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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.
| 1501. |
The raction `2FeCl_(3) + SnCl_(2) rarr 2FeCl_(2) + SnCl_(4)` is an example ofA. first order reactionB. third order reactionC. second order reactionD. Zero order reaction |
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Answer» Correct Answer - B Rate `=K [FeCl_(3)]^(2)[SnCl_(2)]` `n=2+1=3` |
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| 1502. |
Assertion: According to transition state theory, for the formation of an activated complex, one of the vibrational degree of freedom is converted into a translation degree of freedom. Reason: Energy of the activated complex is higher than the energy of the reactant molecules.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but the reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
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Answer» Correct Answer - A `A + B rarr X^(**) rarr` Products The activated complex `X^(**)` has higher energy than reactants. It is a special molecule which decomposes such that one vibrational degree of freedom is converted into a translational degree of freedom along the reaction coordinate. |
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| 1503. |
Assertion (A) : The molecularity of the reaction `H_(2) + Br_(2) rarr 2HBr` is `2`. Reason (R ): The order of the reaction is `3//2`.A. If both `(A)` and `(R )` are correct, and `(R )` is the correct explnation of `(A)`.B. If both `(A)` and `(R )` are correct, but `(R )` is noth the correct explanation of `(A)`.C. If `(A)` is correct, but `(R )` is incorrect.D. If `(A)` is incorrect, but `(R )` is correct. |
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Answer» Correct Answer - B Molecularity is `2` because two molecules of the reactants are involved in the given elementary reactions.Correct Answer - B Molecularity is `2` because two molecules of the reactants are involved in the given elementary reactions. |
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| 1504. |
Assertion (A): For the reaction `RCl + NaOH(g) rarr ROH + NaCl`, the rate of reaction is reduced to half on reducing the cocentration of `RCl` to half. Reason (R ): The rate of the reaction is represented by `k[RCl]`, i.e., it is a first order reaction.A. If both `(A)` and `(R )` are correct, and `(R )` is the correct explnation of `(A)`.B. If both `(A)` and `(R )` are correct, but `(R )` is noth the correct explanation of `(A)`.C. If `(A)` is correct, but `(R )` is incorrect.D. If `(A)` is incorrect, but `(R )` is correct. |
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Answer» Correct Answer - C The rate of reaction is represented by `k[RCl][NaOH]`, i.e., it is a second order reaction.Correct Answer - C The rate of reaction is represented by `k[RCl][NaOH]`, i.e., it is a second order reaction. |
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| 1505. |
During the study of kinetics of chemical or nuclear reaction, `t_(1//3)` can be defined asA. A. One-third of the reaction molecules are left.B. B. Two-third of the reaction molecules are left.C. C. One-third of half of the reaction molecules are left.D. D. One-third of two-thirds of the reaction molecules are left. |
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Answer» Correct Answer - B `t_(1//3)` is the time, in which `(1)/(3)rd` of the initial reaction molecules have been converted into Products. `rArr (2)/(3) rd` of the initial reaction molecules will be left. |
