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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.
| 201. |
Statement: In rate laws, unlike in the expression for equilibrium constants, the exponents for concentration do not necessarity match stoichiometric coefficients. Explanation: It is the mechanism and not the balanced chemical equation for the overall change that governs the reaction rate.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 Reaction rate is experimental quantity and not necessary depends on stoichiometric coefficients. |
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| 202. |
A subtance undergoes first order decomposition involving two parallel first order reaction as `:` The mole percent of B in the products is `:`A. 23.17B. 76.68C. 30.16D. 69.84 |
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Answer» Correct Answer - 2 `(k_(1)xx100)/(k_(1)+k_(2))=(1.28xx10^(-4))/(1.25xx10^(-4)+3.80xx10^(-5))=76.83` |
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| 203. |
The half life period of a subtance in a certain enzyme-catalysed reaction is 138s. The time required for the concentration of the substance to fall from 1.28 mg`L^(-1)` is:A. 414sB. 552 sC. 690 sD. 276 s |
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Answer» Correct Answer - C c) Amount left after n half life peridos `[A]_(0)` `=[A]_(0)/(2^(n))` `2^(n) = [A]_(0)/[A]= (1.28 mg L^(-1))/(0.04 mg L^(-1))` `[2]^(n) = 32=[2]^(5)` No. of `t_(1//2)` periods =5 Time required = `138 xx 5=690s` |
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| 204. |
The half life of a substance in a certain enzyme catalyzed reaction is 138s. The time required for the concentration of the substance to fall from `1.28 mg L^(-1) to 0.04 mg L^(-1)` :A. 267sB. 414sC. 552sD. 690s |
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Answer» Correct Answer - A Enzyme catalyzed reactions follow first order Kinetics, where `t_(1//2)` is constant. At the end of every half life, half of the reactant is consumed. Thus, `1.28mgL^(-1)overset(t_(1//2))(rarr)0.64mgL_(-1)overset(t_(1//2))(rarr)0.32mgL^(-1)`overset(t_(1//2))(rarr)0.16mgL_(-1)overset(t_(1//2))(rarr)0.08mgL_(-1)overset(t_(1//2))(rarr)0.04mgL_(-1)` Since five half-lives are involved to reduce `1.28mhL^(-1)` to `0.04mgL^(-1)` , the total times is `5xx138s=690s` Alternatively `0.04mgL^(-1)=(1.28mgL^(-1))/((2)^(n))` or `2^(n)=(1.28mgL^(-1))/(0.04mgL^(-1))=32=(2)^(5)` Hence, `n=5` , i.e., five half lives are involved. |
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| 205. |
Populatin growth of humans and bacteria follows first order growth kinetics. Suppose 50 bacteria are placed in a flask containing nutrients for the bacteria so that they can multiply . A study at `3.5^(@)C` gave the following results : `{:("Time (minutes)",0,15,30,45,60),("Number of bacteria",100,200,400,800,1600):}` Unit of rate constant for first order growth is:A. `"min"^(-1)`B. `"min"^(-2)`C. `"min"^(-3)`D. unitless |
| Answer» Correct Answer - A | |
| 206. |
Populatin growth of humans and bacteria follows first order growth kinetics. Suppose 50 bacteria are placed in a flask containing nutrients for the bacteria so that they can multiply . A study at `3.5^(@)C` gave the following results : `{:("Time (minutes)",0,15,30,45,60),("Number of bacteria",100,200,400,800,1600):}` The rate constant for the reaction is :A. `0.0462 "min"^(-1)`B. `0.462 "min"^(-1)`C. `4.62 "min"^(-1)`D. `46.2 "min"^(-1)` |
| Answer» Correct Answer - A | |
| 207. |
The half life of a substance in a certain enzyme-catalysed reaction is 138 s . The time required for the concentration of the substance to fall from 1.28 mg `L^(-1)` to `0.04` mg `L^(-1)` isA. 690sB. 276 sC. 414 sD. 552 s |
