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InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1951. |
AN experiment requires minimum beta activity produced at the rate of 346 beta particles per minute. The half- life period of `_(42)Mo^(99)` , which is a beta emitter , is 66.6 h . Find the minimum amount of `_(42)Mo^(99)` required to carry out the experiment in 6.909 h. |
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Answer» Correct Answer - `5.2xx10^(-16)g` |
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| 1952. |
The decompoistion of `N_(2)O_(5)` according to the equation `2N_(2)O_(5)(g) rarr 4NO_(2)(g)+O_(2)(g)` is a first order reaction. After `30 min`, form the start of the decompoistion in a closed vessel, the total pressure developed is found t be `284.5 mm Hg`. On complete decompoistion, the total pressure is `584.5 mm Hg`. Calculate the rate constant of the reaction. |
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Answer» Correct Answer - `5.2xx10^(-3)min^(-1)` |
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| 1953. |
In the Arrhenius for a certain reaction, the value of `A` and `E_(a)` (activation energy) are `4 xx 10^(13) sec^(-1)` and `98.6 "kJ mol"^(-1)`, respectively. If the reaction is of first order, the temperature at which its half-life period is `10` minutes isA. `280 K`B. `290 K`C. `311.35 K`D. `418.26 K` |
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Answer» Correct Answer - C Calculation of `k`: We know that `k = (2.303)/(t) "log"((a)/(a-x))` `(t_(1//2)=10xx60 sec)=1.1558xx10^(-3)` According to Arrehenius equation, `log k = log A - (Ea)/(2.303 RT)` Substituting the various values in the above equation `log 1.155 xx 10^(-3) = log 4 xx 10^(13) - (98.6)/(2.303 xx 8.314 xx 10^(-3) xx T)` On usual calculation, `T = 311.35 K` |
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| 1954. |
A follows first order reaction. `(A) rarr Product` The concentration of `A` changes form `0.1 M` to `0.025 M` in `40 min`. Find the rate of reaction of `A` when the concentration of `A` is `0.01 M`.A. (a) `3.47xx10^(-4) M//min`B. (b) `3.47xx10^(-5) M//min`C. (c ) `1.73xx10^(-4) M//min`D. (d) `1.73xx10^(-5) M//min` |
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Answer» Correct Answer - a For the first order reaction `K=2.303/t"log"a/(a-x)` `a=0.1 M, a-x=0.025 M, t=40 min` `K=2.303/40"log"0.1/0.025` `=2.303/40"log"4=0.0347 min^(-1)` `[A] rarr` product Thus, rate=`K[A]` rate`=0.0347xx0.01 M min^(-1)` `=3.47xx10^(-4) M min^(-1)` |
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| 1955. |
In a Arrhenius equation for a certain reaction, the values of `A and E_(a)` ( energy of activation)are `4xx10^(13)s^(-1)` and `98.6KJ mol^(-1)`, respectively. If the reaction of first order at, what temperature will its life periof be `10 min`.A.B.C.D. |
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Answer» Correct Answer - 311.34K |
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| 1956. |
What is the order of reaction for which rate becomes half if volume of the container having same amount of reactant is doubled? Assume gaseous phase reaction. |
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Answer» Correct Answer - 1 Rate, For `I`: Let a mole of reactant in vessel of `V` litre `:. r_(1)=K[a/V]^(n) …(1)` For `II`: The volume is doubled, rate becomes half. `:. R_(1)/2=K[a/(2V)]^(n) …(2)` By eqs. (1) and (2) `2=2^(n)` or `n=1` Order of reaction =`1` |
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| 1957. |
In a Arrhenius equation for a certain reaction, the values of `A and E_(a)` ( energy of activation)are `4xx10^(13)s^(-1)` and `98.6KJ mol^(-1)`, respectively. If the reaction of first order at, what temperature will its life periof be `10 min`. |
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Answer» Accroding to Arrhenius equation, `k=Ae^(-E_(a)//RT)` or `log_(e)k=log_(e)A-(E_(a))/(RT)` or `2.303 log_(10)k=2.303 log _(10)A-(E_(a))/(RT)` For a first order reaction `t_(1//2)=(0693)/(k)` So, `k=(0.693)/(600) sec^(-1)(t_(1//2)=10min=600 sec)` `=1.1xx10^(-3)sec^(-1)` Hence, log `(1.1xx10^(-3))=log(4xx10^(13))-(98.6xx10^(3))/(2.303xx8.314xxT)` T=310.95K` |