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| 1506. |
The rate constant for forward reaction `A(g) hArr 2B(g)` is `1.5 xx 10^(-3) s^(-1)` at `100 K`. If `10^(-5)` moles of `A` and `100 "moles"` of `B` are present in a `10-L` vessel at equilibrium, then the rate constant for the backward reaction at this temperature isA. `A. 1.50 xx 10^(-4) L mol s^(-1)`B. `B. 1.50 xx 10^(-11) L mol^(-1) s^(-1)`C. `C. 1.50 xx 10^(-10) L mol^(-1) s^(-1)`D. D. None of these |
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Answer» Correct Answer - B `k_(eq) = ([B]^(2))/([A]) = ([100//10]^(2))/([10^(-5)//10]) = 10^(8)` `k_(eq) = (k_(f))/(k_(b)) = rArr k_(b) = (1.5 xx 10^(-3))/(10^(8)) = 1.5 xx 10^(-11) L mol^(-1) s^(-1)` |
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| 1507. |
Consider the following parallel reactions being given by `A(t_(1//2) = 1.386 xx 10^(2) hours)`, each path being first order. If the distribution of `B` in the Product mixture is `50%`, the partical half life of `A` for converison into `B` isA. `A. 346.5 h`B. `B. 131 h`C. `C. 115.5 h`D. `D. 31 h` |
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Answer» Correct Answer - A `(2k_(1))/(2k_(1) + 3k_(2)) = 0.5` and `k_(1) + k_(2) = (0.693)/(1.386 xx 10^(2)) =5 xx 10^(-3)h^(-1)` Solving `(k_(1))/(k_(2)) = (2)/(3)` and `k_(1) = 2 xx 10^(-3) h^(-1)` and `k_(2) = 3 xx 10^(-3) h^(-1)` `t_(1//2(A rarr B)) = (0.693)/(k_(2)) = (0.693)/(3 xx 10^(-3)) = 231 h` |
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| 1508. |
The activity of a radioactive sample initially is 3200 dps and after 8hours activity is 100 dps. What will be the activity after 4.8hr from start in dps?A. 150B. 200C. 300D. 400 |
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Answer» Correct Answer - D |
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| 1509. |
Which type of radioactive decay produces a daughter nucleus with a higher atomic number?A. `alpha`B. `beta^(-)`C. `gamma`D. `beta^(+)` |
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Answer» Correct Answer - B |
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| 1510. |
The rate of radioactive decay of a sample are `3 xx 10^(8) dps` and `3 xx 10^(7) dps` after time `20 min` and `43.03 min` respectively. The fraction of radio atom decaying per second is equal toA. `A. (1)/(600)`B. `B. 1`C. `C. 0.5`D. `D. 0.001` |
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Answer» Correct Answer - A `k=(2.303)/(t_(2)-t_(1))log. (r_(1))/(r_(2))=(2.303)/(23.03)log.(3xx10^(8))/(3xx10^(7))= 0.1min^(-1)` `k = -(dN//N)/(dt) = 0.1 min^(-1) = (1)/(600)s^(-1)` |
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| 1511. |
The rate constant for the radioactive decay of `C-11 is 0.0341 min^(-1)`. How long will it take for a sample of C-11 to decrease to `(1)/(4)` of its original activity?A. 20.3minB. 29.3minC. 40.6minD. 58.6min |
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Answer» Correct Answer - C |
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| 1512. |
What mode of radioactive decay is most likely for the isotope `._(11)^(20)Na?`A. AlphaB. BetaC. GammaD. Electron capture |
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Answer» Correct Answer - D |
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| 1513. |
