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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.
| 901. |
The unit of rate constant in case of zero order reaction is ____.A. Concentration `xx "Time"^(-1)`B. `"Concentration"^(-1) xx "Time"^(-1)`C. Concentration `xx "Time"^(2)`D. `"Concentration"^(-1) xx "Time"` |
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Answer» Correct Answer - a For zero order reaction Velocity constant = `(dx)/(dt) = ("Concentration")/("Time")` Unit = concentration`xx "time"^(-1)`. |
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| 902. |
Starting with one mole of a compound `A`, it is found that the reaction is `3//4` completed in `1 hr`. Find the rate constant if the reaction is of `II` order.A. 2.31 `"min"^(-1)`B. `0.231 "min"^(-1)`C. `0.0231 "min"^(-1)`D. `0.00231 "min"^(-1)` |
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Answer» Correct Answer - c `K =- (2.303)/(t)` log `(a)/(a-x)` `k = (2.303)/(60)` log `(1)/(1-3//4)` `K = (2.303)/(60)` log 4 = `0.0231` . |
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| 903. |
The half life of radioactive sodium is 15 hours its disintegration constant isA. `0.0462 h^(-1)`B. `0.0642 h^(-1)`C. `0.0462 s^(-1)`D. `0.0462 "min"^(-1)` |
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Answer» Correct Answer - a `k = (0.693)/(t_(1//2)) = (0.693)/(15) = 0.0462 hr^(-1)`. |
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| 904. |
A certain endothermic reaction: `Ararr` Product, `DeltaH=+ve` proceeds in a sequence of three elementary steps with the rate constants `K_(1), K_(2)` and `K_(3)` and each one having energy of activation `E_(a), E_(2)` and `E_(3)` respectively at `25^(@)C`. The observed rate constant for the reaction is equal to `K_(3) sqrt(K_(1)/K_(2)). A_(1), A_(2)` and `A_(3)` are Arrhenius parameters respectively. Presence of a catalyst decreases the energy of activation of each path by half the value of `E_(1)`. Assuming the order actors same, the observed energy of activation would be:A. (a) `E_(3)-E_(2)/2`B. (b) `2E_(3)-E_(2)`C. (c ) `E_(2)-E_(3)/2`D. (d) `E_(1)/2+E_(2)/2+E_(3)` |
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Answer» Correct Answer - a `E_(obs)=(E_(1)-E_(2)+2E_(3))/(2)` if each is lowered by `E_(1)//2`, then `E_(obs)=(E_(1)-E_(1)/2-(E_(2)-E_(1)/2)+2(E_(3)-E_(1)/2))/(2)` `=E_(3)-E_(2)/2` |
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| 905. |
Consider the elementary reaction sequence shown in figure. Which of the following equations are correct ? A. `(d[A])/(d t) = -k_(1)[A] + k_(4)[D]`B. `(d[C])/(d t) = k_(2)[B] - k_(3)[C]`C. `(d[D])/(d t) = -k_(4)[D] + k_(3)[D]`D. Nothing can be said about order of reactions in this problem. |
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Answer» Correct Answer - B In given sequence of reaction `(d[C])/(d t) = k_(2)[B] - k_(3)[C]` |
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| 906. |
Unit of the constant of zero order reaction isA. One reaction will be more in itB. Reactants do not participate in itC. Rate of reaction is proportional to velocity of moleculesD. Reactants concentration do not change with time |
| Answer» Correct Answer - d | |
| 907. |
Derive order of reaction for the decomposition of `H_(2)O_(2)` from the following data. `{:("Time (in minutes)",0,10,20,30),("Volume of" KMnO_(4),25,16,10.5,7.09):}` needed for `H_(2)O_(2)` |
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Answer» The volume of `KMnO_(4)` used at anytime anytime is proportional to conc. `H_(2)O_(2)` at that time. `{:(,"At t=0",V=25,:.,a,prop,25),((i),"At t=10",V=16,:.,(a-x),prop,16),((ii),"At t=20",V=10.5,:.,(a-x),prop,10.5),((iii),"At t=30",V=7.09,:.,(a-x),prop,7.09):}` Now use, `K=2.303/t log a/((a-x))` For (i), `K=2.303/10 log 25/16` `=4.46xx10^(-2) min^(-1)` For (ii), `K=2.303/20 log 25/10.5` `=4.34xx10^(-2) min^(-1)` For (iii), `K=2.303/30 log 25/7.09` `4.20xx10^(-2) min^(-1)` The values of `K` come almost constant and thus confirming first order reaction. For velocity constant take average of all value of `K`. |
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| 908. |
