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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.
| 1051. |
The decomposition of a compound is found to follow the first order rate law. If it takes 15 minutes for 20 per cent of the original material to react, calculate i) the specific rate constant ii) the time in which 10 percent of the original material remains unreacted. iii) The time it takes for the next 20 percent of the reactant left to react. |
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Answer» Step. I. Calculation of specific rate constant (k). For the first order reaction, `k=2.303/t log a/(a-x)` `a=100%,x=20%`, (a-x) = (100-20) = 80%, t=15 min `k= 2.303/(15 min) log 100/80 = 2.303/(15 min) log 1.25 = 2.303/(15 min) xx 0.0969 = 0.015 min^(-1)` Step ii) Calculation of time in which `10%` of original substance remains unreacted. `a=100%, (a-x)=10%, k=0.015 min^(-1)` `t=2.303/k log a/(a-x) = 2.303/(0.015 min^(-1))log100/10 = 153.5 min` Step III. Calculation of time in which `20%` of the reactant left to reactant ater first 15 minutes. `a=80%, x=80 xx 20//100 = 16%, (a-x) = (80-16) = 64%, k =0.015 min^(-1)` `t=2.303/k log a/(a-x)= 2.303/(0.015 min^(-1)) log80/64` `=2.303/(0.015min^(-1)) log 1.25 = 2.303/(0.015 min^(-1)) xx 0.969 = 15 min` |
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| 1052. |
A first order reaction takes 10 minutes for `25%` decomposition. Calculate half life period of the reaction. |
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Answer» Calculation of rate constant for the reaction. For a first order reaction, , `k=2.303/t loga/(a-x)` a=100, (a-x) = 100-25 =75, t=10mm `k=2.303/(10 min)log 100/75 = (2.303 xx 0.125)/(10 min)= 0.02879 min^(-1)` Step II) Calculation of half life period for the reaction. `t_(1//2) = 0.693/k = 0.693/(0.02879 min^(-1))` = 24.07 minutes. |
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| 1053. |
75% of a first order reaction occurs in 30 min at `27^(@)C` . 87.5% of the same reaction occurs in 30min at `57^(@)C` . The activation energy of reaction is: [In 2=0.7, In 3=1.1]A. `2.64 kJ//mol`B. `2.64` Kcal/molC. `10.97` Kcal//molD. zero |
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Answer» Correct Answer - B |
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| 1054. |
The rate constant and rate of reaction are same forA. first orderB. zero orderC. second orderD. all are wrong |
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Answer» Correct Answer - B For zero-order reaction, `R = k[A]^(@)`. So `R = k`. |
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| 1055. |
The activation energy for a simple chemical reaction `A rarr B` is `E_(a)` in the forward reaction: The activation of the reverse reactionA. `lt 50 kcal`B. 50 kcalC. `gt 50 kcal`D. Either greater than or less than 50 kcal |
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Answer» Correct Answer - D The energy of activation for the reverse reaction will decided by the fact where the reversible is exothermic or endothermic. For an exothermic reaction `(Delta_(r)H^(Ɵ)is-ve)` thus `(E_(a))_(f)lt(E_(a))_(b)` and for an endothermic reaction `(Delta_(r)H^(Ɵ)is+ve)` Thus `(E_(a))_(f)gt(E_(a))_(b)` . Thus, the energy of activation of reverse reaction is either grater or less than `50Kcal` . |
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| 1056. |
The activation energy for a simple chemical reaction `A rarr B` is `E_(a)` in the forward reaction: The activation of the reverse reactionA. is always double of `E_`B. is negative of `E_(a)`C. is always less than `E_(a)`D. can be less than or more than `E_(a)` |
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Answer» Correct Answer - D The activation energy in exothermic and endothermic reactions will be more and less than `E_(a)` respectively. |
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| 1057. |
For a reaction `pA + qB rarr` Product, the rate law expresison is `r = k[A][B]^(m)`. ThenA. `(p + 1) lt (1 + m)`B. `(p + q) gt (1 + m)`C. `(p + q)` may or may not be equal to `(1 + m)`D. `(p + q) = (1 + m)` |