Which nucleous is not radioactive?A. K-38B. K-39C. K-42D. K-43 |
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Answer» Correct Answer - B |
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| 1514. |
The elements Lawrenciym was first synthesized by the reaction `._(98)^(252)Cf+._(103)^(258)Lr+_-` What products ar needed to balance this equation?A. `(4)/(2)alpha+(1)/(0)n`B. `5_(-1e)^(@)`C. `5_(1e)^(@)`D. `5_((1)/(0))^(@)n` |
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Answer» Correct Answer - D |
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| 1515. |
In the decay `._(32)^(68)Ge+._(-1)^(0)e overset("electron capture")to ._(Z)^(A)?+X-rays`A. `A=67,Z=33`B. `A=68,Z=33`C. `A=67,Z=31`D. `A=68,Z=31` |
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Answer» Correct Answer - D |
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| 1516. |
For a reaction `R to P`, the concentration of a reactant changes from 0.05 M to 0.04 M min 30 minutes. What will be average rate of reaction in minutes?A. `4 xx 10^(-4) M "min"^(-1)`B. `8 xx 10^(-4) "min"^(-1)`C. `3.3 xx 10^(-4) M "mim"^(-1)`D. `2.2 xx 10^(-4) M "mim"^(-1)` |
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Answer» Correct Answer - C Average rate `= (Delta [R])/(Delta t) = - ([R]_(2) - [R]_(1))/(t_(2) - t_(1))` `= - ((0.04 - 0.05))/(30) = (0.01)/(30)` `3.3 xx 10^(-4) M "min"^(-1)` |
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| 1517. |
The rate constant of a zero order reaction is` 0.2 mol L^(-3) h^(-1)`. If the concentration of the reactant after 30 min is 0.05 mol ` dm^(-3)`, then its initial concentration would beA. `0.01 "Mol " dm^(-3)`B. `0.15 "Mol " dm^(-3)`C. `0.25 "Mol"dm^(-3)`D. `4.00 "Mol"dm^(-3)` |
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Answer» Correct Answer - B In 30 minutes 0.1 mol `dm^(-3)` react .Remaining concentration is 0.05 mol `dm^(-3)` therefore , initial is `0.10 +0.05 = 0.15 mol dm^(-3)` |
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| 1518. |
The rate constant of a zero order reaction is` 0.2 mol L^(-3) h^(-1)`. If the concentration of the reactant after 30 min is 0.05 mol ` dm^(-3)`, then its initial concentration would beA. `0.01 "mol dm"^(-3)`B. `0.15 "mol dm"^(-3)`C. `0.25 "mol dm"^(-3)`D. `4.00 "mol dm"^(-3)` |
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Answer» Correct Answer - B For zero order reaction, `x=kt-0.2"mol dm"^(-3)h^(-1)xx(30)/(60)h=0.1"mol dm"^(-3)` Now, concentration `=0.05"mol dm"^(-3)` Hence initial concentration`=0.1+0.05=0.15"mol dm"^(-3)` |
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| 1519. |
for the first order reaction the half life period is ( if K is rate constant and `alpha ` in initial concentration )A. `(l n 2)/(k)`B. `(1) /(k alpha )`C. ` (l n k) /( 2 ) `D. `(log k)/(2)` |
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Answer» Correct Answer - A `(l n2)/(k) =(log_(e)2)/(k)=(2.303xxlog2)/(k)=(0.693)/(k)` |
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| 1520. |
The thermal decomposition of a compound is of first order. If `50%` of a sample of the compound is decomposed in 120 minutes, how long it take for `90%` of the compounds to decompose. |
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Answer» Step I: Calculation of rate constant: (k) `k=(0.693)/t_(1//2)=0.693/(120 "min")` Step II: Calculation of time for 90 percent decomposition `a=100%, (a-x)= 10%, k=0.693//120 min^(-1)` `t=(2.303)/(k) log a/(a-x) = (2.303)/(0.693/120 min^(-1)) log 100/10` `=(2.303 xx 120)/(0.693) = 389.79 min`. |
