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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.
| 851. |
Assertion (A) : The rate constant of a zero order reaction has same units as the rate of reaction. Reason (R ): Rate constant of a zero order reaction does not depend upon the units of concentration.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 - C Correct statement -II: `k = "conc."//"time" = "mol L"^(-1) s^(-1)`, i.e, it depends upon units of concetration. |
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| 852. |
Assertion: In the reaction `N_(2) + 3H_(2) rarr 2NH_(3)`, the rate of reaction is different in terms of `N_(2), H_(2)` and `NH_(3)`. Reason: Rate of disapperenace of `N_(2)` and `H_(2)` and rate of formation of `NH_(3)` are not equal to each other.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 - D Correct statement -I: The rate of reaction is same in terms of different reactants and products. |
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| 853. |
Assertion: Emission of light as result of exposure of `P` to air in night is called phosphorescence. Reason: Phosphorus burns in `O_(2)` to give `P_(2)O_(5)` and conversion of chemical energy into light energy producing cold light and the phenomenon is called chemiluminiscence.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 - D The phenomenon is called chemiluminiscence. |
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| 854. |
Write untis of rate constant for first and second order reactions. |
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Answer» i) For first order reaction, units of k = `mol^(-1)Ls^(-1)` for second order reaction, units of k = `mol^(-1)Ls^(-1)` |
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| 855. |
For a reaction : `H_(2) + CI_(2) overset(hv)(to)2HCI` Rate = k (i) Write the order and molecularity of this reaction. (ii) Write the unit of k. |
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Answer» i) Order = zero, molecularity = 2 ii) Units of k = `mol^(-1)L^(-1)s^(-1)` |
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| 856. |
For the nuclear reaction, `._(Z)^(A)X to ._(Z-4)^(A-8)Y, ` " " `t_(1//2=1600yrs.` If initial activity was `10^(7)` dps, how many `alpha-"particles"` will be emitted per second after 4800yrs?A. `1.25xx10^(6)s^(-1)`B. `2.5xx10^(6)s^(-1)`C. `1.25xx10^(7)s^(-1)`D. `5xx10^(7)s^(-1)` |
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Answer» Correct Answer - B |
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| 857. |
Assertion: The emission of light during burning of `P` in `O_(2)` is called chemiluminescence. Reason: The chemical energy is converted into light energy.A. (a) `S` is correct but `E` is wrongB. (b) `S` is wrong but `E` is correctC. (c ) Both `S` and `E` are correct and `E` is correct explanation of `S`D. (d) Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
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Answer» Correct Answer - C Chemiluminscence refers for emission of light as a result of chemical change. |
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| 858. |
Statement: Photosynthesis in plants involves reaction of `CO_(2)` and `H_(2)O` in presence of light and chlorophyll. Explanation: It is chlorophyll which absorbs light and possess this energy to reactant molecules.A. (a) `S` is correct but `E` is wrongB. (b) `S` is wrong but `E` is correctC. (c ) Both `S` and `E` are correct and `E` is correct explanation of `S`D. (d) Both `S` and `E` are correct but `E` is not correct explanation of `S`. |