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| 1958. |
In Arrhenius equation for a certain reaction , the values of A and `E_(a)` (activation energy are `4xx10^(13) "sec"^(-1) and 98.6 kJ "mol"^(-1)` respectively. At what temeperatue, the reaction will have specific constant `1.1xx10^(-3) "sec"^(-1)`? |
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Answer» Accroding to Arrhenius equation, `k=Ae^(-E_(a)//RT)` or `"log"_(e)k="log"_(e)A-(E_(a))/(RT)"log"_(e)e` `or 2.303 "log"k=2.303"log"_(10)A-(E_(a))/(RT)` or `2.303"log"(1.1xx10^(-3))=2.303"log"K(4xx10)^(13)-(98xx10^(3))/(8.314xxT)` `T=(98.6xx10^(3))/(8.314xx2.303xx16.56)K=310.96K` |
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| 1959. |
Water and oxygen atoms react in upper atmospheric level bimolecularly to form two `OH` radicals having heat of reaction `72 kJ` at `400 K` and energy of activation being `77 kJ mol^(-1)`. Calculate the `E_(a)` for bimolecular combination of two `OH` radicals to form `H_(2)O` and `O`-atom. |
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Answer» Correct Answer - 5 `H_(2)O+O overset(E_(a)=77 kJ)(hArr) 2OH, DeltaH=72 kJ` `2 OH rarr H_(2)O+O, DeltaH=-72 kJ` `DeltaH=E_(a(f))-E_(a(b))` `-72=E_(a(f))-77` or `E_(a(f))=5 KJ` |
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| 1960. |
Under the same reaction conditions, the intial concentration of `1.386 mol dm^(-3)` of a substance becomes half in `40 s` and `20 s` theough first order and zero order kinetics, respectively. The ratio `(k_(1)//k_(0))` of the rate constants for first order `(k_(1))` and zero order `(k_(0))` of the reaction isA. `0.5 mol^(-1)dm^(3)`B. `1.0mol dm^(-3)`C. `1.5mol dm^(-3)`D. `2.0mol^(-1) dm^(3)` |
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Answer» Correct Answer - A For first order reaction `k_(1)=(0.693)/(t_(1//2))=(0.693)/(40s)` For a zero order reaction: `k_(0)=([R]_(0))/(2t_(1//2))=(1.386)/(2xx20_(s))` Therefore `(k_(1))/(k_(0))=(0.693)/(40s)xx(40s)/(1.386)` `=(0.693(s^(-1))/(1.386(moldm^(-3)s^(-1)))` `=0.5mol^(-1)dm^(3)` |
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| 1961. |
Under the same reaction conditions, the intial concentration of `1.386 mol dm^(-3)` of a substance becomes half in `40 s` and `20 s` theough first order and zero order kinetics, respectively. The ratio `(k_(1)//k_(0))` of the rate constants for first order `(k_(1))` and zero order `(k_(0))` of the reaction isA. `0.5 "mol" ^(-1)dm^(3)`B. `1"mol dm"^(-3)`C. `1.5 "mol dm"^(-3)`D. `2 "mol"^(-1) dm^(3)` |
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Answer» Correct Answer - A `t_(1//2)=(0.693)/(k_(1))" " t_(1//2)=(a_(0))/(2k_(0))` `40=(0.693)/(k_(1))," " 20=(1.386)/(2k_(0))=(0.693)/(k_(0))` `(20)/(40)=(0.693//k_(0))/(0.693//k_(1))=(k_(1))/(k_(0))` `(k_(1))/(k_(0))=0.5 (sec^(-1))/("mol dm"^(-3)sec^(-1))=0.5 "mol"^(-1) "dm"^(3)` |
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| 1962. |
Consider the reaction, `Cl_(2)(aq) + H_(2)S(aq) rarr S(s) + 2H^(+) (aq) + 2Cl^(-)(aq)` The rate equation for this reaction is, Rate `=k [Cl_(2)][H_(2)S]` Which of these mechanisms is`//`are consistent with this rate equation ? (I) `Cl_(2) + H_(2)S rarr H^(+) + Cl^(-) + Cl^(+) + HS^(-)` (slow) `Cl^(+) + HS^(-) rarr H^(+) + Cl^(-) + S` (fast) (II) `H_(2)S hArr H^(+) +HS^(-)` (fast equilibrium) `Cl^(+) + HS^(-) rarr 2 Cl^(-) + H^(+) + S` (slow)A. A onlyB. B onlyC. Both A and BD. Neither A nor B |
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Answer» Correct Answer - a For A Rate = `k (Cl_(2)) (H_(2)S)` (By slow step) For B Rate = `k[Cl_(2)][HS^(-)]` `k_(eq) = ([H^(+)][HS^(-)])/([H_(2)S])` (according to equilibrium) Rate = `k[Cl_(2)] (k_(eq)[H_(2)S])/([H^(+)])` Rate = `kk_(eq)[Cl_(2)] ([H_(2)S])/([H^(+)])` . |
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| 1963. |