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| 1521. |
For a first order reaction, show that the time required for `99%` completion is twice the time required for the completion of `90%` of reaction. |
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Answer» For a first order reaction , the time required for `99%` completion is `t_(1) = (2.303)/(k) "log" (100)/(100 - 99)` = `(2.303)/(k) "log" 100` = `2 xx (2.303)/(k)` For a first order reaction , the time required for `90%` completion is `t_(2) = (2.303)/(k) "log" (100)/(100 - 90)` = `(2.303)/(k) "log" 10` = `(2.303)/(k)` Therefore , `t_(1) = 2t_(2)` Hence , the time required for `99%` completion of a first order reaction is twice the time required for the completion of `90%` of the reaction . |
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| 1522. |
Thermal decomposition of a compound is of first order. If `50%` of a sample of the compound is decomposed in `120` minutes how long will it take for `90%` of the compound to decompose?A. `399` minB. `410` minC. `250` minD. `120` min |
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Answer» Correct Answer - A First calculate the values of `K` `K = (0.693)/(T_(1//2)) = (0.693)/(120) = 5.77 xx 10^(3) min-1` Now we know that for a first order reaction `K = (2.303)/(t) "log"((a)/(a-x))` Here the initial concentratiion `a = 100` and `n = 90` `:. 5.77 xx 10^(-3) = (2.303)/(t) "log"(100)/(100-90)` `t = (2.303)/(5.77 xx 10^(-3))"log"((a)/(a-x))` Solving, `t = 399` minute |
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| 1523. |
the concentration of a reactant in a solution falls (i) from 0.2 to 0.1 M in 2 hrs ,(ii ) from 0.2 to 0.05 m in 4 hrs , the order of the hydrolysis of the reactant isA. zeroB. twoC. oneD. half |
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Answer» Correct Answer - C Half life in remaining same . |
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| 1524. |
Thermal decomposition of compound `x` is a first order reaction. If `75%` of `x` is decomposed in `100 min`. How long will it take for `90%` of the compound to decompose ? Given `log 2=0.30`A. `190 min`B. `176.66 min`C. `166.66 min`D. `156.66 min` |
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Answer» Correct Answer - C `K = (1)/(t)ln[(1)/(1-3//4)] = (1)/(100)ln(4)` `K = (ln4)/(100)` and `t._(9) = (1)/(k)ln[10] = (100)/(ln4)ln 10` `= (100)/(.6) = 166.66 min`. |
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| 1525. |
A first order reaction takes 100 min for completion of 60 % of reaction ,The time eequired for completion of 90% of the reaction isA. `150 min `B. `200`minC. `220.9`minD. `246.6` min |
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Answer» Correct Answer - D For the order reaction ,`k=(2.303)/(t)log""(C_(0))/(C_(t))` Case I: `C_(0)=100 M,C_(t) =100 -60 =40, t= 100 min` `k=(2.303)/(100)xxlog""(100)/(40)` `k=(2.303xx0.3979)/(100)` case-II : `C_(0)=100 M,C_(t)=100-90=10M ` `t=(2.303)/(k) log""(100)/(10)` `=(2.303)/(k)` .... substituting K from I `=(2.303xx100)/(2.303xx0.3979)=245.6 min` |
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| 1526. |
If `75%` of a first order reaction completed in `15 min`. Then `90%` of the same reaction completed inA. `20 min`B. `25 min`C. `30 min`D. `150 min` |