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Answer» Correct Answer - a Enzyme catalysed reactions are initially follow first order kinetics when concentration decreases `1.28 mg l^(-1)` to 0.04 mg `l^(-1)` . Then five half-life completed No. of half-lives =5 So , times required = `5 xx 138 = 690 `s |
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| 208. |
In a particular case of bacterial growth following second order kinetics, concentraion of bacteria increases to 4 times initial concentration of 1 M in 24 minutes. What will be the generation time of the bacterial growth, if initial concentration is 2 M.A. 24 minB. 16 minC. 8 minD. 12 min |
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Answer» Correct Answer - C |
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| 209. |
If the half-life period of a first order reaction is 138.6 min, then the value of decay constant for the reaction will be `("in min"^(-1))`A. 5B. 0.5C. 0.05D. 0.005 |
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Answer» Correct Answer - D `lambda=(0.693)/(t_(1//2)),Given, t_(1//2)=138.6min` `therefore" "lambda=(0.693)/(138.6)=0.005"min"^(-1)` |
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| 210. |
In the reaction `A+2B to 3C + 2D`, the rate of disappearance of B is ` 1 xx 10^(-2) mol L^(-1)s^(-1)`. What will be the rate of the reaction and rate of change in concentration of A and C? |
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Answer» `A+2B to 3C + 2D` `Rate = (-d[A])/(dt) = -1/2(d[B])/(dt) = 1/3(d[C])/(dt) = 1/2(d[D])/(dt)` `(d[A])/(dt) = 1/2(d[B])/(dt) = 1/2 xx (1 xx 10^(-2))=0.5 xx 10^(2) mol L^(-1)s^(-1)` `(d[C])/(dt) = 3/2 (d[B])/(dt) = 3/2 xx (1 xx 10^(-2)) = 1.5 xx 10^(-2)mol L^(-1)s^(-1)` |
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| 211. |
In order to experimentally determine order, initial rate method can be used Once order is determined integrated rate law can be obtained by using concentration of reactants Based on this information and data given below, `{:(,"[A] Molarity",,"[B] Molarity",,Rate of Reaction[Msec^(-1)]),("1",10^(-2)M,,3xx10^(-3)M,,2xx10^(-2)),("2",2xx10^(-2)M,,3xx10^(-3)M,,4xx10^(-2)),("3",2xx10^(-2),,6xx10^(-3)M,,8xx10^(-2)):}` Answer the question which follow Reaction: `A(g)+B(g) rarr C(g)` What will be the value of rate constant for the above reaction?A. `2/3xx10^(3)M min^(-1)`B. `2/3xx10^(3)M^(-1)min^(-1)`C. `4xx10^(4)M^(-1)min^(-1)`D. `4xx10^(4)M min^(-1)` |
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Answer» Correct Answer - B |
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| 212. |
Rate of disappearance of the reactant A at two different temperature is given by `A rarr B` `(-d[A])/(dt)=2 xx 10^(-2) "sec"^(-1)[A]-4xx10^(-3)"sec"^(-1)[B] at 300 K` `(-d[A])/(dt)=4 xx 10^(-2) "sec"^(-1)[A]-16xx10^(-4)"sec"^(-1)[B] at 400 K` heat of reaction in the given temperature range, when equilibrium is set up is :A. `(2.303 xx 2 xx 300 xx 400)/(100) "log"50` CalB. `(2.303 xx 2 xx 300 xx 400)/(100) "log"250` CalC. `(2.303 xx 2 xx 300 xx 400)/(100) "log"5` CalD. none of these |
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Answer» Correct Answer - C |
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| 213. |
For a first order reaction, half life is found to be `138.6` min, what will be the rate of disappearance after 1 half life if initial concentraion of reactant is 3M?A. `7.5xx10^(-3)"M min"^(-1)`B. `(3)/(2)"M sec"^(-1)`C. 5xx10^(-3)"M min"^(-1)`D. `15xx10^(-3)"M min^(-1)` |
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Answer» Correct Answer - A |
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| 214. |
For a reaction : 2A(g) `rarr` B(g) + 3C(g), rate constant of disappearance of A is `10^(-3) "M sec"^(-1)`. If initially 2 M of A is taken then what will be concentration of C afterr 5 minutes?A. `0.3M`B. `0.9M`C. `0.45M`D. `1.5xx10^(-2)M` |
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Answer» Correct Answer - C |