The rate constant of a reaction is `3.25 xx 10^(-3) " mol"^(-2) L^(2) " min"^(-1)`. The order of raction isA. zeroB. 1C. 2D. 3 |
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Answer» Correct Answer - D The unit of rate constant for nth order reaction `("mol")^(1-n)L^(n-1)" min"^(-1)` On putting n = 3, we get the unit of given rate constant Hence, the order of reaction is 3. |
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| 909. |
What will be the order of reaction and rate constant for a chemical change having `log t_(50%)` versue log concentration of `(a)` curves as: A. `0, 1`B. `1, 1`C. `2, 2`D. `3, 1` |
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Answer» Correct Answer - a For zero order, `t_(1//2) prop (a)^(1-n)` or `ln t_(1//2)=(1-n)log a+log K` `y=mx+C` Also `1-n=tan 45^(@)=1, :. N=0` |
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| 910. |
The rate constant of a reaction is found to be `3 xx 10^(-3) "mol" L^(-1) "min"^(-1)`. The order of the reaction isA. zeroB. 1C. 2D. 1.5 |
| Answer» Correct Answer - a | |
| 911. |
For a gaseous reaction, following data is given: `ArarrB, k_(1)= 10^(15)e-^(2000//T)``C rarrD, k_(2) = 10^(14)e^(-1000//T)` The temperature at which `k_(1) = k_(2)` isA. 1000KB. 2000 KC. `868.82 K`D. `443.2 K` |
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Answer» When `k_(1)=k_(2)` `10^(15)e^(-2000//T)=10^(14)e^(-1000//T)` `10= e^(1000//T)` `2.303 log 10 =(1000)/(T)` |
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| 912. |
For a gaseous reaction, following data is given: `ArarrB, k_(1)= 10^(15)e-^(2000//T)``C rarrD, k_(2) = 10^(14)e^(-1000//T)` The temperature at which `k_(1) = k_(2)` isA. `1000 K`B. `2000 K`C. `868.82 K`D. `434.2 K` |
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Answer» Correct Answer - D (d) When `k_(1) = k_(2)` `10^(15)e^(-2000//T) =10^(14)e^(-1000//T)` `10= e^(1000//T) rArr ln 10 = (1000)/(T)` `rArr 2.303 log 10 = (1000)/(T)` `T = 434.2 K` |
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| 913. |
For reaction `RX + OH^(-) rarr ROH + X^(-)`, rate expression is `R = 4.7 xx 10^(-5) [RX][OH^(-)] + 2.4 xx 10^(-5)[RX]`. What `%` of reactant react by `S_(N)2` mechanism when `[OH^(-)] = 0.001` molar? |
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Answer» The rate expression involves `S_(N^(1))` and `S_(N^(2))` steps as: `RX+OH^(-)rarrR^(+)+X^(-)overset(OH^(-))rarrROH+X^(-)S_(N^(1))` `RX+OH^(-)rarrHO...R..X^(-)rarrHOR+X^(-)S_(N^(2))` i.e., rate `= 4.7xxunderset(S_(N^(2)))(10^(-5))[RX][OH]+0.24xxunderset(S_(N^(2)))(10^(-5))[RX]` `% rate of S_(N^(2))=[r_(S_(N^(2)))/(r_(S_(N^(1)))+r_(S_(N^(2))))]xx100` `=(4.7xx10^(-5)[RX][OH^(-)])/(4.7xx10^(-5)[RX][OH^(-)]+0.24xx10^(-5)[RX])xx100` `=(4.7xx10^(-5)[OH^(-)])/(4.7xx10^(-5)[OH^(-)]+0.24xx10^(-5))` `=(4.7xx10^(-5)xx0.001)/(4.7xx10^(-5)xx0.001+0.24xx10^(-5))` `=1.9%` |
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| 914. |
For a reaction: `nA rarr` Product, if the rate constant and the rate of reactant are equal what is the order of the reaction?A. `0`B. `1`C. `2`D. `3` |
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Answer» Correct Answer - A For zero order, `k = ["concentration"]^((1-n))t^(-1)` `= mol L^(-1) t^(-1)` Rate of reaction `= (d["Conc"])/(dt) = mol L^(-1) t^(-1)` |
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| 915. |
For an elementary reaction, `2A + B rarr C + D` the molecularity isA. ZeroB. OneC. TwoD. Three |
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Answer» Correct Answer - D `2A +B rarr C + B` Rate `= k[A]^(2)[B]^(1)` `:. O.R. = 2 + 1 = 3` and molecularity is `3[2A + B]`. |
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| 916. |
The rate constant of a reaction is found to be `3 xx 10^(-3) "mol" L^(-1) "min"^(-1)`. The order of the reaction isA. zeroB. `1`C. `2`D. `1.5` |
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Answer» Correct Answer - A `k = (("mol")/("litre"))^(1 - n) = "min"^(-1) = (("mol")/("litre"))^(1) "min"^(-1) n = 0`. |
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| 917. |
For a gaseous reaction, following data is given: `ArarrB, k_(1)= 10^(15)e-^(2000//T)``C rarrD, k_(2) = 10^(14)e^(-1000//T)` The temperature at which `k_(1) = k_(2)` isA. `1000 K`B. `2000 K`C. `868.4 K`D. `434.22 K` |