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Answer» Correct Answer - C Chlorophyll acts as photosensitizer. Neigher `CO_(2)` nor `H_(2)O` absords light. |
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| 859. |
Calculate the half life of the first-order reaction: `C_(2)H_(4)O(g) rarr CH_(4)(g)+CO(g)` The initial pressure of `C_(2)H_(4)O(g)` is `80 mm` and the total pressure at the end of `20 min` is `120 mm`.A. 40 minB. 120 minC. 20 minD. 80 min |
| Answer» Correct Answer - C | |
| 860. |
The correct nature of plot for first order reaction is (are):A. B. C. D. |
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Answer» Correct Answer - A::D `k = (2.303)/(t)log.(c_(0))/(c )` `(k.t)/(2.303) = log c_(0) - log c` `log c = (-(k)/(2.303))t + log c_(0)` Thus, plot of `log c` vs `t` is a straight with negative slope. `t_(1//2) = x` should remain constant. `-(dc)/(dt) = k.c`, `log(-(dc)/(dt)) = log k + log c` So, the plot of `log(-(dc)/(dt))` vs `log c` should be a straight line of the slope equal to unity. |
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| 861. |
Calculate the half life of the first-order reaction: `C_(2)H_(4)O(g) rarr CH_(4)(g)+CO(g)` The initial pressure of `C_(2)H_(4)O(g)` is `80 mm` and the total pressure at the end of `20 min` is `120 mm`.A. `40 min`B. `120 min`C. `20 min`D. `80 min` |
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Answer» Correct Answer - C `{:(,C_(2)H_(4)O(g) ,rarr,CH_(4)(g)+,CO(g)),(t=0,P_(0),,0,0),(t=t,P_(0)-x,,x,2x):}` Total pressure `(P_(t)) = P_(0) - x + x + x = P_(0) + x` `P_(0) = 80 mm` `P_(0) + x = 120 mm` `:. x = 120 - 80 = 40 mm` `k = (2.203)/(20)log.(P_(0))/(P_(0)-x) = (2.3)/(20)log.(80)/(80-40)` `k = (2.3 xx 0.3)/(20) rArr t_(1//2) = (0.69)/(k) = (0.69 xx 20)/(0.69) = 20 min` |
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| 862. |
The rate constant for zero order reaction isA. `A. k = (c_(0))/(2t)`B. `B..k = ((c_(0)-c_(t)))/(t)`C. `C. k = ln((c_(0)-c_(t))/(2t))`D. `D. k = (c_(0))/(c_(t))` |
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Answer» Correct Answer - B For zero order reaction. `(-d[A])/(dt) = k` `underset(c_(0))overset(c_(t))int-d[A] = kunderset(0)overset(t)int dt` `c_(0) - c_(t) = kt rArr k = ((c_(0)-c_(t))/(t))` |
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| 863. |
The converison of vinyl allyl ether to pent-`4`-enol follows first order kinetics. The following plot is obtained for such a reaction: The rate constant for the reaction isA. `4.6 xx 10^(-2) s^(-1)`B. `1.2 xx 10^(-2) s^(-1)`C. `2.3 xx 10^(-2) s^(-1)`D. `8.4 xx 10^(-2) s^(-1)` |
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Answer» Correct Answer - C `k = (2.303)/(t)log.(a)/(a-x)` `(kt)/(2.303) = log.a-log (a-x)` `:. log (a-x) = log a-(kt)/(2.303)` `log(a-x) = log a -(k)/(2.303)t` (Equation of straight line) `y = mx + c` Slope `= (-k)/(2.303) = (-6)/(10 xx 60)` `:. k = 2.303 xx 10^(-2)s^(-1)` |
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| 864. |
Gaseous cyclobutane isomerizes to butadiene in a first order process which has `k` value at `153^(@)C` of `3.3 xx 10^(-4) s^(-1)`. How many minutes would it take for the isomerization to proceeds `40%` to completion at this temperature ?A. `A. 26 min`B. `B. 52 min`C. `C. 13 min`D. D. None of these |
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Answer» Correct Answer - A `t = (2.3)/(3.3 xx 10^(-4) xx 60 min^(-1)) log.((100)/(100-40))` Solve for `t ~~ 26 min`. |
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| 865. |