Consider the reaction, `Cl_(2)(aq) + H_(2)S(aq) rarr S(s) + 2H^(+) (aq) + 2Cl^(-)(aq)` The rate equation for this reaction is, Rate `=k [Cl_(2)][H_(2)S]` Which of these mechanisms is`//`are consistent with this rate equation ? (I) `Cl_(2) + H_(2)S rarr H^(+) + Cl^(-) + Cl^(+) + HS^(-)` (slow) `Cl^(+) + HS^(-) rarr H^(+) + Cl^(-) + S` (fast) (II) `H_(2)S hArr H^(+) +HS^(-)` (fast equilibrium) `Cl^(+) + HS^(-) rarr 2 Cl^(-) + H^(+) + S` (slow)A. (II) onlyB. Both (I) and (II)C. Neither (I) nor (II)D. (I) only |
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Answer» Correct Answer - D Slowest step is the rate determining step. Thus, in case (I), rate law is given as rate `= k[Cl_(2)][H_(2)S]` While for the reaction given in case (II), rate law is given as rate `= k[H_(2)S][Cl_(2)][H^(+)]^(-1)`. Hence, only mechanism (I) is consistent with the given rate law. |
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| 1964. |
`T_(50)` of first order reaction is 10 min. starting with 10 `"mol"^(-1)` rate after 20 min is :A. `0.0693xx"mol" L^(-1)min^(-1)`B. `0.0693xx2.5"mol" L^(-1)min^(-1)`C. `0.0693xx5"mol" L^(-1)min^(-1)`D. `0.0693xx10"mol" L^(-1)min^(-1)` |
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Answer» Correct Answer - B Remainin concetration of reactant after 20 min `=(1)/(4)xx102.5"mol"L^(-1)` `"Rate"=kxx["Reactant"]=(0.693)/(t_(1//2))xx["Reactant"]` `=(0.693)/(10xx2.5=0.0693xx2.5 "mol"L^(-1) min^(-1)` |
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| 1965. |
For a first order reaction , `A to` Products, the concentrations of A changes from 0.1 M to 0.025 M in 40 minutes. The rate of reaction when the concentration of A is 0.01 M is:A. `1.73xx10^(-3)` M/minB. `3.47xx10^(-4)` M/minC. `3.47xx10^(-5)` M/minD. `1.73xx10^(-4)` M/min |
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Answer» Correct Answer - B `k=(2.303)/(t)log.(a)/(a-x)=(2.303)/(40)log((0.1)/(0.025))=0.03465 "min"^(-1)` Rate = `k[A] =0.03465xx0.1=3.47xx10^(-4)M//"min"` |
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| 1966. |
`H_(2)O` and `O` atom react in upper atmosphere bimolecularly to form two `OH` radicals. `DeltaH` for the reaction is `72 kJ` at `500K` and energy of activation is `77kJ mol^(-1)`. Estimate `E_(a)` for bimolecular recombination of two `OH` radicals to form `H_(2)O` and `O` atom. |
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Answer» `H_(2)O+O overset(E_(a))(rarr)2OH, DeltaH=72 kJ` `2OH overset(E_(b))(rarr)H_(2)O+O, DeltaH=-72 kJ` Also `E_(a)-E_(b)=DeltaH` `77-E_(b)=72` `:. E_(b)=5 kJ mol^(-1)` |
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| 1967. |
From the following data, the activation, energy for the reaction, (cal/mol) is : A. `4xx10^(4)`B. `2xx10^(4)`C. `8xx10^(4)`D. `3xx10^(4)` |
| Answer» Correct Answer - A | |
| 1968. |
Under the same reaction conditions, the intial concentration of `1.386 mol dm^(-3)` of a substance becomes half in `40 s` and `20 s` theough first order and zero order kinetics, respectively. The ratio `(k_(1)//k_(0))` of the rate constants for first order `(k_(1))` and zero order `(k_(0))` of the reaction isA. (a) `0.5 mol^(-1) dm^(3)`B. (b) `1.0 mol dm^(-3)`C. (c ) `1.5 mol dm^(-3)`D. (d) `2.0 mol^(-1) dm^(3)` |
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Answer» Correct Answer - a For first order reaction, `K=2.303/t_(1//2)"log"a/(0.5a)` `=2.303/t_(1//2)log_(10)2=0.693/t_(1//2)` `:. t_(1//2)=0.693/K`, `:. K_(1)=0.693/t_(1//2)=0.693/40` for zero order reaction, `:. K_(0)=A_(0)/(2xxt_(1//2))=1.386/(2xx20)` Now, `K_(1)/K_(0)=0.693/40xx40/1.386=0.5 mol^(-1) dm^(3)` |
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| 1969. |
Consider the reaction, `Cl_(2)(aq) + H_(2)S(aq) rarr S(s) + 2H^(+) (aq) + 2Cl^(-)(aq)` The rate equation for this reaction is, Rate `=k [Cl_(2)][H_(2)S]` Which of these mechanisms is`//`are consistent with this rate equation ? (I) `Cl_(2) + H_(2)S rarr H^(+) + Cl^(-) + Cl^(+) + HS^(-)` (slow) `Cl^(+) + HS^(-) rarr H^(+) + Cl^(-) + S` (fast) (II) `H_(2)S hArr H^(+) +HS^(-)` (fast equilibrium) `Cl^(+) + HS^(-) rarr 2 Cl^(-) + H^(+) + S` (slow)A. (a) `2 only`B. (b) `Both 1 and 2`C. (c ) `Neither 1 nor 2`D. (d) `1 only` |