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Answer» Correct Answer - d If `K_(1)=K_(2)`, then `10^(15) exp(-2000/T)=10^(14) exp(-1000/T)` or `15-2000/(Txx2.303)=14-1000/(Txx2.303)` `:. 1000/(Txx2.303) = 1` `:. T=1000/2.303=434.22` |
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| 918. |
The role of a catalyst is to changeA. Gibbs energy of reactionB. enthalpy of reactionC. activation energy of reactionD. equilibrium constant. |
| Answer» Correct Answer - C | |
| 919. |
Analyze the generalized rate data: `RX + M^(ɵ) rarr` Product `|{:("Experiment",[RX] "Substrate",[M^(ɵ)] "Attaking species","Rate"),(I,0.10 M,0.10 M,1.2 xx 10^(-4)),(II,0.20 M,0.10 M,2.4 xx 10^(-4)),(III,0.10 M,0.20 M,2.4 xx 10^(-4)),(IV,0.20 M,0.20 M,4.8 xx 10^(-4)):}|` The value of rate constant for the give experiment data isA. `1.2 xx 10^(-2)`B. `1.2 xx 10^(-4)`C. `1.2 xx 10^(-3)`D. `2.4 xx 10^(-3)` |
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Answer» Correct Answer - A form experimetn I and II, when the concentration of `[RX]` becomes double, rate becomes double, so order w.r.t. `(RX) = 1`. ismilarly, form experiments II and IV, when the concentration of `M^(ɵ)` is doubled, rate also is doubled. So order w.r.t `[M^(ɵ)] = 1`. Hence, rate `= k[RX][M^(ɵ)]` ...(i) So overall second order reaction. form Eq. (i), `1.2 xx 10^(-4) = k[0.1][0.1]` `:. k = 1.2 xx 10^(-2)` |
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| 920. |
It is generalized that a `10^(@)C` increases in temperature casues the rate of reaction to double. Applied to a reaction at `295 K`, what is the value of `E_(a)`?A. `120 "kcal mol"^(-1)`B. `1200 "kcal mol"^(-1)`C. `1.2 "kcal mol"^(-1)`D. `12 "kcal mol"^(-1)` |
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Answer» Correct Answer - D Only reaction whose `E_(a)` falls in the range of `50-55 kJ mol^(-1)` or `12-13 kcal mol^(-1)` are found to double their rate for `10^(@)C` rise in temperature i.e., form `298` to `308 K`. Alternatively Use Arrheinius equation `T_(1) = 295 K, T_(2) = 305 K`. `log.(k_(2))/(k_(1)) = (E_(a))/(2.3 xx 2 xx 10^(-3))((10)/(295 xx 305))` `E_(a) = (log(2) xx 2.3 xx 2 xx 10^(-3) xx 295 xx 305)/(10)` `E_(a) = (0.3 xx 4.6 xx 295 xx 305)/(10^(4))` `= 12.4 kcal mol^(-1)` `~~ 12 kcal mol^(-1)`Correct Answer - D Only reaction whose `E_(a)` falls in the range of `50-55 kJ mol^(-1)` or `12-13 kcal mol^(-1)` are found to double their rate for `10^(@)C` rise in temperature i.e., form `298` to `308 K`. Alternatively Use Arrheinius equation `T_(1) = 295 K, T_(2) = 305 K`. `log.(k_(2))/(k_(1)) = (E_(a))/(2.3 xx 2 xx 10^(-3))((10)/(295 xx 305))` `E_(a) = (log(2) xx 2.3 xx 2 xx 10^(-3) xx 295 xx 305)/(10)` `E_(a) = (0.3 xx 4.6 xx 295 xx 305)/(10^(4))` `= 12.4 kcal mol^(-1)` `~~ 12 kcal mol^(-1)` |
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| 921. |
For reaction `RX + OH^(-) rarr ROH + X^(-)`, rate expression is `R = 4.7 xx 10^(-5) [RX][OH^(-)] + 2.4 xx 10^(-5)[RX]`. What `%` of reactant react by `S_(N)2` mechanism when `[OH^(-)] = 0.001` molar?A. `1.9`B. `66.2`C. `95.1`D. `16.4` |
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Answer» Correct Answer - a The rate is made up of two parts `S_(N^(1))` and `S_(N^(2))` `rate=4.7xx10^(-5)[RX][OH^(-)]+0.24xxunderset(S_(N^(1)))(10^(-5))[RX]` Thus `% S_(N^(2))=[S_(N^(2))/(S_(N^(2))+S_(N^(1)))]xx100` `=[(4.7xx10^(-5)[RX][OH^(-)])/(4.7xx10^(-5)[RX][OH^(-)]+0.24xx10^(-5)[RX])]xx100` `=[(4.7[OH^(-)])/(4.7[OH^(-)]+0.24)]xx100` `=(4.7xx0.001)/(4.7xx0.001+0.24)=1.9%` |
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| 922. |
The rate of reaction `((dx)/(dt))` varies with nature, physical state and concentration of reactants, temperature, exposure to light and catalyst, whereas rate constant `(K)` varies with temperature and catalyst only. The rate constant `K` is given as `K=Ae^(-E_(a)//RT)` where `A` is Arrhenius parameter or pre-exponential factor and `E_(a)` is energy of activation. The minimum energy required for a reaction is called threshold energy and the additional energy required by reactant molecules to attain threshold energy level is called energy of activation. How much faster would a reaction proceed at `25^(@)C` than at `0^(@)C` if the activation energy is `2` cal?A. (a) `2` timesB. (b) `16` timesC. (c ) `11` timesD. (d) Almost at same speed |