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Answer» Correct Answer - C If reaction is complex, then we cannot predict order by stoichiometric coefficients and cannot justify relationship between `(p + q)` an `(l + m)`. |
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| 1058. |
Which of the following statement about the rate of a chemical reaction is (are) not true?A. (a) The rate remains constant throughout the reaction in all order of reactionB. (b) The rate increases as the reaction proceedsC. (c ) The rate decreases as the reaction proceedsD. (d) None of these |
| Answer» Correct Answer - a, b | |
| 1059. |
In which of the following, `E_(a)` for backward reaction is greater than `E_(a)` for forward reaction?A. `Aoverset(E_(a)-50"kcal")rarrB,DeltaH=-10` kcalB. `Aoverset(E_(a)-50"kcal")rarrB,DeltaH=+10` kcalC. `Aoverset(E_(a)-60"kcal")rarrB,DeltaH=+20` kcalD. all of the above |
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Answer» Correct Answer - A `DeltaH=(E_(a))_(f)-(E_(a))_(b)` `(E_(a))_(b)=(E_(a))_(f)-DeltaH=50+10=60"kcal"` |
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| 1060. |
The minimum energy required for molecules to enter into the reaction is calledA. Potential energyB. Kinetic energyC. Nuclear energyD. Activation energy |
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Answer» Correct Answer - d The energy necessary for molecules to undergo chemical reaction is known as Activation energy . |
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| 1061. |
The reaction rate at a given temperature becomes slower thenA. The Free energy of activation is higherB. The free energy of activation is lowerC. The entropy changesD. The initial concentration of the reactants remains constant |
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Answer» Correct Answer - a Slowest reaction rate indicates higher energy of activation . |
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| 1062. |
A large increase in the rate of reaction for a rise in temperature is due toA. The decrease in the number of collisionB. The increase in the number of activated moleculesC. The shortening of the mean free pathD. The lowering of the activation energy |
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Answer» Correct Answer - b The increase in collisions frequency brings in an increase in effective collisions and thus rate of reaction increases . |
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| 1063. |
A large increase in the rate of a reaction for a rise in temperature is due toA. Increase in collision frequencyB. Lowering of activation energyC. Increase in number of effective collisionsD. None of these |
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Answer» Correct Answer - c More is the number of effective collision, more will be rate of reaction. The increase in temperature results in increase in kinetic energy of molecule. |
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| 1064. |
A gaseous reaction `A_(2)(g) rarr B(g) + (1)/(2) C(g)` shows increase in pressure form `100 mm` to `120 mm` in `5 min`. What is the rate of disappearance of `A_(2)` ?A. `4`B. `8`C. `16`D. `2` |
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Answer» Correct Answer - b `A_(2)rarr B(g)+1/2C(g)` `(-d[A_(2)])/(dt)=(2d[C])/(dt)` Thus, `(d[C])/(dt)=(120-100)/(5)= 4 mm min^(-1)` `:. -(d[A_(2)])/(dt)=2xx4 mm min^(-1)` |
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| 1065. |
For the reaction `2NO_(2)+F_(2)rarr 2NO_(2)F`, following mechanism has been provided: `NO_(2)+F_(2) overset("slow")(rarr) NO_(2)F+F` `NO_(2)+Foverset("fast")(rarr) NO_(2)F` Thus rate expression of the above reaction can be writtens as:A. `r=K[NO_(2)]^(2)[F_(2)]`B. `r=K[NO_(2)]`C. `r=K[NO_(2)][F_(2)]`D. `r=K[F_(2)]` |
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Answer» Correct Answer - c Slowest step of mechanism decides the rate expression. |
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| 1066. |