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Answer» Correct Answer - d Rate depends upon the slow step, thus slow step should involve `1` molecule of `Cl_(2)` and `1` molecule of `H_(2)S`. Hence, `1` is the correct mechanism. Rate=`K[Cl_(2)][H_(2)S]` |
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| 1970. |
For a first order reaction `A rarr P`, the temperature `(T)` dependent rate constant `(k)` was found to follow the equation `log k = -2000(1//T) + 6.0`. The pre-exponential factor `A` and the activation energy `E_(a)`, respective, areA. `1xx10^(6)s^(-1) and 9.2 kJ "mol"^(-1)`B. `6s^(-1) and 16.6kJ "mol"^(-1)`C. `1xx10^(6)s^(-1) and 16.6 kJ "mol"^(-1)`D. `1xx10^(6)s^(-1) and 38.3 kJ "mol"^(-1)` |
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Answer» Correct Answer - D `log k= log_(10)A-(E_(a))/(2.303)xx(1)/(T)` |
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| 1971. |
For a first order reaction `A rarr P`, the temperature `(T)` dependent rate constant `(k)` was found to follow the equation `log k = -2000(1//T) + 6.0`. The pre-exponential factor `A` and the activation energy `E_(a)`, respective, areA. `1.0 xx 10^(6)s^(-1)` and `9.2 kJ mol^(-1)`B. `6.0 s^(-2)` and `16.6 kJ mol^(-1)`C. `1.0 xx 10^(6) s^(-1)` and `16.6 kJ mol^(-1)`D. `1.0 xx 10^(6) s^(-1)` and `38.3 kJ mol^(-1)` |
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Answer» Correct Answer - D d) According to Arrhenius equation, `k=Ae^(-Ea//RT)` In k = In `A-(E_(a))/(RT)` `2.303 logk = 2.303 log A - (E_(a))/(RT)` log k `= log A-(E_(a))/(2.303RT)` On comparing with the given equation. `logk = 6.0 - (2000) xx 1/T` log A = 6.0, `A="Antilog"6 = 10^(6)` `E_(a)/(2.303R)= 2000` `E_(a) = 2000 xx 2.303 xx R` `=2000 xx 2.303 xx 8.314 J` =`38.29 kJ`. |
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| 1972. |
For a first order reaction `A rarr P`, the temperature `(T)` dependent rate constant `(k)` was found to follow the equation `log k = -2000(1//T) + 6.0`. The pre-exponential factor `A` and the activation energy `E_(a)`, respective, areA. ` 1.0 xx 10^(6) s^(-1) and 9.2kJ mol^(-1)`B. ` 6.0 s^(-1) and 16.6kJ" "mol^(-1)`C. ` 1.0 xx 10^(6) s^(-1) and 16.6kJ" "mol^(-1)`D. ` 1.0 xx 10^(6) s^(-1) and 38.3kJ" "mol^(-1)` |
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Answer» Correct Answer - d |
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| 1973. |
For the chemical reaction `XrarrY`, it is found that the rate of reaction increases by `2.25` times when the concentration of `X` is increased by `1.5` times, what is the order w.r.t. `X` ?A. `0`B. `1`C. `2`D. `3` |
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Answer» Correct Answer - C `XrarrY` `Rate_(1) = k[X]^(m)`, `Rate_(2) = 2.25 Rate_(1) = k(1.5[X])^(m)` `rArr (1.5)^(m) = 2.25 rArr m=2` |
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| 1974. |
K is rate constant at temp T then value of `underset(Ttooo)lim` log K is equal to :-A. `A//2.303`B. AC. `2.303A`D. log A |
| Answer» Correct Answer - D | |
| 1975. |
In gaseous reaction, important for the understanding of the upper atmosphere `H_(2)O` and O react bimolecularly to from two OH readicals. `DeltaH` for this reaction is 72 kJ at 500 K and `E_(a)` is 77 kJ `"mol"^(-1)`, then `E_(a)` for the bimolecular recombination of two OH readicals to form `H_(2)O` and O is :A. `3 kJ "mol"^(-1)`B. `4 kJ "mol"^(-1)`C. `5 kJ "mol"^(-1)`D. `7 kJ "mol"^(-1)` |
| Answer» Correct Answer - C | |
| 1976. |
For a first order reaction `A rarr P`, the temperature `(T)` dependent rate constant `(k)` was found to follow the equation `log k = -2000(1//T) + 6.0`. The pre-exponential factor `A` and the activation energy `E_(a)`, respective, areA. (a) `1.0xx10^(6) s^(-1)` and `9.2 kJ mol^(-1)`B. (b) `6.0 s^(-1)` and `16.6 kJ mol^(-1)`C. (c ) `1.0xx10^(6) s^(-1)` and `16.6 kJ mol^(-1)`D. (d) `1.0xx10^(6) s^(-1)` and `38.3 kJ mol^(-1)` |