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Answer» Correct Answer - d `K_(1)=Ae^(-2/(RxxT_(1)))=Ae^(-2/(2xx273))` `K_(2)=Ae^(-2/(RxxT_(2)))=Ae^(-2/(2xx298))` `:. K_(1)/K_(2)=e^([-1/298+1/273])=1.0003` `:. r_(1)/r_(2)=1.0003` |
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| 923. |
The role of catalyst is to change:A. Gibbs energy of reactionB. enthalpy of reactionC. activation energy of reaction.D. equilibrium constant. |
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Answer» Correct Answer - C |
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| 924. |
Analyze the generalized rate data: `RX + M^(ɵ) rarr` Product `|{:("Experiment",[RX] "Substrate",[M^(ɵ)] "Attaking species","Rate"),(I,0.10 M,0.10 M,1.2 xx 10^(-4)),(II,0.20 M,0.10 M,2.4 xx 10^(-4)),(III,0.10 M,0.20 M,2.4 xx 10^(-4)),(IV,0.20 M,0.20 M,4.8 xx 10^(-4)):}|` For the reaction under conisderation, `3^(@)` alkyl has been found to be the most favourable alkyl group. Which of the following attacking species `(M^(ɵ))` will give the best yield in the reaction ?A. `(CH_(3))_(2)CH-O^(ɵ)`B. `(CH_(3))_(3)C-O^(ɵ)`C. `overset(ɵ)(OH)`D. `CH_(3)CH_(2)O^(ɵ)` |
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Answer» Correct Answer - B It is clear that reaction is second order, so it can be either `S_(N)2` or `E_(2)`. The reaction under conisderation, `3^(@)RX` is favourable. So reaction Would be `E_(2)` not `S_(N)2`. `3^(@)RX gt 2^(@)RX gt 1^(@)RX rArr Favours E_(2)` `1^(@)RX gt 2^(@)RX gt 3^(@)RX rArr Favours S_(N)2`. Hence `E_(2)` is favoured by strong bulkyl bronsted base. Acidic order: `H_(2)O gt C_(2)H_(5)OH gt (CH_(3))_(2)CH-OH gt (CH_(3))_(3) C-OH`. Baisc order: `overset(ɵ)(OH) lt C_(2)H_(5)O^(ɵ) lt (CH_(3))_(2) CH-O^(ɵ) lt (CH_(3))_(3)C-O^(ɵ)` Therefore, stronger base `(CH_(3))_(3)C-O^(ɵ)` will give the best yield of the reaction. Hence the answer is (b). Note: If the rreaction proceeds via `S_(N)2` mechanism, then `(M^(ɵ))` acts as nucleophilie. Since all of the species have same nuclepohile centre, so stronger the base stronger will be the nucleophile. `:.` Baisc order and nucleophile order should be: `(CH_(3))_(3)C-O^(ɵ) gt (CH_(3))_(2)CH - O^(ɵ) gt C_(2)H_(5)O^(ɵ) gt overset(ɵ)(OH)` So `(CH_(3))_(3)C-O^(ɵ)` should act as a stronger nucleophile. But the actual order of nucleophile is different. `:.` Nucleophile order: `C_(2)H_(5)O^(ɵ) gt (CH_(3))_(2)CH-O^(ɵ)gt(CH_(3))_(3)C-O^(ɵ) gt overset(ɵ)(OH)` This is because of its bulkiness, `(CH_(3))_(3)C-O^(ɵ)` is a poorer nucleophile because of steric hindrance, baiscity, and nucleophilicity may diverge. Thus, if `RX` would have been `1^(@)RX`, then `S_(N)2` reaction would have taken placed and attacking species `(C_(2)H_(5)O(ɵ))` would gove the best yield of the reaction. |
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| 925. |
For the reactions `:` `4KClO _(3) rarr 3KClO_(4)+KCl` if `" "-(d[KClO_(3)])/(dt)=k_(1)[KClO_(3)]^(4)` `(d[KClO_(4)])/(dt)=k_(2)[KClO_(3)]^(4)` `(d[KCl])/(dt)=k_(3)[KClO_(3)]^(4)` the correct relation between `k_(1),k_(2) & k_(3)` is `:`A. `k_(1)=k_(2)=k_(3)`B. `4k_(1)=3k_(2)=k_(3)`C. `3k_(1)=4k_(2)=12k_(3)`D. none of these |
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Answer» Correct Answer - C Rate `=-(1)/(4)(d[KClO_(3)])/(dt)=(1)/(3)(d[KClO_(4)])/(dt)=(d[KCl])/(dt)` or `(k_(1))/(4)[KClO_(3)]^(4)=k_(2)/3[KClO_(4)]^(4)=k_(3)[KClO_(4)]^(4)` or `3k_(1)=4k_(2)=12k_(3)` |
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| 926. |
The rate of reaction `((dx)/(dt))` varies with nature, physical state and concentration of reactants, temperature, exposure to light and catalyst, whereas rate constant `(K)` varies with temperature and catalyst only. The rate constant `K` is given as `K=Ae^(-E_(a)//RT)` where `A` is Arrhenius parameter or pre-exponential factor and `E_(a)` is energy of activation. The minimum energy required for a reaction is called threshold energy and the additional energy required by reactant molecules to attain threshold energy level is called energy of activation. At what conditions exponential factor is zero a reaction?A. (a) Inifinite remperatureB. (b) Free radical combinationC. (c ) Energy of activation = Threshold enerhyD. (d) All of these |
