693/100 s^(-1)` `t(99.9%) = 2.303/k log a/(a-x)` `=(2.303 xx 100s/(0.693)) xx log 100/1` `=(2.303 xx 100 xx 2)/(0.693) = 666.64` s
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Calculate the Materials Consurned Ratio, Employees Benefit Expenses Ratio, Admninistrative Expenses Ratio, Selling and Distribution Expenses Ratio and
[Meterials Consumed Ratio = 50 % , Employees Benefit Expenses Ratio= 15.63 % Administration Expenses Ratio= 14.25 % , Selling and Distribution Expenses Ratio =3.25 %,Other Expenses Ratio = 10...
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Compounds A and B react with a common reagent with first order kinetics in both cases. If 99% of A must react before 1% of B has reacted, what is the
Correct Answer - D The fraction of A and B remaining at time t when 99% of A has reacted, are (A=0.01, B=0.99) `k_A=1/t` In 100 and `k_B=1/t` In `100/99` `:....
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The reaction `Aoverset(k)rarr` Product, is zero order while the reaction `Boverset(k)rarr` Product, is first order reaction. For what intial concentra
Correct Answer - A For zero order reaction, x=kt `:. (a)/(2)=kxxt_(1//2),i.e.,t_(1//2)=(a)/(2k)" ".....(i)` For frist order reaction, `t_(1//2)=(log_(e)2)/(k)" ".....(ii)` From eqs. (i) and (ii), `(a)/(2k)=(log_(e)2)/(k)` `a=log_(e)4M`
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