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Answer» Correct Answer - B `t_(1//2) = (15)/(2)min t_(90%) = (10)/(3) xx t_(1//2) = (10)/(3) xx (15)/(2) = 25 min`. |
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| 1527. |
`80%` of a fisrt order reaction was completed in `70 min`. How much it will take for `90%` completion of a reaction?A. `114 min`B. `140 min`C. `100 min`D. `200 min` |
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Answer» Correct Answer - C `t_(80%) = log.(100)/(100-80) = log.(10)/(2)` `= log 5 = 0.69 ~~ 0.7` `t_(90%) = log.(100)/(100-90) = log 10 = 1` `(t_(80%))/(t_(90%)) = (0.7)/(1) rArr (70)/(t_(90%)) = (0.7)/(1)` `:. t_(90%) = (70)/(0.7) = (700)/(7) = 100 min` |
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| 1528. |
`90%` of a first order reaction was completed in `100 min`. How much time it will take for `80%` completion of a reactionA. `90 min`B. `80 min`C. `70 min`D. `60 min` |
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Answer» Correct Answer - C `t_(90%) = log.(100)/(100-90) = log 10 = 1` `t_(80%) = log.(100)/(100-80) = log 5 = 0.7` `(t_(80%))/(t_(90%)) = (0.7)/(1)` `:. t_(80%) = 100 xx 0.7 = 70 min` |
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| 1529. |
The half-life period of a first order process is 1.6 min. It will be 90% completed inA. 0.8 minB. 3.2 minC. 5.3 minD. 1.6 min |
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Answer» Correct Answer - C For first order reaction, `t_(1//2)=(0.693)/(k)` `implies " "k=(0.693)/(1.6)=0.4332" min"^(-1)` Now, for 90% completion, `k=(2.303)/(t)log.(a)/(a-x)` `implies " "0.4332=(2.303)/(t)log.(100)/(10)implies t=5.3min` |
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| 1530. |
Consider the reaction: `N_(2(g))+3H_(2(g))rarr 2NH_(3(g))`. The equally relationship between `-(d[NH_(3)])/(dt)` and `-(d[H_(2)])/(dt)` is:A. `(d[NH_(3)])/(dt)=-(1)/(3)(d[H_(2)])/(dt)`B. `+(d[NH_(3)])/(dt)=-(2)/(3)(d[H_(2)])/(dt)`C. `+(d[NH_(3)])/(dt)=-(3)/(2)(d[H_(2)])/(dt)`D. `(d[NH_(3)])/(dt)=-(d[H_(2)])/(dt)` |
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Answer» Correct Answer - B for the reactions , `N_(2)(g) +3H_(2)(g) to 2NH_(2)(g)` the rate of reactionn wrt `N_(2)=-(d[N_(2)])/(dt)`[ Rate of disappearance ] the rate of reaction with respect to `H_(2)=-(1)/(3)(d[H_(2)])/(dt)`[rate of disappearance ] the rate of reaction with respect to `NH_(3)=+(1)/(2)(d[NH_(3)])/(dt)`[rate of appearance ] hence at a fixed time `-(d[N_(2)])/(dt)=-(1)/(3)(d[H_(2)])/(dt)=+(1)/(2)(d[NH_(3)])/(dt)` `or +(d[NH_(3)])/(dt)=-(2)/(3)(d[H_(2)])/(dt)` `or +(d[NH_(3)])/(dt)=-(2d[N_(2)])/(dt)` |
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| 1531. |
A first order reaction is `40%` completed in 50 minutes. What is the time required for `90%` of the reaction to complete? |
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Answer» For the first order reaction, `t=(2.303)/t log a/(a-x)` Ist case: `a=100%, x=40%, (a-x)=100-40=60%, t=50` min `t_(40%)=(2.303)/k log(100%)/(60%)` `50"min" = 2.303/k log (100)/(60)=2.303/k log 1.667 = (2.303 xx 0.2219)/k`…………(i) IInd case: `a=100%, x=90%, (a-x)=100-90=10%` `t_(90%) = 2.03/k log (100%)/(10%)=2.303/k log 10=2.303/k`.........(ii) Divide eqn. (ii) by eqn. (i), `t_(90%)/(50"min") = 2.303/(2.303 xx 0.2219) = 4.506 t_(90%)=4.506 xx (50"min") = 2.25 xx 10^(2)`min. |
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| 1532. |
