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| 24351. |
The reaction A to B follows first order kinetics. The time taken for 0.8 mole of A to produce 0.6 mole of B is 1 hour. What is the time taken for conversion of 0.9 mole of A to produce 0.675 mole of B ? |
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Answer» 1 hour |
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| 24352. |
The reaction, A + 2B to C +D obeys rate equation, Rate =k[A]^(x)[B]^(y) What would be the order of the reaction? |
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Answer» |
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| 24353. |
The reaction : 3O_(2)hArr2O3, DeltaH= + 69 , 000 calories is aided by : |
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Answer» high TEMPERATURE and low pressure |
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| 24354. |
The reaction A overset(k)(to) Product, is zero order while the reaction Boverset(k)(to) Product, is first order reaction. For what initial concentration of A are the half lives of the two reactions equal ? |
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Answer» `(log_(e)4)M` `:.(a)/(2)xxkxxt_(1//2)`, i.e. `t_(1//2)=(a)/(2K)`……..`(i)` For first order reaction, `t_(1//2)=(log_(e )2)/(K)` ……….`(ii)` From `(i)` and `(ii)` `(a)/(2k)=(log_(e)2)/(k)` `a=log_(e )4M` |
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| 24355. |
The reaction : A rarr Bfollows first order kinetics . The time taken for 0.8 mol of A to produce 0.6 mol of B is 1 hour . What is the time taken for conversion of 0.9 mol of A to produce 0.675 molof B ? |
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Answer» 1 HOUR same `i.e (0.6)/(0.8) = (3)/(4) ` and `(0.675)/(0.90) = (3)/(4)` `:.` Time taken is same . |
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| 24356. |
The reaction, 3ClO^(-) (aq) rarr ClO_(3)^(-) (aq)+2Cl^(-) (aq), is an example of |
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Answer» OXIDATION reaction |
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| 24357. |
The reaction, A + B to P can never be |
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Answer» SECOND ORDER reaction |
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| 24358. |
The reaction A+BhArrC+D+ heat has reached equlibrium. The reaction may be made to proceed forward by |
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Answer» ADDING more C |
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| 24359. |
The reaction A+BtoC has zero order. What is the rate equation ? |
| Answer» Solution :Rate `=((DX)/(DT))=k[A]^(0)[B]^(0)=k` (rate CONST.) | |
| 24360. |
The reaction, 3Br_(2)+6CO_(3)^(2-)+3H_(2)Oto5Br^(-)+BrO_(3)^(-)+6HCO_(3)^(-) |
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Answer» decolourization of reddish brown colour of `Br_(2)` Addition of `H_(2)SO_(4)` shift reaction in reverse direction. So, bromine is GETTING oxidised as well as reduced. |
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| 24361. |
The reaction3CIO^(-) ""_((aq))to CIO_((aq))^(-) +2CI^(-) ""_((aq))isanexampleof |
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Answer» OXIDATION REACTION |
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| 24362. |
The reaction, 2SO_(2(g))+O_(2(g))leftrightarrowSO_(3(g)) is carried out in a 1 dm^(3) vessel and 2 dm^(3) vessel separately. The ratio of the reaction velocities will be |
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Answer» `1:8` |
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| 24363. |
The reaction |
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Answer» WURTZ REACTION |
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| 24364. |
The reaction, 2SO_(2(g))+O_(2(g))hArr2SO_(3(g)) is carried out in a 1 dm^(3) vessel and 2dm^(3) vessel separtely. The ratio of the reaction velocities will be |
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Answer» `1:8` For `1dm^(3)R=k[SO_(2)]^(2)[O_(2)]` `R=K[(1)/(1)]^(2)[(1)/(1)]=1` For `2dm^(3)R=K[(1)/(2)]^(2)[(1)/(2)]=1/8` So, the RATIO is `8:1` |
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| 24365. |
The reaction, 2SO_2 + O_2 + 2H_2O rarr 2H_2SO_4 is an example of: |
