InterviewSolution
Saved Bookmarks
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.
| 1251. |
The rate of the backward reaction in a reversible reactionA. positiveB. negativeC. Either (a) or (b)D. None of the above |
|
Answer» Correct Answer - A The rate of the bakwrad reactoin in a reversible reaction is positive. Rate of reaction can never be negative. |
|
| 1252. |
A first order reaction is found to have a rate constant, `k = 4.2 xx 10^(-12) s^(-1)`. Find the half-life of the reaction.A. `1.26xx10^(13)s`B. `1.65xx10^(11)s`C. `1.65xx10^(11)s`D. `1.26xx10^(13)s` |
|
Answer» Correct Answer - B For first order reaction, `t_(1//2)=(0.693)/(k)=(0.693)/(4.2xx10^(-12)S^(-1))` `=1.65xx10^(11)s` |
|
| 1253. |
Consider the following, `C_(2)H_(2)(g)+H_(2)(g)toC_(2)H_(6)(g)`. (I) The above reaction is an example of first order kinetics. (II) Rate of the reaction will be given as `Rate =k[C_(2)H_(4)]`. Which of the above stetement(s) `is//are` correct? Choose the correct option.A. Only IB. Only IIC. Both (a) and (b)D. None of the above |
|
Answer» Correct Answer - C I. Given reaction is an example of first order kinetics `C_(2)H_(4)(g)+H_(2)+toC_(2)H_(6)(g)` II. `R=k[C_(2)H_(4)]` Hence, both the statements are true. |
|
| 1254. |
When one reactant is present in excess in a chemical reaction between two substances, then the reaction is known asA. first order reactionB. second order reactionC. zero order reactionD. pseudo first order reaction |
|
Answer» Correct Answer - D Whenone reactant is present in excess in a chemical reaction between two substances , then the reaction is known as pseudo first order reaction. |
|
| 1255. |
The rate law for the reaction `x A + y B to mP + nQ ` is Rate = `k[A]^(c) [D]^(d)` . What is the total order of the reactionA. (x + y)B. (m + n)C. ( c + d)D. x/y |
|
Answer» Correct Answer - c Order is the sum of the powers to which the concentration terms are raised in the rate equation. |
|
| 1256. |
In a gaseous reaction, an increase in the pressure of the reactant gases increases the rate of reaction. |
| Answer» Correct Answer - T | |
| 1257. |
The rate constant for two parallel reactions were found to be `1.0xx10^(-2)dm^(3)mol^(-1)and 3.0xx10^(-2)dm^(3)mol^(-1)s^(-1)`. If the corresponding energies of activation of the correspoing energies of activation of the parallel reaction are 60.0 Kj `mol^(-1)` and 70.0 KJ `mol^(-1)` respectively ,what is the apparent overall energy of activation ?A. `130.0KJmol^(-1)`B. `67.5KJmol^(-1)`C. `100.0KJmol^(-1)`D. `65.0KJmol^(-1)` |
|
Answer» Correct Answer - B |
|
| 1258. |
The rate of law for the reaction `xA+yB=mP+nQ` is Rate `k[A]^(c)[B]^(d)`. What is the total order of reaction ?A. (x+y)B. (m+n)C. (c+d)D. `(x)/(y)` |
|
Answer» Correct Answer - C Order is the sum of the powers to which the concentration terms are raised in the rate equation. Hence, order is (c+d). |
|
| 1259. |
Foe the rate law, =`K[A]^(3//2)[B]^(-1)` the overall order of a reaction isA. zeroB. halfC. oneD. two |
|
Answer» Correct Answer - B Rate =`k[A]^(x)[B]^(y)` order =x+y `therefore "order" ((3)/(2))+(-1)=(1//2)`,i.e half-order. |
|
| 1260. |
Consider a reversible reaction, which Statement about this reaction is correct? (P) The reaction will never reach equilibrium (Q) `IF ((k_(1))/(k_(2)))=2"then at"t=oo,[A]=(2a)/(3)` if initially only A was present and at `t=0,[A]=a.`(R ) At any instant ,If [A] =`(a)/(5)"then" [B] =(2a)/(5), if(k_(1)/(k_(2)=2.`A. Only PB. P and QC. P,Q,RD. Q and R |
|
Answer» Correct Answer - A |
|
| 1261. |
The order of reaction can never be zero. |
| Answer» Correct Answer - F | |
| 1262. |