Consider the reaction: `N_(2(g))+3H_(2(g))rarr 2NH_(3(g))`. The equally relationship between `-(d[NH_(3)])/(dt)` and `-(d[H_(2)])/(dt)` is:A. `(d[NH_(3)])/(dt) = (-d[H_(2)])/(dt)`B. `(d[NH_(3)])/(dt) = -1/3(d[H_(2)])/(dt)`C. `(d[NH_(3)])/(dt) = -2/3 (d[H_(2)])/(dt)`D. `(d[NH_(3)])/(dt) = -3/2 (d[H_(2)])/(dt)` |
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Answer» Correct Answer - C c) is the correct relation. |
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| 1533. |
Consider the reaction: `N_(2(g))+3H_(2(g))rarr 2NH_(3(g))`. The equally relationship between `-(d[NH_(3)])/(dt)` and `-(d[H_(2)])/(dt)` is: |
| Answer» `-1/3(d[H_(2)])/(dt) = +1/2(d[NH_(3)])/(dt)` | |
| 1534. |
In a catalyst experiment involving the Haber process `N_(2) + 3H_(2) rarr 2NH_(3)`, the rate of reaction was measured as Rate `= (Delta[NH_(3)])/(Delta t) = 2.0 xx 10^(-4) mol L^(-1) s^(-1)` What is the rate of reaction expressed in terms of (a) `N_(2)` (b) `H_(2)` ?A. `2.50 xx 10^(-4) mol L^(-1) s^(-1)`B. `1.25 xx 10^(-4) mol L^(-1) s^(-1)`C. `3.75 xx 10^(-4) mol L^(-1) s^(-1)`D. `5.00 xx 10^(-4) mol L^(-1) s^(-1)` |
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Answer» Correct Answer - C `N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g)` Rate `= (-d[N_(2)])/(dt) = (-d[H_(2)])/(3 dt) = (+d[NH_(3)])/(2dt)` `:. (-d[H_(2)])/(dt) = (3)/(2)(d[NH_(3)])/(dt)` `= (3)/(2) xx 2.5 xx 10^(-4) mol L^(-1) s^(-1)` `= 3.75 xx 10^(-4) mol L^(-1) s^(-1)`Correct Answer - C `N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g)` Rate `= (-d[N_(2)])/(dt) = (-d[H_(2)])/(3 dt) = (+d[NH_(3)])/(2dt)` `:. (-d[H_(2)])/(dt) = (3)/(2)(d[NH_(3)])/(dt)` `= (3)/(2) xx 2.5 xx 10^(-4) mol L^(-1) s^(-1)` `= 3.75 xx 10^(-4) mol L^(-1) s^(-1)` |
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| 1535. |
A reaction of first-order completed `90%` in 90 minutes, hence, it is completed `50%` in approximately :A. 50 minB. 54 minC. 27 minD. 62 min |
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Answer» Correct Answer - C |
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| 1536. |
For the reaction, `N_(2) + 3H_(2) rarr 2NH_(3)`, if `(d[NH_(3)])/(d t) = 2 xx 10^(-4) "mol L"^(-1) s^(-1)`, the value of `(-d[H_(2)])/(d t)` would be:A. `1 xx 10^(-4) "mol" L^(-1) s^(-1)`B. `3 xx 10^(-4) "mol" L^(-1) s^(-1)`C. `4 xx 10^(-4) "mol" L^(-1) s^(-1)`D. `6 xx 10^(-4) "mol" L^(-1) s^(-1)` |
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Answer» Correct Answer - B `(1)/(2)(d[NH_(3)])/(d t) = (1)/(3)(d[H_(2)])/(d t)` `:. -(d[H_(2)])/(d t) = (3)/(2) xx (d[NH_(3)])/(d t)` `= (3)/(2) xx 2 xx 10^(-4) = 3 xx 10^(-4)` |
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| 1537. |
For the reaction `N_(2)+3H_(2) rarr 2NH_(3)`, if `(d[NH_(3)])/(dt).=4xx10^(-4)` mol `L^(-1)s^(-1)`, the value of `(-d[H_(2)])/(dt)` would beA. `3xx10^(-4) mol L^(-1)S^(-1)`B. `4xx10^(-4) mol L^(-1)S^(-1)`C. `6xx10^(-4) mol L^(-1)S^(-1)`D. `1xx10^(-4) mol L^(-1)S^(-1)` |
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Answer» Correct Answer - A For the reaction , `N_(2)+3H_(2) to NH_(3)` `rate =-(d[N_(2)])/(dt)=-(1)/(3)(d[H_(2)])/(dt)=+(1)/(2) (d[NH_(3)])/(dt)` `or -(1)/(3)(d[H_(2)])/(dt)=+(d[NH_(3)])/(dty)` `-(d[H_(2)])/(dt)=(3)/(2)xx2xx10^(-04)molL^(-1)S^(-1)` `=3xx10^(-4)molL^(-1)S^(-1)` |
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| 1538. |