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Answer» SYNTHESIS of `H_2SO_4` |
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| 24366. |
The reaction 2ROH+2 Na to RONa+H_(2) suggests that alcohols are |
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Answer» Acidic |
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| 24367. |
The reaction 2O_(3)(g)rarr3O_(2)(g) is believed to occur according to the mechanism given below. Step -I : O_(3)(g)underset(k_(-1))overset(k_(1))hArrO_(2)(g)+O(g) (fast) Step- II : O_(3)(g)+O(g)overset(k_(2))rarr2O_(2)(g) (slow) Write the rate law for the reaction . Determine the order of the reaction with respect toO_(3)(g) and O_(2)(g). |
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Answer» Solution :The slow step of reaction is its rate-determiningstep, becauserate of this step determinesthe overall rate of the reaction. So, the rate of the slow step will be the rate of the reaction. Thus, rate `=k_(2)[O_(3)][O]""...[1]` The rate equation contains concentration term of reaction INTERMEDIATE O, which does not appear in the overall balanced equation of the reaction. But the rate law of a reaction shouldnot appear in theoverall balanced equation. So, we REQUIRE to eliminate [O] from equation (1). This can be done by making use of the equilibrium CONDITION of step - 1.For this equilibrium, we can write equilibrium constant EXPRESSION as `k_(1) = ([O_(2)][O])/([O_(3)])` This gives`[O]=k_(1)[O_(3)]//[O_(2)]` Substituting the expression for [O] into equation (1)gives rate `k_(2)K_(1) = ([O_(3)]^(2))/([O_(2)])=k[O_(3)]^(2)[O_(2)]^(-1)[k=k_(2)K_(1)` = constant] This equation represents the rate law for the GIVEN reaction . It appears from the equation that the reaction has an order 2 with respect to`O_(3)and -1` with respect to`O_(2)`. |
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| 24368. |
The reaction 2R-Z+2Na overset("dry ether") to R-R+2NaX is an example of |
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Answer» Williamson'ssynthesis |
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| 24369. |
The reaction , 2O_(3) to 3O_(2) , is assigned the following mechanism. (i) O_(3) hArrO_(2)+O (ii) O_(3)+O overset("slow")(to)2O_(2) The ratelaw of the reaction will be |
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Answer» `R PROP[O_(3)]^(2)[O_(2)]` |
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| 24370. |
Thereaction2NO_((g)) + C1_(2(g))to 2NOC1_((g))is firstorderin C1_(2) andsecondorderin NOWritethe ratelawfor thereaction |
| Answer» Solution :RATELAW : RATE = `k[NO]^(2) [C1_(2)]` | |
| 24371. |
The reaction 2NO_(2)(g) to 4NO_(2)(g)+O_(2)(g) follows first order kinetics. The pressure of a vessel containing only N_(2)O_(5) was found to increase from 50mm Hg to 87.5mm Hg in 30min. The pressure exerted by the gases after 60min will be (assume temperature remains constant): |
| Answer» Answer :A | |
| 24372. |
The reaction 2NO_(2)Cl to 2NO_(2)+Cl_(2) , has experimental data that Rate =k[NO_(2)Cl]. Propose the mechanism. |
| Answer» Solution :`{:(NO_(2)Cloverset("slow")(to)NO_(2)+Cl),(UL(NO_(2)Cl+Cloverset("FAST")(to)NO_(2)+Cl_(2))),(2NO_(2)Cl to2NO_(2)+Cl_(2)):}` | |
| 24373. |
The reaction 2NO + 2H_(2) to N_(2) +2H_(2)O follows the mechanism: I) NO + NO_(2) ltimplies N_(2)O_(2) (fast) II) N_(2)O_(2) + H_(2) to N_(2)O + H_(2)O (slow) III) N_(2)O + H_(2) to N_(2) + H_(2)O (fast) the rate constant for the slow step (II) is 1.2 xx 10^(-4) mol^(-1)Lmin^(-1) while the equilibrium constant for step (I) is 1.4 xx 10^(-2) what is the rate of reaction when the concentration of NO and H_(2) each is 0.5 mol^(-1)L? |
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Answer» `2.1 xx 10^(-7) mol L^(-1)min^(-1)` `=1.2 xx 10^(-4)mol^(-1)L min^(-1) xx (1.4 xx 10^(-2))` `=1.68 xx 10^(-6) mol^(-1)L min^(-1)` RATE =`k_("obs")[NO]^(2)[H_(2)]` `=(1.68 xx 10^(-6) mol^(-1)L min^(-1)) xx (0.5 molL^(-1))^(3)` |
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| 24374. |
The reaction, 2NO+br_(2)to2NOBris supposed to have the meachanism Step -I NO+Br_(2)hArrNOBr_(2) Step-II NOBr_(2)+NOoverset("slow")(to)2NOBr Predict the rate expression for the reaction |