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Answer» Correct Answer - d `log k=-(2000)1/T+6.0 …(1)` and, ln k=ln `A-E_(a)/(RT)` `2.303log k=2.303 log A-E_(a)/(RT)` `log k=(-E_(a))/(2.303R)xx1/T+log A …(2)` By eqs. (1) and (2), `(-E_(a))/(2.303R)=-2000` `:. E_(a)=2.303xx8.314xx2000=38.29 kJ` and `log A=6, A=10^(6)` |
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| 1977. |
Rate of a reaction can be expressed by Arrhenius equation as: `k = Ae^(-E_(a)//RT)` In this equation, `E_(a)` represents:A. (a) The energy above which all the colliding molecules will reactB. (b) The energy below which colliding molecules will not reactC. (c ) The total energy of the reactant molecules at a temperature `T`D. (d) the fraction of molecules with enrgy greater than the activation energy of the reaction` |
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Answer» Correct Answer - b `E` is activation energy. Thus order with respect to NO is `2`. If `E_(a)+E_(R)=E_("Threshold enrgy")`, then only molecule may react. |
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| 1978. |
The catalytic decomposition of `H_(2)O` was studied by titrating it at different intervals with `KMnO_(4)` and the following data were obtained : `{:("t (second)",0,600,1200),("V of KMmO"_(4)(mL)",22.8,13.8,8.3):}` Calculate the velocity constant for the reaction assuming it to be a first order reaction. |
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Answer» Correct Answer - `8.4xx10^(-3)s^(-1)` Volume of `KMnO_(4)` used `prop` concentration of `H_(2)O_(2)` Hence, `k=(2.303)/(t)log_(10).(V_(0))/(v_(t))` |
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| 1979. |
For a reaction , `2N_(2)O_(5)rarr4NO_(2)+O_(2),` the rate is directly proportional to `[N_(2)O_(5)].` At `45^(@)C, 90%` of the `N_(2)O_(5)` react in 3600 s. The value of the rate constant isA. `3.2xx10^(-4)s^(-1)`B. `6.4xx10^(-4)s^(-1)`C. `8.5xx10^(-4)s^(-1)`D. `12.8xx10^(-4)s^(-1)` |
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Answer» Correct Answer - B Reaction is first order. So, `In(a)/(a-x)=kt` Where, a=initial concentration =100 (left) a-x =concentration at time t=10 `(becausex=90%)` `In(100)/(10)=kxx3600" "(t=3600s)` `rArr2.303log10=kxx3600as(log10=1)` `k=(2.303)/(3600)=6.4xx10^(-4)s^(-1)` |
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| 1980. |
The accompanying figure depicts a change in concentration of species `A` and `B` for the reaction `ArarrB`, as a function of time. The point of inter section of the two curves represents A. `T_(1//2)`B. `T_(3//4)`C. `T_(2//3)`D. data insufficient to predict |
| Answer» Correct Answer - A | |
| 1981. |
For a first order reaction `A rarr P`, the temperature `(T)` dependent rate constant `(k)` was found to follow the equation `log k = -2000(1//T) + 6.0`. The pre-exponential factor `A` and the activation energy `E_(a)`, respective, areA. `1.0xx10^(6)s^(-1) and 9.2 kJ mol^(-1)`B. `6.0 s^(-1) and 16.6kJ mol^(-1)`C. `1.0xx10^(6)s^(-1) and 16.6 kJ mol^(-1)`D. `1.0xx10^(6) s^(-1) and 38.3 kJ mol^(-1)` |
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Answer» Correct Answer - D According to Arrhenius equation `K=Ae^(-E_(a)//RT)` Taking logarithm of both the sides, we get log `k=logA-(E_(a))/(2.303RT)` We are given log `k=6.0-(2000)(1)/(T)` comparing the two, we obtain log `A=6.0` or `A=10^(6)sec^(-1)` `(E_(a))/(2.303R)=2000` or `E_(a)=(2000)(2.303)(8.14JK^(-1)mol^(-1))` `=38.3KJmol^(-1)` |
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| 1982. |
Conisder a reaction `aG+bH rarr` Products. When concentration of both the reactants `G` and `H` is doubled, the rate increases eight times. However, when the concentration of `G` is doubled, keeping the concentration of `H` fixed, the rate is doubled. The overall order of reaction isA. `0`B. `1`C. `2`D. `3` |