For the reaction `N_(2)(g) +3H_(2)(g) to 2NH_(3)(g)` If `Delta[NH_(3)]//Deltat= 4 xx 10^(-8)mol L^(-1)s^(-1)`, what is the value of `Delta[H_(2)]//Deltat`=? |
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Answer» Rate of reaction = `(-d[N_(2)])/(dt) = -1/3 (d[H_(2)])/(dt) = 1/2(d[NH_(3)])/(dt)` `(d[H_(2)])/(dt) = -(3d)/2[NH_(3)]/(dt)` `=-3/2(d[NH_(3)])/(dt) = -3/2 xx (4 xx 10^(-4) mol L^(-1)s^(-1))` `-6 xx 10^(-4)mol L^(-1)s^(-)`. |
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| 1539. |
For the reaction `N_(2) + 3H_(2) to 2NH_(3)` if `(Delta[NH_(3)])/(Deltat) = 2 xx 10^(-4) mol L^(-1)s^(-1)`, the value of `(-Delta[H_(2)])/(Deltat)` would beA. `1 xx 10^(-3) mol L^(-1)s^(-1)`B. `3 xx 10^(-4)molL^(-1)s^(-1)`C. `4 xx 10^(-4) mol L^(-1)s^(-1)`D. `6 xx 10^(-4)molL^(-1)s^(-1)` |
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Answer» Correct Answer - B b) `(-Delta[H_(2)])/(Deltat) = 3//2 xx 2 xx 10^(-4) mol L^(-1)s^(-1)` `=3 xx 10^(4) mol L^(-1)s^(-1)` |
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| 1540. |
If 60 % of a first order reaction was completed in 60 minutes, 50 % of the same reaction would be completed in approximately [log = 4 = 0.60, log 5 = 0.69].A. 45 minutesB. 60 minutesC. 40 minutesD. 50 minutes. |
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Answer» Correct Answer - A a) For first order reaction `k=(2.303)/(60 "min") log 100/40` `=(2.303)/(60"min") log 2.5` `=(2.303 xx 2.5)/(60 "min") = 0.0153` `t-(1//2) = 0.693/k = 0.693/(0.0153)=45.3` min. |
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| 1541. |
For the reaction, `N_(2) + 3H_(2) rarr 2NH_(3)`, if `(d[NH_(3)])/(d t) = 2 xx 10^(-4) "mol L"^(-1) s^(-1)`, the value of `(-d[H_(2)])/(d t)` would be:A. `4 xx 10^(-4) mol L^(-1)s^(-1)`B. `6 xx 10^(-4) mol L^(-1)s^(-1)`C. `1 xx 10^(-4) mol L^(-1)s^(-1)`D. `3 xx 10^(-4) molL^(-1)s^(-1)` |
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Answer» Correct Answer - D d) Rate =`-1/3 (d[H_(2)])/(dt) = (-d[N_(2)])/(dt)` `=-1/2(d[NH_(3)])/(dt)` Given `(d[NH_(3)])/(dt) = 2 xx 10^(-4) mol L^(-1)s^(-1)` `(-d[H_(2)])/(dt) = 3/2 (d[NH_(3)])/(dt) = 3/2 xx (2 xx 10^(-4))` `3 xx 10^(-4) mol L^(-1)s^(-1)` |
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| 1542. |
The activation energy of a reaction can be determined from the slope of which of the following graphs?A. In `k " vs " 1/T`B. `T/In k " vs " 1/T`C. In `k " vs " T`D. In `(k)/T " vs " T` |
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Answer» Correct Answer - A a) It is the correct answer. |
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| 1543. |
What is the activation energy for a reaction if the rate is doubled when the temperature is raised from `20^(@)` C to `35^(@)`C (R=`8.314 J mol^(-1)K^(-1))`A. 15.1 kJ `mol^(-1)`B. `342 kJ mol^(-1)`C. `269 kJmol^(-1)`D. `34.7 kjmol^(-1)` |
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Answer» Correct Answer - D d) `k_(2)/k_(1) = 2, T_(1)=20+273=293 K`, `T_(2)=35 + 273=308`K `k=8.314 JK^(-1)mol^(-1)` `log k_(2)/k_(1) =E_(a)/(2.303 ) [1/T_(1)-1/T_(2)]` `log2= (E_(a))/(2.303 xx 8.314(JK^(-1)mol^(-1))[1/(293K)-1/(308K)]` `0.3010= (E_(a) xx 15)/(2.303 xx 8.314 (Jmol^(-1)) xx 293 xx 308)` `E_(a) = (0.3010 xx 2.303 xx 8.314 x 293 xx 308)/(15) (Jmol^(-1))` `=34.673 J mol^(-1) = 34.673 kJ mol^(-1)` |
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| 1544. |