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Answer» Correct Answer - d `K=Ae^(-E_(a)//RT)` If `Trarr oo` then `K=A` Also `E_(a)=0` for free radical combination. |
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| 927. |
The reaction `A(g) + 2B(g) rarr C(g) + D(g)` is an elementary process. In an experiment involvig this reaction, the initial partial pressure of `A` and `B` are `p_(A) = 0.60 atm` and `p_(B) = 0.80 atm`, respectively. When `p_(C ) = 0.20 atm`, the rate of reaction relative to the initial rate isA. `(1)/(6)`B. `(1)/(48)`C. `(1)/(4)`D. `(1)/(24)` |
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Answer» Correct Answer - 1 Initial Rate`=k[A][B]^(2)` Rate `1=k[0.60][0.80]^(2)" " ….(i)` Rate `2=k[0.60][0.80]^(2)" "….(ii)` Given `Rxxn` `{:(A(g),+,2B(g),rarr,C(g),+,D(g),),(0.60atm,,0.80atm,,0,,0,t=0),((0.60-0.2),,(0.80-0.40),,0.20,,0.20atm,),(0.40atm,,0.40atm,,0.20atm,,0.20atm,t=1):}` ltBRgt Put value of pressure of `t=t` in rate equation `(II)` Rate `2=k(40][40]^(2)` `("Rate2")/("Rate 1")=(k[40][40]^(2))/(k[60][80]^(2))` `(Rate2)/(Rate1)=(2)/(3)xx(1)/(4)=(1)/(6)` |
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| 928. |
The reaction `A(g) + 2B(g) rarr C(g) + D(g)` is an elementary process. In an experiment involvig this reaction, the initial partial pressure of `A` and `B` are `p_(A) = 0.60 atm` and `p_(B) = 0.80 atm`, respectively. When `p_(C ) = 0.20 atm`, the rate of reaction relative to the initial rate isA. `1//6`B. `1//12`C. `1//36`D. `1//18` |
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Answer» Correct Answer - A `A(g) + 2B(g) rarr C(g)+D(g)` `{:(t=0,0.6,0.8,-,-),(t=t,0.4,0.4,0.2,0.2):}` `r_(0) = k xx 0.6 xx (0.8)^(2)` `r_(1) = k xx 0.4 xx (0.4)^(2)` ltbr. `rArr (r_(1))/(r_(0)) = (0.4 xx (0.4)^(2))/(0.6 xx (0.8)^(2)) = (1)/(6)` |
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| 929. |
In the respect of the equation `k=Ae^(-E_(a)//RT)` is chemical kinetics, which one of the following statement is correct ?A. k is equilibrium constantB. A is adsorption factorC. `E_(a)` is energy of activationD. R is Rydberg constant |
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Answer» Correct Answer - C |
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| 930. |
For an elementary reaction : A + 2B `rarr` C when `1 M` A was taken with `10^(-4)M` B, time taken for B to reduce to half was found to be 10 seconds. Calculate `t_(1//2)` when 1 M A is reacted with `10^(-5)M` B. [`t_(1//2)`=time for B to reduce to 50% of original]A. 10 secondsB. 1 secondsC. 100 secondsD. 20 seconds |
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Answer» Correct Answer - C |
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| 931. |
The rate of reaction `((dx)/(dt))` varies with nature, physical state and concentration of reactants, temperature, exposure to light and catalyst, whereas rate constant `(K)` varies with temperature and catalyst only. The rate constant `K` is given as `K=Ae^(-E_(a)//RT)` where `A` is Arrhenius parameter or pre-exponential factor and `E_(a)` is energy of activation. The minimum energy required for a reaction is called threshold energy and the additional energy required by reactant molecules to attain threshold energy level is called energy of activation. The quantity `-E_(a)//RT` in `-Ae^(-E_(a)//RT)` is referred as:A. (a) Boltzmann factorB. (b) Frequency factorC. (c ) Activation factorD. (d) None of these |
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Answer» Correct Answer - a It is a fact |
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| 932. |
The rate of reaction `((dx)/(dt))` varies with nature, physical state and concentration of reactants, temperature, exposure to light and catalyst, whereas rate constant `(K)` varies with temperature and catalyst only. The rate constant `K` is given as `K=Ae^(-E_(a)//RT)` where `A` is Arrhenius parameter or pre-exponential factor and `E_(a)` is energy of activation. The minimum energy required for a reaction is called threshold energy and the additional energy required by reactant molecules to attain threshold energy level is called energy of activation. For a reversible reaction, `A+B hArr C+D, DeltaH=-A kcal`. If energy is `B` kcal, the energy of activation for backward reaction in kcal is:A. (a) `-A+B`B. (b) `A+B`C. (c ) `A-B`D. (d) `-A-B` |