The isomerization of cyclopropane to form propane is a first order reaction. At `760 K, 85%` of a sample of cyclopropane change to propane in `79 min`. Calculate the value of the rate constant.A. `A. 3.66 xx 10^(-2) min^(-1)`B. `B. 1.04 xx 10^(-2) min^(-1)`C. `C. 2.42 min^(-1)`D. `D. 2.40 xx 10^(-2) min^(-1)` |
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Answer» Correct Answer - D `:. k = (2.3)/(79 min) log.((100)/(100-85))` Solve for `k ~~ 2.40 xx 10^(-2) min^(-1)` |
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| 866. |
The rate constant for an isomerization reaction, `A rarr B` is `4.5 xx 10^(-3) min^(-1)`. If the initial concentration of `A` is `1 M`, calculate the rate of the reaction after `1 h`.A. `0.763 M`B. `1.763 M`C. `2.763M`D. `0.076M` |
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Answer» Correct Answer - 1 From the unit of rate constant, it follows that the given isomerization reaction is a first order reaction. The concentration of A after `1h` as determined from the expression. `log.([A])/([A]_(0))=-(k)/(2.303)t` `log.([A])/([A]_(0))=-(4.5xx10^(-3)mi n^(-1))/(2.303)(60 mi n)=-0.1172 mol L^(-1)" "[ As[A_(0)]=1M]` `:. " "[A]=0.7634M` |
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| 867. |
The rate constant for an isomerization reaction, `A rarr B` is `4.5 xx 10^(-3) min^(-1)`. If the initial concentration of `A` is `1 M`, calculate the rate of the reaction after `1 h`. |
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Answer» We know, `Ararr B , K=4.5xx10^(-3) min^(-1)` `[A]_(0)=1 M` For I order reaction, `K=2.303/t"log"_(10) a/((a-x))` at `t=60 min, (a-x)=?` `4.5xx10^(-3)=2.303/60"log"_(10) 1/((a-x))` `:. (a-x)=0.7634` Thus, rate after `60` minute=`K (a-x)` `=4.5xx10^(-3)xx0.7634=3.4354xx10^(-3)` |
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| 868. |
The rate constant `k`, for the reaction `N_(2)O_(5)(g) rarr 2NO_(2) (g) + (1)/(2) O_(2)(g)` is `2.3 xx 10^(-2) s^(-1)`. Which equation given below describes the change of `[N_(2)O_(5)]` with time ? `[N_(2)O_(5)]_(0)` and `[N_(2)O_(5)]_(t)` correspond to concentration of `N_(2)O_(5)` initially and at time, `t` ?A. `[N_(3)O_(5)]_(0)=[N_(2)O_(5)]_(t)e^(kt)`B. `log_(e).([N_(2)O_(5)]_(0))/([N_(2)O_(5)])=kt`C. `log_(10)[N_(2)O_(5)]_(t)=log_(10)[N_(2)O_(5)]_(0)-kt`D. `[N_(2)O_(5)]_(t)=[N_(2)O_(5)]_(0)+kt` |
| Answer» Correct Answer - B | |
| 869. |
Cyclopropane rearranges to form propene `to CH_(3) - CH = CH_(2)` This follows first order kinetics. The rate constant is `2.714 xx 10^(-3) sec^(-1)`. The unitial concentration of cyclopropane is 0.29 M. What will be the concentraion of cyclopropane after 100 sec?A. 0.035 MB. 0.22 MC. 0.145 MD. 0.0018 M |
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Answer» Correct Answer - B `k = (2.303)/(t) "log" (a)/((a - x))` (a - x) is the concentration left after 100 sec. `2.7 xx 10^(-3) = (2.303)/(100) "log" (0.29)/((a - x))` `implies (0.27)/(2.303) = "log" (0.29)/((a -x)) implies 0.117 = "log" (0.29)/((a - x))` `implies (a - x) = 0.22 M` |
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| 870. |
Statement-1 : Conservation of a `gamma` photon into an electron and a positron is an example of pair production. Statement-2: Pair production refers to the creation of an elementary particle and its antiparticle , usually when a photon interacts with a nucleus.A. Statement-1 is True, statement-2 is True, Statement-2 is a correct explantion for Statement-1.B. Statement-1 is True, Statement-2 is True, Statement-2 is NOT a correct explanation for Statement-1C. Statement-1 is True, Statement-2 is False.D. Statement-1 is False, Statement-2 is True. |