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| 215. |
For the reaction `A + 2B rarr` products (started with concentration taken in stoichiometric proportion), the experimentally determined rate law is : `-(d[A])/(d t) = ksqrt([A])sqrt([B])` The half life time of the reaction would be :A. `(0.693)/(k)`B. `(0.693)/(1//k)`C. `(0.693)/(sqrt(2k))`D. not defined |
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Answer» Correct Answer - C `{:(A,+,,2B rarr,"products"),(a-x,,,2a - 2x,):}` `- (d[A])/(d t) = ksqrt([A])sqrt([B])` Reactant are in their stoichiometric proportional `rArr -(d(a-x))/(d t) = k = sqrt((a-x)) sqrt(2(a-x))` `(d x)/(d t) = sqrt(2)k (a-x)" "t_(1//2) = (0.693)/(sqrt(2)k)` |
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| 216. |
AT 373 K, a gaseous reaction A `rarr` 2B + C is found to be of first order. Starting with pure A, the total pressure at the end of 10 min. was 176 mm and after a long time when A was completely dissociated, it was 270 mm. The pressure of A at the end of 10 minutes was :A. 94 mmB. 47 mmC. 43 mmD. 90 mm |
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Answer» Correct Answer - B |
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| 217. |
For the following reaction : A(g) +B(g) `rarr` 2C(g) + D(g), the rate law is given as rate of disappearance `A=K[A]^(2//3)[B]^(1//3)`. If initial concentration of A and B are 2M each and no C and D were present initially, then the time at which total concentration (sum of concentration A, B, C, and D) will become `5.5` M, is : (Given : K = `1.386xx10^(-1)min^(-1)`A. 5 minB. 10 secC. 600 secD. `(1)/(2)`min |
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Answer» Correct Answer - C |
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| 218. |
Taking the reaction, `A +2B rarr` Products to be of second order, which of the following is/are the correct rate law expresisons `(s)`?A. `(dx)/(dt) = k[A]^(2)`B. `(dx)/(st) = k[A][B]^(2)`C. `(dx)/(dt) = k[A][B]`D. `(dx)/(dt) = k_(1)[A]+k_(2)[B]^(2)` |
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Answer» Correct Answer - A::C For second order reaction, order is `2`. |
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| 219. |
Consdier a first order gas phase reaction, A(g) `rarr` 2B(g) + C(g) The initial pressure taking only A in a rigid vessel was found to be P atme. After a lapse of 10 minutes, the pressure of the system increases by x units and became `P_(10)` atm. The rate constant for the reaction is given by :A. `k=(2.303)/(10)"log"(P)/(P+x)"min"^(-1)`B. `k=(2.303)/(10)"log"(P)/(P-(x)/(2))"min"^(-1)`C. `k=2.303xx"6 log"(P)/(P_(10)-(P))"hr"^(-1)`D. `k=2.303xx"6 log"(2P)/(3P-P_(10))"hr"^(-1)` |
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Answer» Correct Answer - B |
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| 220. |
In a first order reaction, `75%` of the reactants disappeared in `1.386 hr`. What is the rate constant ?A. `3.6 xx 10^(-3)s^(-1)`B. `7.2 xx 10^(-3)s^(-1)`C. `2.7 xx 10^(-4)s^(-1)`D. `1.8 xx 10^(-3)s^(-1)` |
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Answer» Correct Answer - C Let original amount `N_(0) = 100` Since, disappeared amount `= 75%` `:.` left amount, `N = 100 - 75 = 25` `:.` Since, `(N_(0))/(N) = 2^(n)` where `n =` no. of hlaf lives or, `(100)/(25) = 2^(n)` or, `2^(2) = 2^(2)`, here `n = 2` `because` Total time `= n xx t_(1//2)` `1.386` hours `= 2 xx t_(1//2)`, `t_(1//2) = 0.693` hours `= 2494.8 sec`. `k = (0.693)/(2494.8) = 2.7 xx 10^(-4) sec^(-1)` |
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| 221. |
For the first order reaction `A(g) rarr 2B(g) + C(g)`, the initial presuure is `P_(A) = 90 m Hg`, the pressure after `10` minutes is found to be `180 mm Hg`. The rate constant of the reaction isA. `1.15 xx 10^(-3) sec^(-1)`B. `2.3 xx 10^(-3) sec^(-1)`C. `3.45 xx 10^(-3) sec^(-1)`D. `6 xx 10^(-3) sec^(-1)` |