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Answer» Correct Answer - D Doubling the concentration of both `G` and `H` increases the rate of the reaction by eight times. Doubling the concentration of only `G` also doubles the rate of the reaction, so the order with respect to `G` is one. It means that the rate of the reaction is also dependent on the second power of the concentration of `H`. So the rate law is `r = k[G][H]^(2)` and the order is `3`.Correct Answer - D Doubling the concentration of both `G` and `H` increases the rate of the reaction by eight times. Doubling the concentration of only `G` also doubles the rate of the reaction, so the order with respect to `G` is one. It means that the rate of the reaction is also dependent on the second power of the concentration of `H`. So the rate law is `r = k[G][H]^(2)` and the order is `3`. |
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| 1983. |
Rate of a reaction can be expressed by Arrhenius equation as: `k = Ae^(-E_(a)//RT)` In this equation, `E_(a)` represents:A. the fraction of molecules with energy greater than the activation energy of the reaction.B. the total energy of the reacting molecules at a temperature, TC. the energy below which colliding molecules will not reactD. the energy above which all the coliding molecules will react |
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Answer» Correct Answer - C `E_(0)` is energy of activation, the energy required to form activated complex let or transition state.Thus, below this energy, colliding molecules will react. |
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| 1984. |
The accompanying figure depicts a change in concentration of species `A` and `B` for the reaction `ArarrB`, as a function of time. The point of inter section of the two curves represents A. `t_(1//2)`B. `t_(3//4)`C. `t_(2//3)`D. Data insufficient to predict |
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Answer» Correct Answer - A The intersection point indicates that half life to the reactant `A` is converted to `B`. |
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| 1985. |
For a first order reaction : `ArarrB` with initial concentration =aA. `t_(1//2)=(k)/(a)`B. `t_(3//4)=2t_(1//3)`C. `t_(1//2)=(0.693)/(k)`D. `t_(1//2)=kxx0.693` |
| Answer» Correct Answer - B::C | |
| 1986. |
In reaction, `ArarrB +C` the following data were obtained. `{:("t in seconds",0,900,180),("Concentration of A",50.8,19.7,7.62):}` Prove that it is a first order reaction. |
| Answer» Calculate the value of K in both cases by using first order equation. | |
| 1987. |
The reaction `2N_(2)O_(5)(g)rarr4NO_(2)(g)+O_(2)(g)` is found to be firt order with respect to `N_(2)O_(5)(g)`. Which of the following is correct ?A. `log(p_(N_(2)O_(5)))` versus time with -ve slopeB. `(p_(N_(2)O_(5)))^(-1)` versus timeC. `(p_(N_(2)O_(5)))` versus timeD. `log(p_(N_(2)O_(5)))` versus tiem with +ve slope |
| Answer» Correct Answer - A | |
| 1988. |
Conisder a reaction `aG+bH rarr` Products. When concentration of both the reactants `G` and `H` is doubled, the rate increases eight times. However, when the concentration of `G` is doubled, keeping the concentration of `H` fixed, the rate is doubled. The overall order of reaction is |
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Answer» Correct Answer - D `aG+bHrarr`products The rate law is `Rate_(1)=k[G]^(x)[H]^(y)` Rate^(2)=k[2G]^(x)[2H]^(y)` Dividing rate by we get (Rate_(2))/(Rate_(1))=(8Rate_(1))/(Rate_(1))=(k[2g]^(x)[2H]^(y))/(k[G]^(x)[H]^(y))` `8=2^(x).2^(y)` `2^(3)2^(x+y)` :. Overall order is `3` . We need second information only if we asked about individual orders. |
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| 1989. |
The accompanying figure depicts a change in concentration of species `A` and `B` for the reaction `ArarrB`, as a function of time. The point of inter section of the two curves represents A. `t_(1//2)`B. `t_(3//4)`C. `t_(2//3)`D. `t_(1//4)` |
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Answer» Correct Answer - a The intersection point indicates that half of the reaction `X` is converted into Y`. |