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Answer» Solution :The RATE expression is derived by step-II of the reaction mechanism which is a slow step.thus Rate `=K[NOBr_(2)][NO]`………`(i)` HOWEVER, `NOBr_(2)` is an intermediate and thus `[NOBr_(2)]` should be eliminated from Eq. `(i)` Also for, step `(i)` Equilibrium constant `K_(c )` is `K_(c )=([NOBr_(2)])/([NO][Br_(2)])`...........`(ii)` `:.[NOBr_(2)]=K_(c )[NO][Br_(2)]` Then from Eqs. `(i)` and `(ii)` Rate `=k.K_(c )[NO]^(2)[Br]^(2)` Rate `=K.[NO]^(2)[Br_(2)]` |
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| 24375. |
The reaction 2NO+Br_2to2NOBr follows the mechanism : (1)NO+Br_2hoverset("Fast")hArrNOBr_2 (2)NOBr_2+Nooverset("Slow")to2NOBr Which of the following is/are true regarding this : |
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Answer» The order of the REACTION with respect to NO is two :The molecularity of the steps (1) and (2) are two each |
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| 24376. |
The reaction 2NO + Br_2rarr 2NOBr, obey.s the following mechanism: |
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Answer» `R=[NO]^2[Br_2]` |
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| 24377. |
The reaction : 2N_(2)O_(5)(g) to 2NO_(2)(g) + (O_(2))_(g) was studied and the following data were collected. Determine: (i) the order, (ii) the rate law and (iii) rate constant for the reaction. |
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Answer» Solution :Let the order of the REACTION = n `Rate propto [N_(2)O_(5)]^(n)` `(34 xx 10^(-5)) propto (1.13 xx 0^(-2))^(n)` `(25 xx 10^(-5)) propto (0.84 xx 10^(-2))^(n)` DIVIDING eqn. (i) by eqn. (ii), we have `(34 xx 10^(-5))/(25 xx 10^(-5)) = (1.13 xx 10^(-2))^(n)/(0.84 xx 10^(-2))^(n), (1.36)^(1) = (1.35)^(n)` or n=1 i) The order of the reaction = 1, ii) Rate law, The rate of reaction depends UPON one concentration team of `N_(2)O_(5)` because of the reaction is 1. ii) Rate law. The rate of reaction depends upon one concentration term of `N_(2)O_(5)` because order of the reaction is 1. Reaction Rate = `k[N_(2)O_(5)]` III) Rate constant, `k = "Reaction rate"/([N_(2)O_(5)])= (18 xx 10^(-5) mol L^(-1)min^(-1))/(0.62 xx 10^(-2)mol L^(-1))= 0.0.29 min^(-1)` |
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| 24378. |
The reaction: 2MnO_(4)^(-) + 3Mn^(2+)+ 2H_(2)O rarr 5MnO_(2) + 4H^(+) is catalysed by….. |
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Answer» `NiSO_(4)` |
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| 24379. |
The reaction 2H_(2)O_(2)to 2H_(2)O + O_(2) shows that |
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Answer» it ACTS as reducing agent |
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| 24380. |
The reaction 2KI+Cl_(2) to 2KCl+I_(2) |
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Answer» is possible |
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| 24381. |
The reaction 2H_(2) O_(2) to 2 H_(2)O+ O_(2) is a |
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Answer» ZERO order reaction |
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| 24382. |
The reaction, 2H_2O_2rarr 2H_2O + O_2, shows that H_2O_2: |
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Answer» Is decomposed |
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| 24383. |
The reaction 2ClCH_2 CH_2 OH overset((i) KOH) underset((ii) "Conc." H_2 SO_4) to will produce |
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Answer» `CH_(2)=CH_(2)OH`
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| 24384. |
The reaction 2CH_(3)-C-underset(O)underset(||)(C)C_(2)H_(5) overset(C_(2)H_(5)ON a)toCH_(3)-underset(O)underset(||)(C)-CH_(2)-underset(O)underset(||)(C)-OC_(2)H_(5)+C_(2)H_(5)OH is called |
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Answer» Etard reaction |
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| 24385. |
The reaction 2C_6H_5I+2Na overset(ether, Delta)(to) C_6H_5-C_6H_5+2NaI |
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Answer» FITTING REACTION |
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| 24386. |
The reaction 2C_(6)H_(5)CHOoverset(50% NaOH)rarrC_(6)H_(5)COONA+C_(6)H_(5)CH_(2)OHis an example of |
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Answer» perkine REACTION |
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| 24387. |