What is the activation energy for a reaction if its rate doubles when the temperature is raised from `20^(@)C` to `35^(@)C`? `(R = 8.314 J "mol K"^(-))`A. 342`kj mol^(-1)`B. `269kjmol^(-1)`C. `34.7kj Mol^(-1)`D. `15.1 kj Mol^(-1)` |
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Answer» Correct Answer - C given initial temperature `T_(1)=20+273=293K` final temperature `T_(2) = 35+273=308K` ` R= 8314mol^(-1) K^(-1)` since ,rate becomes double on raising temperature , `therefore I_(2) = 2r_(1) or (I_(2))/(I_(1))=2` As rate constant `k prop I` `therefore (K_(2))/(K_(1))=2` from arrherius equation , we know that `log (K_(2))/(K_(1))=-(E_(a))/(2.303R) [(T_(1)-T_(2))/(T_(1)T_(2))]` `log 2 =-(E_(a))/(2.303xx8.314)[(293-308)/(293xx308)]` ` 0.3010=-(E_(a))/(2.303xx8314)[(-15)/(293xx308)]` ` therefore E_(a) =(0.3010xx2.303xx8.314xx293xx308)/(15)` ` 34673.48 mol^(-1)=34.7 kj mol^(-1)` |
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| 1545. |
For a certain gaseous reactions a `10^(@)C` rise of temp, from `25^(@)C` to `35^(@)C` doubles the rate of reaction. What is the value of activation energy ?A. `10/(2.303Rxx298xx308)`B. `(2.303xx10)/(298xx308R)`C. `(0.693xx10)/(290xx308)`D. `(0.693Rxx298xx308)/10` |
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Answer» Correct Answer - D |
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| 1546. |
The rate constants `k_(1)` and `k_(2)` for two different reactions are `10^(16) * e^(-2000//T)` and `10^(15) * e^(-1000//T)` , respectively . The temperature at which `k_(1) = k_(2)` isA. 2000 KB. `(1000)/(2.303) K`C. 1000 KD. `(2000)/(2.303)K` |
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Answer» Correct Answer - b `10^(16) e^((-2000)/(T)) = 10^(15) e^((-1000)/(T))` `10 = e^((-1000)/(T))/(e^(-2000)/(T)) implies 10 e^((1)/(T) xx 1000) implies "ln" 10 = (1000)/(T)` `implies 2.303` log 10 = `(1000)/(T) implies T = (1000)/(2.303) K` . |
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| 1547. |
Two different first order reactions have rate consants `k_(1) and k_(2) "at" T_(1)(k_(1)gtk_(2))` . If temperature is increased form `T_(1) "to"T_(2)` ,then new consant become `k_(3) and k_(4)` respeectively. Which among the following relartions is correctA. `k_(1)gtk_(2)=k_(3)=k_(4)`B. `k_(1)ltk_(3) and k_(2)ltk_(4)`C. `k_(1)=k_(3)=k_(4)`D. `k_(1)gtk_(2)gtk_(3)gtk_(4)` |
| Answer» Correct Answer - B | |
| 1548. |
The rate constant of a reaction depends onA. temperature or the reactionB. extent of the reactionC. initial concentration of the reactantsD. the time of completion of reaction. |
| Answer» Correct Answer - A | |
| 1549. |
For this reaction `N_(2(g))+3H_(2(g))to2NH_(3(g))` The relation between `(d(NH_(3)))/(dt)` and `-(d(H_(2)))/(dt)` is :-A. `-(1)/(3) (d[H_(2)])/(dt) = + (1)/(2) (d[NH_(3)])/(dt)`B. `-(1)/(2) (d[H_(2)])/(dt) = + (1)/(3) (d[NH_(3)])/(dt)`C. `+(1)/(2) (d[H_(2)])/(dt) = - (1)/(3) (d[NH_(3)])/(dt)`D. `+(1)/(3) (d[H_(2)])/(dt) = - (1)/(2) (d[NH_(3)])/(dt)` |
| Answer» Correct Answer - A | |
| 1550. |
For a reaction, `2N_(2)O_(5)(g) to 4NO_(2)(g) + O_(2)(g)` rate of reaction is:A. Rate `= - (d[N_(2)O_(5)])/(dt) = - (1)/(4) (d[NO_(2)])/(dt) = (1)/(2)(d[O_(2)])/(dt)`B. Rate `= - (1)/(2) (d[n_(2)O_(5)])/(dt) = (1)/(4)(d[NO_(2)])/(dt) = (d[O_(2)])/(2)`C. Rate `= - (1)/(4) (d[N_(2)O_(5)])/(dt) = (1)/(2) (d[NO_(2)])/(dt) = (d[O_(2)])/(dt)`D. Rate `= - (1)/(2) (d[N_(2)O_(5)])/(dt) = (1)/(2) (d[NO_(2)])/(dt) = (1)/(2) (d[O_(2)])/(dt)` |
| Answer» Correct Answer - B | |