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Answer» Correct Answer - b `E_(a_(f))-E_(a_(b))=DeltaH` `B-E_(a_(b))=-A` `:. E_(a_(b))=A+B` |
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| 933. |
The rate of reaction `((dx)/(dt))` varies with nature, physical state and concentration of reactants, temperature, exposure to light and catalyst, whereas rate constant `(K)` varies with temperature and catalyst only. The rate constant `K` is given as `K=Ae^(-E_(a)//RT)` where `A` is Arrhenius parameter or pre-exponential factor and `E_(a)` is energy of activation. The minimum energy required for a reaction is called threshold energy and the additional energy required by reactant molecules to attain threshold energy level is called energy of activation. For an endothermic reaction, which one is true if `DeltaH` is heat of reaction and `E_(a)` is energy of activation?A. (a) `E_(a) gt DeltaH`B. (b) `E_(a) lt DeltaH`C. (c ) `E_(a)_(lt)^(gt) DeltaH`D. (d) `E_(a) = 0` |
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Answer» Correct Answer - a For endothermic reaction, `E_(a) gt DeltaH`, For exothermic, `E_(a) _(gt)^(lt)DeltaH` |
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| 934. |
The rate of reaction `((dx)/(dt))` varies with nature, physical state and concentration of reactants, temperature, exposure to light and catalyst, whereas rate constant `(K)` varies with temperature and catalyst only. The rate constant `K` is given as `K=Ae^(-E_(a)//RT)` where `A` is Arrhenius parameter or pre-exponential factor and `E_(a)` is energy of activation. The minimum energy required for a reaction is called threshold energy and the additional energy required by reactant molecules to attain threshold energy level is called energy of activation. For a chemical reaction: `Ararr` Product, the rate of disappearance of `A` is given by: `(-dC_(A))/(dt)=K_(1) (C_(A))/(1+K_(2)C_(A))`. At low `C_(A)`, the order of reaction and rate constants are respectively:A. (a) `I, K_(1)/K_(2)`B. (b) `I, K_(1)`C. (c ) `II, K_(1)/K_(2)`D. (d) `II, K_(1)/(K_(1)+K_(2))` |
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Answer» Correct Answer - b `-(dC_(A))/(dt)=(K_(1)C_(A))/(1+K_(2)C_(A)), if C_(A)= low` `1+K_(2)C_(A)=1` `:. (dC_(A))/(dt)=K_(1)C_(A)` |
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| 935. |
For a first order reaction rate is given by , R=K[Reactant] where `K=a.e^(-E_(alpha)//RT)`. Under what conditions will the rate of the reaction be smallest?A. Low concentration of reactant, high temperature and low activation energy.B. High concentration of raectant, low temperature and low activation energy.C. Low cocentration of reactants, low temperature and high activation energy.D. High concentration of reactant, low temperature and high activation energy. |
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Answer» Correct Answer - C |
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| 936. |
The reaction `N_(2)O_(5)` (in `C Cl_(4)` solution) `rarr 2NO_(2)` (solution) `+(1)/(2) O_(2)(g)` is of first order in `N_(2)O_(5)` with rate constant `6.2 xx 10^(-1) s^(-1)`. What is the value of rate of reaction when `[N_(2)O_(5)] = 1.25` mole ?A. `7.75 xx 10^(-1) "mol l"^(-1) s^(-1)`B. `6.35 xx 10^(-3) "mol l"^(-1) s^(-1)`C. `5.15 xx 10^(-5) "mol l"^(-1) s^(-1)`D. `3.85 xx 10^(-1) "mol l"^(-1) s^(-1)` |
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Answer» Correct Answer - A Rate `= K(N_(2)O_(5)) = 6.2 xx 10^(-1) xx 1.25` `= 7.75 xx 10^(-1) mol^(-1) s^(-1)` |
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| 937. |
For an elementary reaction the variation of rate constant (k) with temperature is given by the following equation `log_(10) k=5.4 -100/T` Where, T is temperature on Kelvin scal and k is in terms of `sec^(-1)` Identify the incorrect options.A. There is no finite temperature at which rate constant can be `4xx10^(6) sec^(-1)`B. Fraction of activated for the reactions will be any temperatureC. Activation energy for the reaction will be approx 460.6 cal.D. Rate of reaction will vary linearly with concentration of reactant. |