The reaction 2C + O_2 rarr 2CO_2 is carried out by taking 24 g carbon and 96 gm o_2 which one is limiting reagent : |
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Answer» C |
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| 24388. |
The reaction 2A_((g))+B_((g))hArr3C_((g))+D_((g)) is begun with the concentrations of A and B both at an initial value of 1.00 M. when equilibrium is reached, the concentration of D is measured and found to be 0.25 M. The value for the equilibrium constant for this reaction is given by the expression |
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Answer» `"["(0.75)^(3)(0.25)"]"DIV"["(1.00)^(2)(1.00)"]"` `{:("INITIAL",1,1,0,0),(Eq.,1-0.50,1-0.25,0.75,0.25):}` `K=((0.75)^(3)(0.25))/((0.50)^(2)(0.75))` |
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| 24389. |
The reaction, 2ArarrB+Cfollow zero order kinetics. The differential rate equation for the reaction is: |
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Answer» `(DX)/dt=K[A]^0` |
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| 24390. |
The reaction , 2AB(g) + 2C(g) to A_(2(g)) + 2BC_((g)) proceeds according to the mechanism . I. 2AB hArr A_(2)B_(2)(fast) II. A_(2)B_(2)+C to A_(2)B + BC(slow ) III. A_(2)B+C to A_(2)+BC (fast) what will be the initial rate taking [AB] = 0.2 M and [C] = 0.5 M ?The K_(c)for the step I is10^(2) M^(-1)and rate constant for the step II is3.0 xx 10^(-3) mol^(-1) min^(-1) |
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Answer» `0.0716 M MIN^(-1)` |
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| 24391. |
The reaction, 2A to B+C follows zero order kinetics. The differential rate equation for the reaction is |
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Answer» `(dx)/(dt) =k[A]^0` |
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| 24392. |
The reaction , 2 NO(g) + O_(2) (g) to 2NO_(2) (g) , is second order with respect to NO and first order with respect to O_(2) . If the volume of reactants is suddenly reduced to half value , the rate of reaction would be |
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Answer» One-fourth of ORIGINAL VALUE |
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| 24393. |
The reaction 2" NO"(g)+O_(2)(g)to2" NO"_(2)(g)" and "2" CO "(g)+O_(2)(g)to2" CO"_(2)(g) look to be similar. Yet the former is faster than the latter at the same temperature. Explain why. |
| Answer» SOLUTION :The former has LOWER ACTIVATION ENERGY than the latter REACTION. Lower the activation energy, faster is the reaction. | |
| 24394. |
The reaction ._(1)H^(2) + ._(1)H^(3) rarr ._(2)He^(4) + ._(0)n^(1)+ energy represents |
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Answer» Nuclear fission `._(1)H^(2) + ._(1)H^(3) RARR ._(2)He^(4) + ._(0)^(1) N +` energy |
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| 24395. |
The reaction 2 Fe Cl_(3) + SnCl_(2) to 2 FeCl_(2) + SnCl_(4) is an example of |
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Answer» First order REACTION |
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| 24396. |
The reaction ._(13)Al^(27) + ._(2)He^(4) rarr ._(14)Si^(30) + ._(1)H^(1) is of the type |
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Answer» NUCLEAR fusion |
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| 24397. |
The reaction |
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Answer» WURTZ reaction |
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| 24398. |
The reaction : ""_(1)^(2)D+""_(1)^(3)Tto""_(1)^(2)He+""_(0)^(1)n is an example of |
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Answer» NUCLEAR FISSION |
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| 24399. |
The reaction, (1//2)H_(2)(g) + AgCl(s) to H^(+)(aq) + Cl^(-)(aq) + Ag(s), occurs in the galvanic cell: |
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Answer» `Ag|AgCl(s) |KCl(soln)|AgNO_(3)(soln)|Ag` |
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| 24400. |
The reaction ,1//2H_(2(g)) + AgCl_((s)) to H_((aq))^(+) + Cl_((aq))^(-) + Ag_((s)) occurs in the galvanic cell : |
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Answer» `Ag | AgCl_((s)) |KCl_((So LN))|AgNO_(3(so ln))|Ag` |
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