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Answer» Correct Answer - A `{:("The reaction is",A_((g)),rarr,B_((g)),+, C_((g))),("at t=0 pressure",PA,,0,,0),("Pressure after",P_(4)-x,,2x,,x),("10 minutes",,,,,):}` Total pressure after `10` minutes, `P_(T) = P_(A) + P_(B) + P_( C)` or, `P_(T) = (P_(A) - x) + 2x + x` `= P_(A) + 2x = 180` `because P_(A) = 90mm` `90 + 2x = 180` or, `X = 45 mm` `:. P_(A) - x = 90 - 45 = 45 mm` Since, `kt = 2.303 "log"(P_(A))/(P_(A) - x)` or, `k xx 10 = 2.303 "log"(90)/(90 - 45)` `k xx 10 = 0.693` or, `k(0.693)/(10) = 0.0693 "min"^(-1)` `= (0.693)/(10 xx 60 sec) = 1.155 xx 10^(-3) sec^(-1)` |
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| 222. |
For a first order reaction, the ratio of time for the completion of `99.9%` and half of the reaction isA. `8`B. `10`C. `9`D. `12` |
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Answer» Correct Answer - B For reaction first order, `t_(1//2) = (2.303)/(k) "log"(a)/(a-(a)/(2))` `= (2.303)/(k) log2 = (2.303)/(k) = 0.3010` `t_(99.9%) = (2.303)/(k)"log"(a)/(a-0.999 a)` `= (2.303)/(k)log10^(-3) = (2.303)/(k) xx 3` `(t_(99.9%))/(t_(1//2)) = (3)/(0.3010) ~~ 10`. |
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| 223. |
A first order reaction is one-fifth completed in 40 minutes. The time reuired for its `100%` completion is :A. 100 minutesB. 200 minutesC. 350 minutesD. infinity |
| Answer» Correct Answer - D | |
| 224. |
The reaction `N_(2)O_(5) to 2NO_(2)+I//2O_(2)` is of first order in `N_(2)O_(5)`. Its rate constant is `6.2 xx 10^(-6)s^(-1)`. If the beginning `N_(2)O_(5)` is 15 `mol L^(-1)`, calculate the rate of reaction in the beginning. |
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Answer» Rate of reaction( R) = `k[N_(2)O_(5)]=6.2 xx 10^(-6)s^(-1)xx 15 mol L^(-1)` `=93 xx 10^(-6)mol L^(-1)s^(-1)= 9.3 xx 10^(-5) molL^(-1)s^(-1)` |
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| 225. |
The gas phase reaction `2NO_(2)+O_(3)rarrN_(2)O_(5)+O_(2)` has the rate constant `k = 2.0xx10^(-4) dm^(3) mol^(-1) s^(-1)` at 300 K. The order of the reaction is |
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Answer» Correct Answer - C The units of rate constant depend on th eoverall reaction order. Units of rate constant, we can decide the overall reaction order. If `n` is the overall order of th ereaction, then units of rate constant are given as `["Concentration"]^(^(1-n))[sec]^(-1)` `[moldm^(-3)]^(1-n)[sec]^(-1)` Equating the units of given rate constant with the general expression we get, `dm^(3)mol^(-1)s^(-1)=[moldm^(-3)]^(1-n)[sec]^(-1)` `n=2` |
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| 226. |
For the reaction, `N_(2)O_(5) rarr 2NO_(2)+1/2O_(2), Given` `-(d[N_(2)O_(5)])/(dt)=K_(1)[NO_(2)O_(5)]` `(d[NO_(2)])/(dt)=K_(2)[N_(2)O_(5)] and (d[O_(2)])/(dt)=K_(3)[N_(2)O_(5)]` The relation in between `K_(1), K_(2)` and `K_(3)` is:A. `2K_(1)=K_(2)=4K_(3)`B. `K_(1)=K_(2)=K_(3)`C. `2K_(1)=4K_(2)=K_(3)`D. None of these |
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Answer» Correct Answer - a `(-d[N_(2)O_(5)])/(dt)=1/2(d[NO_(2)])/(dt)=(2d[O_(2)])/(dt)` `:. K_(1)[N_(2)O_(5)]=K_(2)/2[N_(2)O_(5)]=2K_(3)[N_(2)O_(5)]` |
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| 227. |
Consider the reaction ` 2NO_((g) ) +O_(2(g) ) to 2NO_(2(g)).` if ` (d [NO_(2)])/(dt)=0.052 M//s ` then `-(d[O_(2)])/(dt)` will beA. `0.052 m//s`B. `0.114m//s`C. `0.026m//s`D. `-0.026 m//s` |
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Answer» Correct Answer - C `"Rate"=-(1)/(2)(d[NO])/(dt)=-(d[O_(2)])/(dt)=+(1)/(2) (d[NO_(2)])/(dt)` `-(d[O_(2)])/(dt)=(1)/(2)xx0.052 =0.026 m//s` |
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| 228. |