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Answer» Correct Answer - a |
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| 871. |
A lump of coal burns at moderate rate in air while coal dust burns explosively. Explain. |
| Answer» In case of coal dust, greater surface area is available as compared to the solid coal This means that coal dust will react faster than the lump of coal. Since combustion is an exothermic reaction, greater amount of heat evolved in the combustion of coal dust leads to explosion also. | |
| 872. |
For the reaction `N_(2)O_(5) rarr 2NO_(2) + (1)/(2) O_(2)`, the rate of disappearance of `N_(2)O_(5)` is `6.25 xx 10^(-3) "mol L"^(-1) s^(-1)`. The rate of formation of `NO_(2)` and `O_(2)` will be respectively.A. `1.25xx10^(-2)mol L^(-1)s^(-1) and 6.25xx10^(-3)mol^(-1)L^(-1)s^(-1)`B. `6.25xx10^(-3)mol L^(-1)s^(-1) and 6.25xx10^(-3)mol^(-1)L^(-1)s^(-1)`C. `1.25xx10^(-2)mol L^(-1)s^(-1)and 3.125xx10^(-3) mol^(-1)L^(-1)s^(-1)`D. `6.25xx10^(-3)mol L^(-1)s^(-1) and 3.125xx10^(-3)mol^(-1)L^(-1)s^(-1)` |
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Answer» Correct Answer - C For the given reaction `N_(2)O_(5)(g)rarr2NO_(2)(g)+(1)/(2)O_(2)(g)` the rate can be expressed in terms of any of the three species (reactant or products): Rate `=(-d[N_(2)O_(5)])/(dt)=(d[NO_(2)])/(2dt)=(d[O_(2)])/(1//2dt)` Thus, rate of information of `NO_(2)` is given as: `(d[NO_(2)])/(dt)=2((-d[N_(2))_(5)])/(dt))` `=2.(6.25xx10^(-3)molL^(-1)s^(-1))` `=12.5xx10^(-3)molL^(-1)s^(-1)` `=1.25xx10^(-2)molL^(-1)s^(-1)` The rate of formation of `O_(2)` is given as `(d[O_(2)])/(dt)=(1)/(2)(-(d[N_(2)O_(5)])/(dt))` `=(1)/(2)(6.25xx10^(-3)molL^(-1)s^(-1))` `=3.125xx10^(-3)molL^(-1)s^(-1)` |
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| 873. |
which one of the following statements for the order of a raction is incorrect ?A. order is not influenced by stoichliometric coefficient of the reactantsB. Order of reaction is sum of power to the concentration terms of reactants to express the rate of reactionC. Order of reaction is always whole numberD. Order can be determined only experimentally |
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Answer» Correct Answer - C Order of reaction may be zero whole number or fraction number . |
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| 874. |
Which one of the following statements for the order of a reaction is incorrect ?A. Order of reaction is always whole numberB. Order can be determined only experimentallyC. Order is not influenced by stoichiometric coefficient of the reactantsD. Order of reaction is sum of power to the concentration terms of reactants to express the rate of reaction |
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Answer» Correct Answer - A Order of reaction may be fractional. |
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| 875. |
During the kinetic study of the reaction `2A +B rarr C + D` following results were obtained. `{:(, Run[A],[B] i n M,"Initial rate of fo rmation of D in "m s^(-1),),(I,0.1,0.1,6.0 xx 10^(-3),),(II,0.3,0.2,7.2xx10^(-2),),(III,0.3,0.4,2.88xx10^(-1),),(IV,0.4,0.1,2.40xx10^(-2),):}` On the basis of above data which one is correct ?A. `Rate =k[A][B]^(2)`B. `Rate = k[A]^(2)[B]`C. `Rate = k[A][B]`D. `Rate = k[A]^(2)[B]^(2)` |