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| 1990. |
The reaction `2N_(2)O_(5)(g)rarr4NO_(2)(g)+O_(2)(g)` follows first order kinetics. The pressure of a vessel containing only `N_(2)O_(5)` was found to increase from 50 mm Hg to `87.5` mm Hg in 30 mim. The pressure exerted by the gases after 60 min. will be (Assume temperature remains constant):A. `106.25` mm HgB. `116.25` mm HgC. 125 mm HgD. `150 mm Hg |
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Answer» Correct Answer - A |
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| 1991. |
The decomposition of `N_(2)O_(5)` occurs as , `2N_(2)O_(5) to 4 NO_(2) + O_(2)` , follows `I^(st)` order kinetics ,henceA. The reaction is unimolecularB. The reaction is bimolecularC. `T_(1//2) prop a^(0)`D. None of these |
| Answer» Correct Answer - c | |
| 1992. |
Assertion: Order of reaction can never be fractional for an elementary reaction. Reason: An elementary reaction takes place by one step mechanism.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but the reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A AN elementary reaction is one step reaction and in such reaction order of reaction and molecularity are same. Note that molecularity can never be fractional. |
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| 1993. |
The rate and mechamical reaction are studied in chemical kinetics. The elementary reactions are single step reaction having no mechanism. The order of reaction and molecularity are same for elementary reactions. The rate of forward reaction `aA + bBrarr cC+dD` is given as: `rate =((dx)/(dt))=-1/a(d[A])/(dt)=-1/b(d[B])/(dt)=1/c(d[C])/(dt)=1/d(d[D])/(dt)` or expression can be written as : rate `=K_(1)[A]^(a)[B]^(b)-K_(2)[C]^(c )[D]^(d)`. At equilibrium, rate =`0`. The constants `K, K_(1), K_(2)` are rate constants of respective reaction. In case of reactions governed by two or more steps reaction mechanism, the rate is given by the slowest step of mechanism. For a hypothetical reaction `aA+bBrarr` Product, the rate law is: rate `=K[A]^(x)[B]^(y)`, then:A. (a) `(a+b)=(x+y)`B. (b) `(a+b) lt (x+y)`C. (c ) `(a+b) gt (x+y)`D. (d) Any of these |
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Answer» Correct Answer - d Order of reaction does not depend upon stoichiometry. It is an experimental value. |
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| 1994. |
The rate and mechamical reaction are studied in chemical kinetics. The elementary reactions are single step reaction having no mechanism. The order of reaction and molecularity are same for elementary reactions. The rate of forward reaction `aA + bBrarr cC+dD` is given as: `rate =((dx)/(dt))=-1/a(d[A])/(dt)=-1/b(d[B])/(dt)=1/c(d[C])/(dt)=1/d(d[D])/(dt)` or expression can be written as : rate `=K_(1)[A]^(a)[B]^(b)-K_(2)[C]^(c )[D]^(d)`. At equilibrium, rate =`0`. The constants `K, K_(1), K_(2)` are rate constants of respective reaction. In case of reactions governed by two or more steps reaction mechanism, the rate is given by the slowest step of mechanism. At the point of intersection of the two curve shown for the reaction: `Ararr nB` the concentration of `B` is given by: A. (a) `(nA_(0))/2`B. (b) `(A_(0))/(n-1)`C. (c ) `(nA_(0))/(n+1)`D. (d) `((n-1))/((n+1))A_(0)` |
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Answer» Correct Answer - c At intersection, `[A]_("left")=[B]_("formed")` `underset(A_(0))(A)rarr underset(0)(nB)` `A_(0)(1-x)=nA_(0)x` `:. X=1/(n+1)` `:. [B]=(n.A_(0))/((n+1))` |
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| 1995. |
Calculating the average reaction rate : Calculate the average rate of decomposition of `N_(2)O_(5)` by the reaction `2N_(2)O_(5)(g)rarr4NO_(2)(g)+O_(2)(g)` during the time interval from `t=600s` `t=1200s`, using the following data: Time , `[N_(2)O_(5)]` `600s` , `1.24xx10^(-2)M` `1200s` , `0.93xx10^(-2)M` Strategy : Calculater a `Delta` quantity in concentration, `[N_(2)O_(5)]`, by taking the final value minus the value. Then, divide it by the time interval, `Delta t`. |