A second order reaction requires `70 min` to change the concentration of reactants form `0.08 M` to `0.01 M`. How much time will require to become `0.04 M` ?A. `10 min`B. `20 min`C. `30 min`D. `40 min` |
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Answer» Correct Answer - A For second order, when `(a-x) = 0.01 M` `k_(2) = (1)/(t).(x)/(a(a-x))` `k_(2) = (1)/(70). (0.07)/(0.08 xx 0.01)` …(i) When `(a-x) = 0.04 M` `k_(2) = (1)/(t).(0.04)/(0.08 xx 0.04)` …(ii) form Eqs. (i) and (ii), `(0.07)/(70 xx 0.08 xx 0.01) = (0.04)/(t xx 0.08 xx 0.04)` `rArr t = 10 min` |
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| 1067. |
The hydrogenation of vegetable ghee at `25^(@)C` reduces the pressure of `H_(2)` form `2 atm` to `1.2 atm` in `50min`. Calculate the rate of reaction in terms of change of (a) Pressure per minute (b) Molarity per secondA. `1.09xx10^(-6)`B. `1.09xx10^(-5)`C. `1.09xx10^(-7)`D. `1.09xx10^(-9)` |
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Answer» Correct Answer - b The change in molarity `=n/V=(DeltaP)/(RT)` `=0.8/(0.0821xx273)=0.0327` `:.` Rate of reaction=`0.0327/(50xx60)` `=1.09xx10^(-5) mol litre^(-1) sec^(-1)` |
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| 1068. |
t some temperature, the rate constant for the reaction of the type `2A rarr` Products is `0.08 Ms^(-1)`. The time it takes for the concentration of A to drop from `1.50M to 0.30M` isA. `7.5s`B. `9.5s`C. `11.5s`D. `13.5s` |
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Answer» Correct Answer - A From the unit of rate constants `(molL^(-1)s^(-1))` , we can conclude that the reaction is of zero order. For a reaction, `aArarr`products, that is zero order, the integater rate law is `[A]_(t)=[A]_(0)-akt` or `t=[A]_(0)-[A]_(t)//ak` Substituting the valus, we get `t=([(1.50)-(0.30)]M)/((2)(0.08Ms^(-1)))` `=7.5s` |
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| 1069. |
The hydrogenation of vegetable ghee at `25^(@)C` reduces the pressure of `H_(2)` form `2 atm` to `1.2 atm` in `50min`. Calculate the rate of reaction in terms of change of (a) Pressure per minute (b) Molarity per secondA. `1.09xx10^(-5)mol L^(-1)s^(-1)`B. `2.67xx10^(-4)mol L^(-1)s^(-1)`C. `8.94xx10^(-7)mol L^(-1)s^(-1)`D. `3.25xx10^(-3)mol L^(-1)s^(-1)` |
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Answer» Correct Answer - A `-(Deltap_(H_(2))/(Deltat)=-((1.2-2.0)atm)/(50min)xx(min)/(60s)` `=2.67xx10^(-4)s^(-1)` Assuming ideal gas behavior, we have `PV=nRT` `P=(n)/(V)RT=CRT` Thus, `DeltaP=DeltaCRT` Diving both sides by time interval, `Deltat` , we get `(Deltap)/(DeltaT)=(Deltac)/(Ddeltat)RT` or `(Deltac)/(Deltat)=(Deltap//Deltat)/(RT)` `=(2.67xx10^(-4)atms^(-1))/((0.0821(Latm)/(mol.K))(298K))` `=1.09xx10^(-5)molL^(-1)s^(-1)` |
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| 1070. |
The activation energy for a reaction at the temperature T K was found to be `2.303RT J mol^(-1)`. The ratio of the rate constant to Arrhenius factor isA. `10^(-1)`B. `10^(-2)`C. `2 xx 10^(-3)`D. `2 xx 10^(-2)` |
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Answer» Correct Answer - A a) According to Arrhenius equation, log k = log A`-E/(2.303R) xx 1/T` `2.303 log k = 2.303 log A-(E_(a))/(RT)` Substituting the value of `E_(a)` as 2.303 RT and dividing by 2.303 on both sides, we get. log k`-log A=-1` `log(k/A)=log 10^(-1)` Taking anti log. `(k/A)=10^(-1)` |
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| 1071. |
A first order reaction has a rate constant of `5 xx 10^(-3) s^(-1)`. How long will `5.0 g` of this reaction take to reduce to `3.0 g` ?A. 34.07 sB. 7.57 sC. 10.10 sD. 15 g |
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Answer» Correct Answer - A `t = (2.303)/(k) "log" (a)/((a - x))` or `t = (2.303)/(15 xx 10^(-3)) "log" (5)/(3) = 34.07s` |
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| 1072. |
The time required for 100 percent completion of a zero order reaction is:A. `(2k)/a`B. `a/(2k)`C. `a/k`D. `ak` |