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Answer» Correct Answer - A From the run `(I)~ and `(IV)` , we observe that when the concentration of `A` is increased `4` times keeping the times. Thus, the reaction is of first order with respect to `A` . From the run `(II)` and `(III)` , we motice that when the concentration od `B` is doubbled, the rate is increased four times Thus this reaction is of second order with respect to `B` i.e. Rate `=K[A][B]^(2)` Alternatively Rate `=k[A]^(x)[B]^(y)` Considering the run `(I)` and `(IV)` , we have `(2.40xx10^(-2))/(6.0xx10^(-3))=((0.4)^(x)(0.1)^(y))/((0.1)^(x)(1.0)^(y))` `4=4^(x)` implies `4^(1)=4^(x)` or `x=1` Considering the run `(II)` and `(III)` , we have `(2.88xx10^(-1))/(7.2xx10^(-2))=((0.3)^(x)(0.4)^(y))/((0.3)^(x)(0.2)^(y))` `4=2^(y)` `2^(2)=2^(y)` Hence, `y=2` |
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| 876. |
During the kinetic study of the reaction , `2A + B to C+ D ,` following results were obtained Based on the above date which one of the following is correct ?A. `Rate =k[A]^(2)[B]`B. `Rate =K[A][B}`C. `Rate k [A]^(2) [B]^(2)`D. `Rate =k [A][B]^(2)` |
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Answer» Correct Answer - D Let the of reaction with respect to A is x and with respect to B is Y thus , rate `=k|a|^(X)|B|^(y)` ( x and y are stoichiometric coefficient ) for the given cases , `I, Rate=K(0,1)^(X)(0,1)^(y) =6.0xx10^(-3)` `II. rate =K (0,3)^(X)(0.2)^(y)=7.2xx10^(-2)` `III. rate =K(0.3)^(X)(0.40)^(y)=2.88xx10^(-1)` `IV, rate = K(0,4)^(X) (0.1) ^(y) =2.40xx10^(-2)` dividing Eq. (i) by Eq.(iv) we get `((0.1)/(0.4))^(X)((0.1)/(0.1))^(Y)=(6.0xx10^(-3))/(2.4xx10^(-2))` `or ((1)/(4))^(x)=((1)/(4))^(1)` ` therefore x=1` On dividing Eq.(II) by Eq.(ii) we get `((0.3)/(0.3))^(x)((0.2)/(0.4))^(y)=(7.2xx10^(-2))/(2.88xx10^(-1))` `or ((1)/(2))^(y)=(1)/(4)` `or ((1)/(2))^(y) =((1)/(2))^(2)` `therefore y=2` thus rata law is , rate = `K[A]^(1) [B]^(2)` `=k|A|[B]^(2)` |
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| 877. |
For the reaction `N_(2)O_(5) rarr 2NO_(2) + (1)/(2) O_(2)`, the rate of disappearance of `N_(2)O_(5)` is `6.25 xx 10^(-3) "mol L"^(-1) s^(-1)`. The rate of formation of `NO_(2)` and `O_(2)` will be respectively.A. `6.25xx10^(-3)molL^(-1)S^(-1)and 6.25xx10^(-3)molL^(-1)S^(-1)`B. `1.25xx10^(-2)molL^(-1)S^(-1)and 3.125xx10^(-3)molL^(-1)S^(-1)`C. `6.25xx10^(-3)molL^(-1)S^(-1)and 3.125xx10^(-3)molL^(-1)S^(-1)`D. `1.25xx10^(-2)molL^(-1)S^(-1)and 6.25xx10^(-3)molL^(-1)S^(-1)` |
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Answer» Correct Answer - B key idea rate of disappearance of reactant = rate of appearance of product `or (1)/(" Stoichiometric coefficient of reactant")(d"[ reactant]")/(dt)` `=+(1)/(" Stoichiometric coefficient of reactant")(d"[product]")/(dt)` for the reaction , `N_(2) O_(5)(g) to 2NO_(3) (g) +(1)/(2)O_(2)(g)` ` (-d[N_(2)O_(5)])/(dt)=+(1)/(2)(d[NO_(2)])(dt)=+(2d[O_(2)])/(dt)` `therefore (d[NO_(2)])/(dt) =-2(d[N_(2)O_(5)])/(dt)` `=2xx6.25xx10^(-3)molL^(-1)S^(-1)` `=12.5xx10^(-3)mol L^(-1)S^(-1)` `= 1.25xx10^(-2)mol L^(-1)S^(-1)` `(d[O_(2)])/(dt) =- (d[N_(2)O_(5)])/(dt ) xx(1)/(2 ) ` `=(6.25xx10^(-3)mol L^(-1)S^(-1))/(2) ` ` =3.125xx10^(-3)molL^(-1) S^(-1)` |
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| 878. |
During the kinetic study of the reaction `2A +B rarr C + D` following results were obtained. `{:(, Run[A],[B] i n M,"Initial rate of fo rmation of D in "m s^(-1),),(I,0.1,0.1,6.0 xx 10^(-3),),(II,0.3,0.2,7.2xx10^(-2),),(III,0.3,0.4,2.88xx10^(-1),),(IV,0.4,0.1,2.40xx10^(-2),):}` On the basis of above data which one is correct ?A. `r = k[A]^(2)[B]`B. `r = k[A][B]`C. `r = k[A]^(2)[B]^(2)`D. `r = k[A][B]^(2)` |
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Answer» Correct Answer - D Use: `r = K[A]^(m)[B]^(n)` |