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Answer» Average rate of decomposition of `N_(2)O_(5)` `= -(Delta[N_(2)O_(5)])/(Delta t)` `= - ((0.93-1.24)xx10^(-2)M)/((1200-600)s)` `= -(-0.31xx10^(-2)M)/(600s)= 5.2xx10^(-6)M//s` |
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| 1996. |
The decomposition of `N_(2)O_(5)` occurs as, `2N_(2)O_(5) rarr4NO_(2)+O_(2)` and follows `I` order kinetics, hence:A. The reaction is bimolecularB. The reaction is unimolecularC. `t_(1//2) prop a^(0)`D. None of these |
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Answer» Correct Answer - c For I order reaction, half-life is independent of concentration. |
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| 1997. |
Conisder a reaction `X + Y rarr` Products. If the initial concentration of `X` increased to four times of its original value, keeping the concentration of `Y` constant, the rate of reaction increases four-fold. When the concentration of both `X` and `Y` becomes four times their original values the rate of reaction becomes `16` times its original values. The observed rate law isA. `k[X]^(2)[Y]^(2)`B. `k[X]^(1)[Y]^(2)`C. `k[X]^(1)[Y]^(1)`D. `k[X]^(2)[Y]^(1)` |
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Answer» Correct Answer - C `r_(1) = k[X]^(m)[Y]^(n)` …(i) `r_(2) = 4r_(1) = k[4X]^(m)[Y]^(n)` …(ii) `r_(3) = 16r_(1) = k[4X]^(m)[4Y]^(n)` …(iii) Divide Eq. (ii) by Eq. (i), `4 = (4)^(m) rArr (4)^(1) = (4)^(m) rArr m = 1` Divide Eq. (iii) by (ii), `4 = (4)^(n) rArr (4)^(1) = (4)^(n) rArr n = 1` The observed rate law `= k[X]^(1)[Y]^(1)` |
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| 1998. |
The number of molecules that react with each in an elementary step gives a measure of theA. order of the reactionB. molecularity of the reactionC. stoichiometry of the reactionD. activation energy of the reaction |
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Answer» Correct Answer - B Elementary reactions are classfied according to their molecularity. The number of reacting species (atoms, molecules or ions) taking part in an elementary reaction, which must colllide simultaneously in order to bring about a chemical reaction is called molecularity of a reaction. |
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| 1999. |
In a first order reaction, the concentration of the reactant decreases form `0.8 M` to `0.4 M` in `15 min`. The time taken for the concentration to change form `0.1 M` to `0.025 M` isA. `7.5` minutesB. `15` minutesC. `30` minutesD. `60` minutes |
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Answer» Correct Answer - C The concentration of the reactants decrease from `0.8` to `0.4` in `15 min` i.e., `T_(1//2) = 15 min`, concentration from `0.1 m` to `0.025` will fall in `2` half lives so total time taken `= 2 xx T_(1//2) = 2 xx 15 = 30 min`. |
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| 2000. |
The reaction `Ararr B` follows first order kinetics. The time taken for `0.8 mol` of `A` to produce `0.6 mol` of `B` is `1 hr`. What is the time taken for the conversion of `9.0 mol` of `A` to Product `0.675 mol` of `B` ?A. `1 hr`B. `0.5 hr`C. `0.25 hr`D. `2 hr` |
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Answer» Correct Answer - A For the reaction `2A + B rarr 3C + D` Rate constant of first-order reaction `k = (2.303)/(t) "log"_(10)((A)_(0))/((A)_(t))` or `k = (2.303)/(1) xx"log"_(10) (0.8)/(0.2)` …(i) (because `0.6` mole of `B` is formed) Suppose `t_(1)` and hour is required for changing the concentration of `A` from `0.9` mole to `0.675` mole of `B`. Remaining mole of `A 0.9 - 0.675 = 0.225` `:. k = (2.303)/(t) "log"_(10)(0.9)/(0.225)` Form Eqs (i) and (ii), we get `k = (2.303)/(t) "log"_(10)(0.8)/(0.2) = (2.303)/(t_(1)) "log"_(10)(0.9)/(0.225)` `2.303 log_(10) 4 = (2.303)/(t) log_(10) 4` `t_(1) = 1 hr` |
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