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Answer» Correct Answer - C c) For zero order reaction, `k=([A]_(0)-[A])/(t)` For 100 percent completion [A]= zero `k=([A]_(0)/t)` or `t=(A)_(0)/k` `t=a/k`, where a is initial concentration. |
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| 1073. |
Derive an expression to calculate time required for completion of zero order reaction.A. `t = ([R_(0)])/(k)`B. `t = [R] - [R_(0)]`C. `t = (k)/([R_(0)])`D. `t = ([R_(0)] - [R])/([R_(0)])` |
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Answer» Correct Answer - A `[R] = [R_(0)] - kt` For completion of reaction [R] = 0 or `t = ([R_(0)])/(k)` |
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| 1074. |
A first order reaction is 50% complete in 20 minutes. What is its rate constant? |
| Answer» Rate constant (k) = `(0.693)/(20 min)= 0.3465 min^(-1)`. | |
| 1075. |
The time required for 100 percent completion of a zero order reaction is:A. akB. `(a)/(2k)`C. `(a)/(k)`D. `(2k)/(a)` |
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Answer» Correct Answer - C For a zero order reaction, `x = kt` For 100% completion of the reaction, x = a Therefore, a = kt or `t = (a)/(k)` |
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| 1076. |
The half life of radioactive sodium is 15 hours. How many hours would it take for 64 g of sodium to decay to one-eight of its original value?A. 3 hoursB. 15 hoursC. 30 hoursD. 45 hours |
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Answer» Correct Answer - D d) `t=2.303/k log (No)/(Nt)` `t=(2.303 xx(15 "hours"))/(0.693)log64/8` `=(2.303 xx 15 xx 0.9030)/(0.693) = 45` hours |
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| 1077. |
Pieces of wood burn faster than a log of wood of the same mass becauseA. surface area of log of wood is larger and needs more time to burnB. pieces of wood have larger surface areaC. all pieces of wood catch fire at the same timeD. log of wood has higher density than pieces of the same wood |
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Answer» Correct Answer - B More the surface area of reactant, more will be the rate of reaction. |
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| 1078. |
The radioisotpe, N-13,has a half-life of 10.0 minutes. What is the rate constant for the radioactive decay of N-13?A. `0.0310 min^(-1)`B. `0.0693 min^(-1)`C. `0.100 min^(1)`D. `6.93min^(-1)` |
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Answer» Correct Answer - B |
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| 1079. |
A reaction is `50%` complete in 2 hours and `75%` complete in 4 hours. What is the order of reaction? |
| Answer» It is given that `t_(1//2)=2h`. Since `75%` completion of the reaction consumes two half lives, this shows that `t_(1//2)` is independent of the initial concentration. The reaction is therefore, of first order. | |
| 1080. |
A reaction is 50% complete in 2 hours and 75% complete in 4 hours . The order of reaction isA. 1B. 2C. 3D. 0 |
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Answer» Correct Answer - a For a first order reaction , `t_(75%) = 2 xx t_(50%)` |
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| 1081. |
Gaseous `N_(2)O_(5)` decomposes according to the following equation: `N_(2)O_(5)(g) rarr 2NO_(2)(g) + (1)/(2)O_(2)(g)` The experimental rate law is `-Delta [N_(2)O_(5)]//Delta t = k[N_(2)O_(5)]`. At a certain temerature the rate constant is `k = 5.0 xx 10^(-4) sec^(-1)`. In how seconds will the concentration of `N_(2)O_(5)` decrease to one-tenth of its initial value ?A. `2.0 xx 10^(3)s`B. `4.6 xx 10^(3)s`C. `2.1 xx 10^(2)s`D. `1.4 xx 10^(3) s` |
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Answer» Correct Answer - B `t = (2.303)/(5 xx 10^(-4)) "log"(A_(0))/(A_(0)//10)` `= 4.6 xx 10^(3) sec`. |
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| 1082. |
In a second order reaction, `20%` of a substance is dissociated in `40 min`. The time taken by `80%` of its dissociation isA. `160 min`B. `640 min`C. `200 min`D. `320 min` |