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| 879. |
For the reaction `N_(2)O_(5) rarr 2NO_(2) + (1)/(2) O_(2)`, the rate of disappearance of `N_(2)O_(5)` is `6.25 xx 10^(-3) "mol L"^(-1) s^(-1)`. The rate of formation of `NO_(2)` and `O_(2)` will be respectively.A. `6.25 xx 10^(-3) "mol L"^(-1) s^(-1)`and `6.25 xx 10^(-3) "mol L"^(-1) s^(-1)`B. `1.25 xx 10^(-2) "mol L"^(-1) s^(-1)` and `3.125 xx 10^(-3) "mol L"^(-1) s^(-1)`C. `6.25 xx 10^(-3) "mol L"^(-1) s^(-1)` and `3.125 xx 10^(-3) "mol L"^(-1) s^(-1)`D. `1.25 xx 10^(-3) "mol L"^(-1)` and `6.25 xx 10^(-3) "mol L"^(-1) s^(-1)` |
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Answer» Correct Answer - B `-(d[N_(2)O_(5)])/(d t) = (1)/(2)(d[NO_(2)])/(d t) = (2d[O_(2)])/(d t) = R` |
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| 880. |
A chemical reaction has catalyst X. Hence XA. Reduce enthalpy of the reactionB. Dercreases rate constant of the reactionC. increasees activation energy of the reactionD. does not affect equilibrium constant of the reaction |
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Answer» Correct Answer - D Although a cateyst speeds up the reaction but it does not shift the position of equilibrium this is due to the fact that the presence fo catalyst redues the height of barrier by providing an alternative path for the reaction and lowers the activation energy However the lowering in activation energy is to the the same extent for the forward as well as the backward reaction . |
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| 881. |
Calculate half life period for a first order reaction having k=`4min^(-1)` |
| Answer» For first order reactions: `k=0.693/t_(1//2) therefore t_(1//2) = 0.693/k 0.693/(4 min^(-1))=0.173` min. | |
| 882. |
Calculate the half-life of the first order reaction, `C_(2)H_(4)O(g)rarrCH_(4)(g)+CO(g)`. If the initial pressure of `C_(2)H_(4)O(g)` is 80 mm and the total pressure at the end of 20 min is 120 mm.A. 40 minB. 120 minC. 20 minD. 80 min |
| Answer» Correct Answer - C | |
| 883. |
For the reaction , `A+3Brarr2C+d` which one of the following is not correct ?A. Rat or disappearance of A= Rate of formation of DB. Rate of formation of C =`(2)/(3)xx` Rate of disappearane of DC. Rate of formation of D =`(1)/(3)xx` Rate of disappearance of BD. Rate of disappearance of `A=2xx` Rate of formation of C |
| Answer» Correct Answer - D | |
| 884. |
With increase in temperature , rate of reactionA. increasesB. decreasesC. remains sameD. may increase |
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Answer» Correct Answer - A With increase in temperature rate of reaction , increase due to increase in number of molecules having threshold energy. |
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| 885. |
Units of rate constant of first and zero order reactions in terms of molarity `M` are respectively:A. (a) `s^(-1), M s^(-1)`B. (b) `s^(-1), M`C. (c ) `M s^(-1), s^(-1)`D. (d) `M, s^(-1)` |
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Answer» Correct Answer - a `K=(dx)/(dt[A])` for I order, unit `=sec^(-1)` `K=(dx)/(dt)` for zero order, unit= mol `litre^(-1) sec^(-1)` |
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| 886. |
Units of rate constant of first and zero order reactions in terms of molarity `M` are respectively:A. `sec^(-1),"M sec"^(-1)`B. `sec^(-1), M`C. `"M sec"^(-1),sec^(-1)`D. `"M sec"^(-1)` |
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Answer» Correct Answer - A |
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| 887. |