Formation of `NO_(2)F` from `NO_(2)` and `F_(2)` as per the reaction `2NO_(2)(g)+F_(2)(g)rarr2NO_(2)F(g)` is a second order reaction, first order with respect to `NO_(2)` and first order with respect to `F_(2)`. If `NO_(2)` and `F_(2)` are present in a closed vessel in ratio `2:1` maintained at a constant temperature with an initial total pressure of 3 atm, what will be the total pressure in the vessel after the reaction is complete?A. 1 atmB. 2 atmC. `2.5` atmD. 3 atm |
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Answer» Correct Answer - B |
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| 229. |
For the reaction. `H_(2)(g) +2ICI(g) rarr 2HCl(g) +I_(2)(g)` one proposed mechanism is `H_(2)(g) +ICI(g) rarr HICI(g) +H(g)" "` (fast) `H(g) +ICl(g)rarr HCl(g) +I(g) " "`(fast) `HICl(g) rarr HCl(g) +I(g)" "` (fast) `I(g) +I(g) rarr I_(2)(g)" "`(fast) Intermediates in this reaction include which of the following?A. HICl onlyB. I onlyC. HICl and H onlyD. HICl, H and I |
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Answer» Correct Answer - D |
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| 230. |
An increse in the concentration of the reactants of a reaction leads to change in:A. Heat of reactionB. Threshold energyC. collision frequencyD. Activation energy |
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Answer» Correct Answer - C Collision theory |
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| 231. |
The term `-dx//dt` in the rate expresison refers to theA. The concentration of the reactantsB. Increase in the concentration of the reactantsC. The instantaneous rate of the reactionD. The average rate of the reaction |
| Answer» Correct Answer - C | |
| 232. |
For the reaction `NO_(2)+COrarrCO_(2)+NO`, the experimental rate expresison is `-dc//dt = k[NO_(2)]^(2)`. Find the number of molecules of `CO` involved in the slowest step.A. `0`B. `1`C. `2`D. `3` |
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Answer» Correct Answer - A Since there is no `[CO]` involved in rate expression therefore `CO` is involved in fast step. |
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| 233. |
The rate of chemical reactionA. increase as the reaction proceedsB. decreases as the reaction prodceedsC. may increase or decrease during the reactionD. reamains constant as the reaction proceeds |
| Answer» Correct Answer - B | |
| 234. |
The term `-dx//dt` in the rate expresison refers to theA. instantaneous rate of reactionB. average rate of reactionC. increasing in the concentration of reactantsD. comcentration of reactants |
| Answer» Correct Answer - A | |
| 235. |
For the hypothetical reaction `2A rarr3C` , the reaction rate r in terms of the rate of change of the concentration is given byA. `r=- (d[A])/(dt)`B. `=-(1)/(2)(d[A])/(dt)`C. `r=- (1)/(3) (d[A])/(dt) `D. `r= (d[A])/(dt)` |
| Answer» Correct Answer - B | |
| 236. |
For the reaction A `rarr` B that is first-order in A, The rate constant is `2.08xx10^(-2)s^(-1)`. How long would it take for [A] to change from `0.100M` to `0.0450M` ?A. `0.0166s`B. `16.7s`C. `38.4s`D. 107 s |
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Answer» Correct Answer - C |
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| 237. |
The rate of disapperance of `SO_(2)` in the reaction `2SO_(2) + O_(2) to 2SO_(3)` is `1.28 xx 10^(-3) ` g/sec then the rate of formation of `SO_(3)` isA. `0.64 xx 10^(-3)` g /secB. `0.80 xx 10^(-3)` g /secC. `1.28 xx 10^(-3)` g /secD. `1.60 xx 10^(-3)` g /sec |
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Answer» Correct Answer - c The rate of formation of `SO_(3)` is `1.28 xx 10^(-3)` g/sec. |
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| 238. |
Why in the redox titration of `KMnO_(4)` vs oxalic acid, we heat oxalic acid solution before starting the titration? |
| Answer» The reaction is very slow at room temperature. By raising the temperature, the reaction rate increases. Normally in the reaction, `KMnO_(4)` acidified with dilute `H_(2)SO_(4)` reacts with oxalic acid at a temperature of about `50^(@)`C. | |