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Answer» Correct Answer - B |
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| 938. |
For a first order reaction, the half-life period is independent ofA. initial concentrationB. cube root of initial concentrationC. first power of final concentrationD. square root of final concentration |
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Answer» Correct Answer - A For nth order reaction `t_(1//2)prop(1)/(a^(1-1))` `t_(1//2)prop(1)/(a^(@))` Since `a^(@)=1` , we can conclude that for a first order reaction of the reacting species. |
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| 939. |
The rate of reaction `((dx)/(dt))` varies with nature, physical state and concentration of reactants, temperature, exposure to light and catalyst, whereas rate constant `(K)` varies with temperature and catalyst only. The rate constant `K` is given as `K=Ae^(-E_(a)//RT)` where `A` is Arrhenius parameter or pre-exponential factor and `E_(a)` is energy of activation. The minimum energy required for a reaction is called threshold energy and the additional energy required by reactant molecules to attain threshold energy level is called energy of activation. The temperature coefficient of reaction `I` is `2` and reaction `II` is `3`. Both have same speed at `25^(@)C` and show `I` order kinetics. The rati of rates of reaction of these two at `75^(@)C` is:A. (a) `7.6`B. (b) `5.6`C. (c ) `6.6`D. (d) `8.6` |
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Answer» Correct Answer - a For `I` order reaction, `r_(1)=K[]^(1)` `:. R_(1)/r_(2)=K_(1)/K_(2)`=temperature coefficient Let the rate of reaction for `I` at `25^(@)C` be `R_(1)` and the rate of reaction for `II` at `25^(@)C` be `R_(2)` `{:(,"Also"",",,,,R_(1)=R_(2)),(,,,,,"Rate of reaction"),( :.,At 25^(@)C,,R_(1),,R_(2)),(,35^(@)C,,2R_(1),,3R_(2)),(,45^(@)C,,(2)^(2)R_(1),,(3)^(2)R_(2)),(,55^(@)C,,(2)^(3) R_(1),,(3)^(2)R_(2)),(,65^(@)C,,(2)^(4)R_(1),,(3)^(4)R_(2)),(,75^(@)C,,(2)^(5)R_(1),,(3)^(5)R_(2)):}` :. Temperature coefficient for `I` reaction `=K_(35)/K_(25)=R_(35)/R_(25)=2` i.e., for each `10^(@)C` rise in temperature, rate becomes `2` times. Similarly for `II` reaction it becomes `3` times. `therefore 75^(@)C`, `("rate of reaction for II")/("rate of reaction for I")=((3)^(5)R^(2))/((2)^(5)R^(1))` `(because R_(1)=R_(2))` =7.5937 |
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| 940. |
Which of the following does not affect the rate of reaction?A. amount of the reactants takenB. physical state of the reactantsC. `Delta H `of reactionD. Size of the vessel |
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Answer» Correct Answer - C Factors affecting rate of reaction |
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| 941. |
The reaction `N_(2)O_(5)` (in `C Cl_(4)` solution) `rarr 2NO_(2)` (solution) `+(1)/(2) O_(2)(g)` is of first order in `N_(2)O_(5)` with rate constant `6.2 xx 10^(-1) s^(-1)`. What is the value of rate of reaction when `[N_(2)O_(5)] = 1.25` mole ?A. `7.75 xx 10^(-1) "mole"^(-1)s^(-1)`B. `6.35 xx 10^(-3) "mole"^(-1)s^(-1)`C. `5.15 xx 10^(-5) "mole"^(-1)s^(-1)`D. `3.85 xx 10-1 "mole"^(-1)s^(-1)` |
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Answer» Correct Answer - A Rate `= K(N_(2)O_(5)) = 6.2 xx 10^(-1) xx 1.25` `= 7.75 xx 10^(-1) mol^(-1) s^(-1)` |
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| 942. |
The energies of activation forward and revers reaction for `A_(2)+B_(2)to2AB` are 180 kJ `mol^(-1)` and 200 kJ `mol^(-1)` respectively. The pressure of a catalyst lowers the activation energy of both (forwar and reverse) reactions by `100 kJ mol^(-1)`. The magnitude of enthalpy change of the reaction `(A_(2)+B_(2)to2AB)` in the presence of catalyst will be (in Kj `mol^(-1)`):A. 280B. 20C. 300D. `-20` |
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Answer» Correct Answer - D |
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| 943. |