For the first order reaction `2N_(2)O_(5)(g) rarr 4NO_(2)(g) + O_(2)(g)`A. (a) The concentration of the reactant decreases exponentially with timeB. (b) the half-life of the reaction decreases with increasing temperatureC. (c ) the half-life of the reaction depends on the initial concentration of the reactantD. (d) the reaction proceeds to `99.6%` completion in eight half-life duration |
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Answer» Correct Answer - a, b, d For first order reaction `[A]=[A]_(0)e^(-kt)` Hence concentration of `[NO_(2)]` decreases exponentially. Also, `t_(1//2)=0.693/K`. Which is independent of concentration and `t_(1//2)` decreases with the increase of temperature. `t_(99.6)=2.303/K"log"(100/0.4)` `t_(99.6)=2.303/K(2.4)=8xx0.693/K=8t_(1//2)` |
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| 888. |
Units of rate constant of first and zero order reactions in terms of molarity `M` are respectively:A. `sec^(-1),Msec^(-1)`B. `sec^(-1)M`C. `M sec^(-1), sec^(-1)`D. `M, sec^(-1)` |
| Answer» Correct Answer - A | |
| 889. |
Units of rate constant of first and zero order reactions in terms of molarity `M` are respectively:A. `s^(-1),Ms^(-1)`B. `s^(-1)`,MC. `Ms^(-1)`D. `M,s^(-1)` |
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Answer» Correct Answer - A For first order reaction , Unit of k = `s^(-1)` ,brgt For zero order reaction k, Unit of k = mol `L^(-1) s^(-1) = Ms^(-1)` |
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| 890. |
Units of rate constant of first and zero order reactions in terms of molarity `M` are respectively:A. `s^(-1), M L^(-1) s^(-1)`B. `s^(-1), M`C. `Ms^(-1), s^(-1)`,D. `M, s^(-1)`, |
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Answer» Correct Answer - A `K = (dx)/(d t[A])` for `I` order `= sec^(-1)` `k = (d x)/(d t)` for zero order, `= "mol litre"^(-1) sec^(-1)` or `k = ["mol"^(-1) L^(-1)]^(+n) sec^(-1)` `rArr n = 1` `k = sec^(-1)` Zero order, `n = 0` `k = "mol L"^(-1) sec^(-1)` |
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| 891. |
Which of these does not influence the rate of reaction?A. Nature of the reactantsB. Concentration of the reactionC. Temperature of the reactionD. Molecularity of the reaction |
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Answer» Correct Answer - D Nature and concentration of the reactants and temperature of the reaction influence the rate of reaction . But molecularity does not affect the rate of reaction . |
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| 892. |
The order of reaction is decided byA. temperatureB. mechanism of reaction as well as relative concentration of reactantsC. molecularityD. pressure. |
| Answer» Correct Answer - B | |
| 893. |
Order of a reaction is decided byA. PressureB. TemperatureC. MolecularityD. Relative concentration of reactants |
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Answer» Correct Answer - d Order of a reaction is decided by relative concentration of reactants |
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| 894. |
Which of these does not influence the rate of reactionA. Nature of the reactantsB. Concentration of the reactantsC. Temperature of the reactionD. Molecularity of the reaction |
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Answer» Correct Answer - d Molecularity of the reaction does not influence the rate of reaction . |
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| 895. |