| 239. |
If for a reaction in which `A(g)` converts to `B(g)` the reaction carried out at const. `V` & `T` results into the following grapg. A. then the reaction must be `A(g) rarr 3B (g)` and is a first -order reaction.B. then the reaction must be `A(g) rarr 3B(g)` and is a second-order reaction.C. then the reaction must be `A(g) rarr 3B(g)` and is a zero-order reaction.D. then the reaction must be `A(g) harr 3B(g)` and is first-order reaction. |
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Answer» Correct Answer - C From the given graph it is clear that the concentration decrease linearly with time zero-order reaction and `A rarr 3B` type. |
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| 240. |
Two reaction, `(I)A rarr` Products and `(II) B rarr` Products, follow first order kinetics. The rate of reaction `(I)` is doubled when the temperature is raised form `300 K` to `310K`. The half life for this reaction at `310K` is `30 min`. At the same temperature `B` decomposes twice as fast as `A`. If the energy of activation for reaction `(II)` is twice that of reaction `(I)`, (a) calculate the rate of constant of reaction `(II)` at `300 K`. |
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Answer» Correct Answer - `0.03267 min^(-1)` For reaction `(I)` `A rarr` Products `T_(1) = 300 K, T_(2) = 310 K` `(k_(1))_(A)` i s at `300 K` and `(k_(2))_(A)` is at `310 K` Given: `[(k_(2)(310 K))/(k_(1)(300 K))]_(A) = 2` …(i) Uisng Arrhenius equation for reaction `(I)` : `[(k_(2)(310 K))/(k_(1)(300 K))]_(A) = (E_(a(A)))/(2.303 R)((T_(2)-T_(1))/(T_(1)T_(2)))` ...(ii) `log 2 = (E_(a(A)))/(2.303R) ((T_(2)-T_(1))/(T_(1)T_(2)))` ...(iii) (For reaction `II`) `B rarr` Products Since at `310 K , B` decomposes twice as fast as `A`. `because [k_(2)(310 K)]_(A) = 0.0231 min^(-1)` `:. [k_(2)(310 K)]_(B) = 2 xx 0.0231 min^(-1)` `= 0.0462 min^(-1)` We have to calculate `[k_(1) (300 K)]_(B) =` ? Uisng Arrhenius equation for reaction `(II)` `log[(k_(2)(310 K))/(k_(1)(300 K))]_(B) = (E_(a(B)))/(2.303 R) ((T_(2)-T_(1))/(T_(1)T_(2)))` ...(iv) Given: `(E_(a))_(B) = (1)/(2)(E_(a))_(A)` ...(v) Substitute the value of Eq. (v) in Eq. (iv), `log[(k_(2)(310 K))/(k_(1)(300 K))]_(B) = (1)/(2) xx (E_(a(A)))/(2.303 R)((T_(2)-T_(1))/(T_(1)T_(2)))` (vi) Comparing Eqs. (iii) and (vi), we get `log[(k_(2)(310 K))/(k_(1)(300 K))]_(B) = (1)/(2) xx log 2` `:. (0.0462)/([k_(1)(300 K)]_(B)) = (2)^((1)/(2)) = sqrt(2) = 1.414` ....(vii) `:. [k_(1) (300 K)]_(B) = (0.0462)/(1.414) = 0.03267 min^(-1)` Correct Answer - `0.03267 min^(-1)` For reaction `(I)` `A rarr` Products `T_(1) = 300 K, T_(2) = 310 K` `(k_(1))_(A)` i s at `300 K` and `(k_(2))_(A)` is at `310 K` Given: `[(k_(2)(310 K))/(k_(1)(300 K))]_(A) = 2` …(i) Uisng Arrhenius equation for reaction `(I)` : `[(k_(2)(310 K))/(k_(1)(300 K))]_(A) = (E_(a(A)))/(2.303 R)((T_(2)-T_(1))/(T_(1)T_(2)))` ...(ii) `log 2 = (E_(a(A)))/(2.303R) ((T_(2)-T_(1))/(T_(1)T_(2)))` ...(iii) (For reaction `II`) `B rarr` Products Since at `310 K , B` decomposes twice as fast as `A`. `because [k_(2)(310 K)]_(A) = 0.0231 min^(-1)` `:. [k_(2)(310 K)]_(B) = 2 xx 0.0231 min^(-1)` `= 0.0462 min^(-1)` We have to calculate `[k_(1) (300 K)]_(B) =` ? Uisng Arrhenius equation for reaction `(II)` `log[(k_(2)(310 K))/(k_(1)(300 K))]_(B) = (E_(a(B)))/(2.303 R) ((T_(2)-T_(1))/(T_(1)T_(2)))` ...(iv) Given: `(E_(a))_(B) = (1)/(2)(E_(a))_(A)` ...(v) Substitute the value of Eq. (v) in Eq. (iv), `log[(k_(2)(310 K))/(k_(1)(300 K))]_(B) = (1)/(2) xx (E_(a(A)))/(2.303 R)((T_(2)-T_(1))/(T_(1)T_(2)))` (vi) Comparing Eqs. (iii) and (vi), we get `log[(k_(2)(310 K))/(k_(1)(300 K))]_(B) = (1)/(2) xx log 2` `:. (0.0462)/([k_(1)(300 K)]_(B)) = (2)^((1)/(2)) = sqrt(2) = 1.414` ....(vii) `:. [k_(1) (300 K)]_(B) = (0.0462)/(1.414) = 0.03267 min^(-1)` |