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Answer» Correct Answer - B For second order reaction, When `x = 20%, (a-x) = 80%, a = 100%` `k_(2) = (1)/(t).(x)/(a(a-x))` `= (1)/(40) xx (20)/(100 xx (100 - 20))` `= (1)/(20 xx 100 xx 80)` …(i) When `x = 80%, (a-x) = 20%` `k_(2) = (1)/(t) xx (80)/(100xx20) = 1//25t` ...(ii) form Eqs. (i) and (ii), `t = (2 xx 100 xx 80)/(25) = 640 min` |
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| 1083. |
The half-life of `Tc^(99)` is 6.0 hr. The delivery of a sample of `Tc^(99)` from the reactor to the nuclear medicine lab of a certain hospital takes 3.0 hr. What is the minimum amount of `Tc^(99)` that must be shipped in order for the lab to receive 10.0 mg?A. 20.0 mgB. 15.0 mgC. 14.1 mgD. 12.5 mg |
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Answer» Correct Answer - b |
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| 1084. |
For a first order reaction , rate constant is `0.6932 "hr"^(-1)` , then half-life for the reaction isA. `0.01` hrB. 1hrC. 2hrD. 10 hr |
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Answer» Correct Answer - b `t_(1//2) = (0.693)/(k) = (0.693)/(0.6932 hr^(-1)) = 1`hr . |
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| 1085. |
Which of the following curve represent zero order reaction of `A rarr` products ?A. B. C. D. |
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Answer» Correct Answer - D For zero-order reaction `-(d[A])/(d t) = k` or `[A]_(0) - [A] = kt` or `[A] = [A]_(0) - kt` It means concentration of `A` linearly decreases with `t`. |
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| 1086. |
Unit of K for third order reaction isA. `(("litre")/("mole")) sec`B. `(("mole")/("litre")) sec`C. `(("litre")/("mole"))^(-1) sec^(-1)`D. `(("mole")/("litre"))^(-1) sec^(-1) ` |
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Answer» Correct Answer - d Unit of rate constant for a `n^(th)` order reaction is `(("lit")/("mol"))^(n-1) sec^(-1)` When n = 3 , the unit of rate constant is `(("lit")/("mol"))^(2) sec^(-1)` or `(("mol")/("lit"))^(-2) sec^(-1)`. |
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| 1087. |
`t_(1//2) =` constant confirms the first order of the reaction as one `a^(2)t_(1//2) =` constant confirms that the reaction is ofA. Zero orderB. First orderC. Second orderD. Third order |
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Answer» Correct Answer - D Given `t_(1/2) prop (1)/(a^(2))` `:. t_(1/2) prop (1)/(a^(n-1))" "n - 1 = 2" "n = 3` |
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| 1088. |
The half life period for a first order reaction isA. Proportional to concentrationB. Independent of concentrationC. Inversely proportional to concentrationD. Inversely proportional to the square of the concentration |
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Answer» Correct Answer - b `t_(1//2) = (0.693)/(K)`, i.e, half life is independent of concentration |
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| 1089. |
In presence of `HCl`, sucrose gets hydrolysed into glucose and fructose. The concentration of sucrose was found to reduce form `0.4 M` to `0.2 M` in `1` hour and `0.1 M` in `2` hours. The order of the reaction isA. ZeroB. OneC. TwoD. None of these |
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Answer» Correct Answer - B `t_(1//2) prop (1)/((C_(0))^(n-1))` `:.` reaction is of first order. |
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| 1090. |
Integrated velocity equation for first order reaction isA. `[A]_(o) = [A]e^(-kt)`B. `K = [A]_(o) e^(-A//t)`C. `Kt = 2.303 log ([A]_(o))/([A])`D. log `([A]_(o))/([A]) = -2.303 Kt` |
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Answer» Correct Answer - c Integrated velocity equation for first order reaction is `k = (2.303)/(t)` log `((A)_(0))/((A))`. |
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| 1091. |
Which of the following rate laws has an overall order of `0.5` for reaction involving substances `x, y` and `z` ?A. Rate `= K(C_(x))(C_(y))(C_(z))`B. Rate `= K(C_(x))^(0.5)(C_(y))^(0.5)(C_(z))^(0.5)`C. Rate `= K(C_(x))^(1.5)(Cy)^(-1)(Cz)^(0)`D. Rate `= K(C_(x))(C_(z))^(n)//(C_(y))^(2)` |