Which of the following is an example of pseudo unimolecular reaction ?A. `CH_(3)COOCH_(3) + H_(2)O overset(H^(+))rarrCH_(3)COOH+CH_(3)OH`B. `CH_(3)COOCH_(3) + H_(2)O overset(OH^(-))rarrCH_(3)COOH+CH_(3)OH`C. `2FeCl_(3) +SnCl_(2)rarr SnCl_(4) + 2FeCl_(2)`D. `NaOH +HCl rarr NaCl + H_(2)O` |
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Answer» Correct Answer - A When in any chemical reaction, one of the reactant is present in large excess, then the second-order reaction becomes first-order reaction and is known as pseudo unimolecular reaction e.g., `CH_(3)COOCH_(3) + H_(2)O overset(H^(+))(rarr) CH_(3) COOH + CH_(3) OH` in this reaction molecularity is `2` but order of reaction is found to be first order experimentally, so it is an example of pseudo unimolecular reaction. |
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| 896. |
Which one of the following is wrongly matchedA. Saponification of `CH_(3)COOC_(2)H_(5)` - Second order reactionB. Hydrolysis of `CH_(3)COOCH_(3)` - First order reactionC. Decomposition of `H_(2)O_(2)`- First order reactionD. Combination of `H_(2)` and `Br_(2)` to give HBr- Zero order reaction |
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Answer» Correct Answer - d `H_(2) + Br_(2) hArr 2 HBr ` is a 1.5 order reaction i.e, `k= [H_(2)][Br_(2)]^(1//2)` . |
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| 897. |
Which of the following is wrongly matched ?A. Saponification of `CH_(3)COOCH_(3)`- Second order reactionB. Hydrolysis of `CH_(3)COOCH_(3)`- Pseudo unimolecular reactionC. Decomposition of `H_(2)O_(2)`-First order reactionD. combination of `H_(2)` and `B_(2)` to give `HBr`- Zero order reaction |
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Answer» Correct Answer - D `H(2)+Br_(2) hArr 2HBr` is a `1.5` order reaction i.e., `K = [H_(2)][Br_(2)]^(1//2)` |
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| 898. |
Show by uisng rate laws how much the rate of reaction `2NO(g) + O_(2)(g) rarr 2NO(g)` will change if the volume of the reaction vessel is diminished to `1//3` of its initial volume.A. `1//3` timesB. `2//3` timesC. `3` timesD. `6` times |
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Answer» Correct Answer - C For foloowing reaction, `2NO(g) + O_(2)(g) rarr 2NO_(2) (g)` when the volume of vessel change into `(1)/(3)`, then concentration of reactant become three times `(because c alpha (1)/(v))` The rate of reaction for first order reaction `prop` concentration. So rate of reaction will increases three times. |
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| 899. |
Show by uisng rate laws how much the rate of reaction `2NO(g) + O_(2)(g) rarr 2NO(g)` will change if the volume of the reaction vessel is diminished to `1//3` of its initial volume. |
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Answer» Rate of reaction (r ) = `k[NO]^(2)[O_(2)]` If the volume of the reaction vessel is diminshed to one-third of its volume, the concentration of the reacting species will become three times. Rate of reaction under the new conditions will become twenty seven (27) times. |
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| 900. |
Correct order for a first order reaction isA. `t_(1 //2) prop C^(-1)`B. `t_(1//2) prop C`C. `t_(1//2) prop C^(0)`D. `t_(1//2) prop C^(1//2)` |
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Answer» Correct Answer - c The relation between half-life period and initial concentration (c) for a `n^(th)` order reaction is given by `t_(1//2) prop (1)/(C^(n-1))` for first order reaction (n = 1) . `t_(1//2) prop (1)/(C^(1-1))` or `t_(1//2) prop C^(0)`. |
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