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| 241. |
The gas phase decompoistion of dimethyl ether follows first order kinetics. `CH_(3)-O-CH_(3)(g)rarrCH_(4)(g)+CO(g)` The reaction is carried out in a constant volume container at `500^(@)C` and has a half life of `14.5 min`. Initially, only dimethyl ether is present at a pressure `0.40 atm`. What is the total pressure of the system after `12min`? (Assume ideal gas behaviour) |
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Answer» Correct Answer - `0.7492 atm` `{:(CH_(3)-,O-CH_(3)to,CH_(4)+,H_(2),+CO),("Inital",0.4,0,0,0),("Final",(0.4-x),x,x,x):}` For first order reaction, Half life `=(0.693)/(k)` or `14.5=(0.693)/(k)` or `k=(0.693)/(14.5)= 4.78xx10^(-2) min^(-1)` Also, `k=(2.303)/(t) log.(A_(0))/(A_(t)) = (2.303)/(t) log.(P_(0))/(P)` or `(0.693)/(14.5) = (2.303)/(12) log.(0.40)/((0.40-x))` or `x = 0.1746 atm` Total pressure of container after `12 min` is `(0.40 -x)+x+x+x = 0.4 + 2x` `= 0.4 + 2xx 0.1746` `= 0.7492 atm` Correct Answer - `0.7492 atm` `{:(CH_(3)-,O-CH_(3)to,CH_(4)+,H_(2),+CO),("Inital",0.4,0,0,0),("Final",(0.4-x),x,x,x):}` For first order reaction, Half life `=(0.693)/(k)` or `14.5=(0.693)/(k)` or `k=(0.693)/(14.5)= 4.78xx10^(-2) min^(-1)` Also, `k=(2.303)/(t) log.(A_(0))/(A_(t)) = (2.303)/(t) log.(P_(0))/(P)` or `(0.693)/(14.5) = (2.303)/(12) log.(0.40)/((0.40-x))` or `x = 0.1746 atm` Total pressure of container after `12 min` is `(0.40 -x)+x+x+x = 0.4 + 2x` `= 0.4 + 2xx 0.1746` `= 0.7492 atm` |
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| 242. |
Higher order `(gt3)` reaction are rare due to :A. Shifting of equilibrium towards reactants due to elastic collisionsB. Loss of active species on collisionsC. Low probability of simultaneous collisons of all the reaction species.D. Increase in entropy and activation energy as more molecule are involved. |
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Answer» Correct Answer - C c) it is the correct answer. |
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| 243. |
The decompoistion of `Cl_(2)O_(7)` at `400 K` in gas phase to `Cl_(2)` and `O_(2)` is a first order reaction. a. After `55 s` at `400 K`, the pressure of `Cl_(2)O_(7)` falls form `0.062` to `0.044 atm`. Calculate `k`. b. Calculate the pressure of `Cl_(2)O_(7)` after `100 s` of decompoistion. |
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Answer» As pressuer `prop` concentration, `k=(2.303)/(t)"log"_(10)(P_(i)("initial pressure"))/(P_(t)("pressure after time t"))` `=(2.303)/(55)"log"_(10)(0.062)/(0.044)=6.2xx10^(-3)s^(-1)` (ii) Again applying the first order kinetic eqution, `k=(2.303)/(t)"log"_(10)(P_(i)("intial pressure"))/(P_(t)("pressure after time t"))` `6.2xx10^(-3)=(2.303)/(100)"log"_(10)(0.062)/(P_(t)` `or(6.2xx10^(-3)xx100)/(2.303)="log"0.062-log_(10)(P_(t))` `or 0.2692=log_(10)0.062-0.2692` `or log_(10)(P_(t))=log_(10)0.062-0.2692` `=(bar(2).7924-0.2692)` `P_(t) =0.033` atmoshpere Pressure after 100 sec =0.033 atm |
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| 244. |
Higher order `(gt 3)` reactions are rare due toA. Low probability of simultaneous collision of all the reacting speciesB. Increase in entropy and activation energy as more molecules are involvedC. Shifting of equilibrium towards reactants due to elastic collisionsD. Loss of active species on collision |
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Answer» Correct Answer - a Higher order `(gt3)` reactions are rare due to low probability of simultaneous collision of all the reacting species . |
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| 245. |
For a reaction in which `A` and `B` form `C`, the following data were obtained from three experiments: What is the rate equation of the reaction and what is the value of rate constant ? |