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Answer» Correct Answer - C Order of reaction is sum of the power raised on concentration terms to express rate expression. |
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| 1092. |
If the rate of reaction between A and B is given by rate `=k[A][B]^(2)` , then the reaction is :A. first orfer in AB. second order in BC. third order oveallD. all are correct |
| Answer» Correct Answer - D | |
| 1093. |
Which of the following rate laws has an overall order of 0.5 for reaction involving substances x , y and zA. Rate = `K (C_(x)) (C_(y)) (C_(z))`B. Rate = `K (C_(x))^(0.5) (C_(y))^(0.5) (C_(z))^(0.5)`C. Rate = `K(C_(x))^(1.5) (C_(y))^(-1) (C_(z))^(0)`D. Rate = `K(C_(x)) (C_(z))^(n) // (C_(y))^(2)` |
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Answer» Correct Answer - c Order of reaction is sum of the power raised on concentration terms to express rate expression. |
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| 1094. |
Velocity constant of a reaction at 290 K was found to be `3.2xx 10^(-3)` . At 310 K it will be aboutA. `1.28 xx 10^(-2)`B. `9.6 xx 10^(-3)`C. `6.4 xx 10^(-3)`D. `3.2 xx 10^(-4)` |
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Answer» Correct Answer - a For 10 K rise in temperature , the rate of reaction nearly doubles . |
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| 1095. |
In presence of HCl , sucrose gets hydrolysed into glucose and fructose . The concentration of sucrose was found to reduce form 0.4 M to 0.2 M in 1 hour and 0.1 M in 2 hours . The order of the reaction isA. ZeroB. OneC. TwoD. None of these |
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Answer» Correct Answer - b `t_(1//2) prop (1)/((C_(0))^(n-1)) therefore` reaction is of first order . |
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| 1096. |
The value of rate constant for a first order reaction `2.303 xx 10^(-2) sec^(-1)` .what will be time required to reduce the concentration to `(1)/(10)` th its initial concentrationA. 10 secondB. 100 secondC. 2303 secondD. 230.3 second |
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Answer» Correct Answer - b `t = (2.303)/(K)` log `([A_(0)])/([A])` `t = (2.303)/(2.303 xx 10^(-2)) "log" (100)/(10) implies t = 10^(2)` log 10 `implies t = 100` sec |
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| 1097. |
Which of the following rate laws has an overall order of `0.5` for reaction involving substances `x, y` and `z` ?A. Rate =`k[C_(x)][C_(y)][C_(z)]`B. Rate = `k[C_(x)]^(0.5)[C_(y)]^(0.5)[C_(z)]^(0.5)`C. Rate = `k[C_(x)]^(1.5)[C_(y)]^(-1)[C_(z)]^(0)`D. Rate = `k[C_(x)][C_(y)]^(-2)[C_(z)]^(0)` |
| Answer» Correct Answer - C | |
| 1098. |
Which of the following statement is false in relation to enzymeA. pH affects their functioningB. Temperature in concentration of the reactant with timeC. They always increase activation energyD. Their reactions are specific |
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Answer» Correct Answer - c Enzymes does not always increase activation energy . |
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| 1099. |
For a reaction `A to B ` , the rate of reaction quadrupled when the concentration of A is doubled . The rate expressions of the reaction is r = `K (A)^(n)` , when the value of n isA. 1B. 0C. 3D. 2 |
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Answer» Correct Answer - d Rate of reaction is quadrupled on doubling the concentration. Thus r `prop [A]^(2)` |
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| 1100. |
For reaction of zero order isA. `k=([A_(0)])/(t) `B. `kt = [A_(0)]=[A]`C. `Kt =[A] -[A_(0)]`D. ` k=(2.303)/(t ) l n ([A_(0)])/([A])` |
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Answer» Correct Answer - B `-(d[A])/(dt)=k ` ` d[A]=-kt ,` integrating `[A]=-k t +C ` where ` C= [A_(0)] at t=0` ` k t = [A_(0)]-[A]` |
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