Consider the reaction `2N_(2)O_(5)rarr 4NO_(2)+O_(2)` If the conc. Of `N_(2)O_(5)` is reduced from `2.33 M to 2.08M` after 184 minutes, the rate of production of `NO_(2)` during this period will be ------- mol `L^(-1) min^(-1)`.A. `2.72xx10^(-3)`B. `1.36xx10^(-3)`C. `0.68xx10^(-3)`D. `8.16xx10^(-3)` |
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Answer» Correct Answer - A A verage rate of concentration of `N_(2)O_(3)` : `-(Delta[N_(2)O_(5)])/(Detat)=-((2.08-2.33)molL^(-1))/(184min) =1.36xx10^(-3)molL^(-1)min^(-1)` The stoichiometry of the reation show that the rate of appearance of `NO_(2)` is rwist the rate of appearnce of `NO_(2)` is twice the rate of disappearnce of `N_(2)O_(3)` . Thus `(Delta[NO_(2)])/(Deltat)=2(-(Delta[N_(2O_(5)])/(Deltat))` `2(1.36xx10^(-3)mole^(-1)mim^(-1))` `=2.73xx10^(-3)molL^(-1)mim^(-1)` |
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| 944. |
The reaction `2A +B to B to D + E ` involves the mechanism , ` A to B `(fast) `B to C `(slow) ` A+C to D +E ` the rate expressio would be ,A. `k[A]^(2)[B]`B. `k[B]`C. `k[A]`D. `k[A][B]` |
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Answer» Correct Answer - B Slowest step is the rate determining step , hence rate =k[B] |
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| 945. |
A certain endothermic reaction: `Ararr` Product, `DeltaH=+ve` proceeds in a sequence of three elementary steps with the rate constants `K_(1), K_(2)` and `K_(3)` and each one having energy of activation `E_(a), E_(2)` and `E_(3)` respectively at `25^(@)C`. The observed rate constant for the reaction is equal to `K_(3) sqrt(K_(1)/K_(2)). A_(1), A_(2)` and `A_(3)` are Arrhenius parameters respectively. The observed Arrhenius parameter for the reaction is:A. (a) `(2A_(1)-A_(2)+2A_(3))/(2)`B. (b) `sqrt(A_(1)/A_(2)).A_(3)`C. (c ) `A_(1)A_(2)A_(3)`D. (d) `(A_(1).A_(3))/(A_(2))` |
| Answer» Correct Answer - b | |
| 946. |
Oxalic acid is oxidised by a acidified `KMnO_(4)` as follows: `2MnO_(4)^(-)+16H^(+)+5C_(2)O_(4)^(2-)to2Mn^(2+)+10CO_(2)+8H_(2)O` The rate of this reaction increases with time becauseA. `CO_(2)` formed escapesB. of presence of sulphuric acid `[H^(+)]`C. Of formation of `Mn^(2+)` which acts an anto catalystD. `KMnO_(4) ` is a strong oxidizing agent |
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Answer» Correct Answer - C `Mn^(2+)` acts as self catalyst |
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| 947. |
For the reaction `N_(2) + 3 H_(2) to 2 NH_(3)` The rate of change of concentration for hydrogen is `0.3 xx 10^(-4) Ms^(-1)` The rate of change of concentration of ammonia is :A. `-0.2xx10^(-4)`B. `0.2xx10^(-4)`C. `0.1xx10^(-4)`D. `0.3xx10^(-4)` |
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Answer» Correct Answer - B `Rate =-(1)/(3)(d[H_(2)])/(dt)=+(1)/(2) (d[NH_(2)])/(dt)` |
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| 948. |
The rate of the reaction `C Cl_(3)CHO + NO rarr CHC_(3) + NO + CO` is given by Rate `= K[C Cl_(3)CHO][NO]`. If concentration is expressed in moles`//`litre, the units of `K` areA. `"litre"^(2) "mole"^(-2) "sec"^(-1)`B. mole `"litre"^(-1) sec^(-1)`C. `"litre mole"^(-1) sec^(-1)`D. `sec^(-1)` |
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Answer» Correct Answer - C It is a second-order reaction and the unit of `k` for second order reaction is litre `"mol"^(-1) sec^(-1)`. |
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| 949. |
Rate constant for a reaction `H_(2) + I_(2) rarr 2HI` is `49`, then rate constant for reaction `2HI rarr H_(2) + I_(2)` isA. `7`B. `1//49`C. `49`D. `21` |
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Answer» Correct Answer - B For a reversible reaction rate constant is also reverse. |
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| 950. |
In many reaction the reaction proceeds in a sequence of steps so the over all rate is determined byA. order of different stepsB. slowest stepC. molecular stepD. Fastest step |
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Answer» Correct Answer - B Slowest step is rate determing step is rate determining step . |
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