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Answer» Let the rate equation be `k[A]^(x)[X]^(y)` From `"From expt"(1),0.3xx10^(-4)=[0.03]^(x)[0.03]^(y)" "....(i)` `"From expt"(2),1.2xx10^(-4)=[0.06]^(x)[0.06]^(y)" "....(ii)` `(1.2xx10^(-4))/(0.3xx10^(-4))=([0.06]^(x)[0.06]^(y))/([0.03]^(x)[0.03]^(y))` `" " =2^(x)xx3^(x)=4 " "....(iii)` Similarly, from expt. (1) and expt (3) `" " 2^(x)xx3^(y)=9 " " ....(iv)` Solving eqs. (ii) and (iv), `" " x=0,y=2` Rate equation `" "` Rate =`k[B]^(2)` Consider eq. (i) again, `k=(0.3xx10^(-4))/([0.03]^(2))=3.33xx10^(-2)"mol"^(-1)L s^(-1)` |
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| 246. |
For the non-stoichiometre reaction `2 A + B to C + D` , the following kinetic data were obtained in three separate experiments , all at 298 K The rate law for the formation of C isA. `(dc)/(dt) = k[A] [B]`B. `(dc)/(dt) = k [A]^(2) [B]`C. `(dc)/(dt) = k [A] [B]^(2)`D. `(dc)/(dt) = k [A]` |
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Answer» Correct Answer - d `1.2 xx 10^(-3) = K (0.1)^(x) (0.1)^(y)` `1.2 xx 10^(-3) = K (0.1)^(x) (0.2)^(y)` `2.4 xx 10^(-3) = K (0.2)^(x) (0.1)^(y) implies R = K (0.1)^(x) (0.2)^(y)` `2.4 xx 10^(-3) = K (0.2)^(x) (0.1)^(y) implies R = K[A]^(1) [B]^(0)` |
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| 247. |
The rate of a reaction doubles when its temperature changes from 300 K to 310 K . Activation energy of such a reaction will be : (R = `8.314 JK^(-1) mol^(-1)` and log2 = 0.301)A. `53.6 kJ mol^(-1)`B. 48.6 kJ `mol^(-1)`C. 58.5 kJ `mol^(-1)`D. 60.5 kJ `mol^(-1)` |
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Answer» Correct Answer - a log `(K_(2))/(K_(1)) = (-E_(a))/(2.030R) ((1)/(T_(2)) - (1)/(T_(1)))` `(K_(2))/(K_(1)) = 2 , T_(2) = 310 K " " T_(1) = 300` K `implies ` log `2 = (-E_(a))/(2.303 xx 8.314 ) ((1)/(310) - (1)/(300))` `implies E_(a) = 53598.6` J/mol = 53.6 KJ/mol |
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| 248. |
For the elementary reaction `M to N` , the rate of disappearance of M increases by a factor of 8 upon doubling the concentration of M . The order of the reaction with respect to M isA. 4B. 2C. 2D. 1 |
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Answer» Correct Answer - b `M to N` `r = K[M]^(x)` as [M] is doubled rate increases by a factor of 8 . i.e. 8r = `K [2M]^(x) implies 8 (2)^(x) implies x = 3`. |
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| 249. |
The rate of a reaction doubles when its temperature changes form `300 K` to `310 K`. Activation energy of such a reaction will be: `(R = 8.314 JK^(-1) mol^(-1) and log 2 = 0.301)`A. `48.6 kJ mol^(-1)`B. (b) `58.5 kJ mol^(-1)`C. `60.5 kJ mol^(-1)`D. `53.6 kJ mol^(-1)` |
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Answer» Correct Answer - D As per Arrhenius equation: `log. (K_(2))/(K_(1)) = (E_(a))/(2.3R)[(T_(2)-T_(1))/(T_(2)T_(1))]` `2.3 log2 = (E_(a))/(8.314)[(10)/(300 xx 310)]` `:. E_(a) = 53.6 kJ mol^(-1)`Correct Answer - D As per Arrhenius equation: `log. (K_(2))/(K_(1)) = (E_(a))/(2.3R)[(T_(2)-T_(1))/(T_(2)T_(1))]` `2.3 log2 = (E_(a))/(8.314)[(10)/(300 xx 310)]` `:. E_(a) = 53.6 kJ mol^(-1)` |
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| 250. |
In the reaction , `P + Q to R + S` The time taken for 75% reaction of P is twice the time taken for 50% reaction of P . The concentration of Q varies with reaction time as shown in the figure . The overall order of the reaction is A. 2B. 3C. 0D. 1 |
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Answer» Correct Answer - d For P , if `t_(50%) = x ` , Then `t_(75%) = 2x` This happens only in first order reaction . so , order with respect to F is 1. For Q , the graph shows that concentration of Q decreases linearly with time . So rate , with respect to Q , remains constant . Hence , it is zero order wrt Q. So overall order is 0 + 1 = 1 |
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