For a zero order reaction a graph of conc. (along Y axis) and time (along X-axis) is linear withA. a zero intercept and a+ ve slopeB. a zero intercept and a - ve slopeC. a non-zero intercept and a- ve slopeD. a non-zero intercept and a + ve slope |
|
Answer» Correct Answer - C For a zero order reaction the graph between concentration and time is a graph with negative slope and non-zero Y-intercept. |
|
| 1263. |
For a chemical reaction ……. Can never be a fractionA. half-lifeB. milecularityC. OrderD. rate constant |
|
Answer» Correct Answer - B For a chemical reaction, molecularity can never be a fraction. |
|
| 1264. |
The formation of `SO_(3) ` from ` SO_(2) and O_(2) ` takes place in the following steps : `(i) SO_(2)+2NO_(2) to 2SO_(3)+2NO` `(ii) 2NO +O_(2) to 2NO_(2)`A. `NO_(2) ` is intermediateB. `NO ` is catalystC. `NO_(2) ` is catayst and `NO ` is intermediateD. `NO` is catalyst and `NO_(2) ` is intermediate |
|
Answer» Correct Answer - C The reactants first combine with the catalyst to form an intermediate complex which is short lived and decomposes to form the product and regenerating the catalyst hence NO is intermediate complex and `NO_(2)` is the catalyst . |
|
| 1265. |
Define rate constant, Write units of rate constant for fist and second order reactions. |
| Answer» For the defination of rate constant as well as the units: | |
| 1266. |
The reaction between NO and `Cl_(2)` takes place in the following two steps : I `{:(" "k_(1)), (NO + Cl_(2) hArr NOCl_(2)), (" "k_(2)), (" ""fast"):}` II. `NOCl_(2)underset("fast")overset(k_(2))rarr2NOCl` The rate law of overall reaction, `2NO+Cl_(2)to2NOCl` can be given by:A. rate= `k[NO]^(2)[Cl_(2)]`B. rate= `k[NO][Cl_(2)]`C. rate= k[NOCl][NO]D. rate = `k[NO][Cl_(2)]^(2)` |
| Answer» Correct Answer - A | |
| 1267. |
Which of the following illustrates the influence of surface area of a solid reactant on reaction rate ?A. Powered sugar dissolves faster than crystallune sugarB. Finely powdered potassium permanganate reacts violently with glycerol than crystalline potassium permanganateC. Explosion occurs when an inflammable substance is finely dispersed in a roomD. All of these |
|
Answer» Correct Answer - D All these statements are correct. In heterogeneous reactions the reactants are present in more than one phase. The rate of such reactions depends upon the surface area of contact between the reactants. The principle is used in the manufacture of exp,osives and rocket fules. A reaction will proceed faster if the reactants are mixed in the liqiud phase (or solution phase) than when carried out under heterogeneous conditions provided the concentrations are the same. Increasesing the fineness of the solid is an attempt to decrease the heterogeneity of the reaction mixture. This implies that reaction rate depends upon the physical states of reactants. It is common Knowledge that one stirs tea to dissolve the sugar quickly. The same is true when ew mix two reactants because stirring increases dispersion of one reactant species in the other. |
|
| 1268. |
i) Write the mathematics expressions relating the variation of the rate constant of a reaction with temperatures. ii) How can you graphically find the activation energy of the reaction from the above expession? iii) The slope of the line in the graph of log k(k=rate constant) versus `1//T` is `-5841`. Calculate the activation energy of the reaction. |
|
Answer» iii) Slope =`(E_(a))/(2.303R)` or `E_(a) = -("slope") xx 2.303 R` `E_(a) = -(-5814) xx (2.303 xx 8.314 J mol^(-1))` |
|
| 1269. |
Which of the following is a very fast reaction ?A. `Fe(s)overset(O_(2)//H_(2)O)rarr Fe_(2)O_(3) . XH_(2)O`B. `Na_(2)SO_(4)(aq.)+BaCI_(2)(aq.)rarrBaSP_(4)(S)+2NaCI(aq.)`C. `C_(12)H_(22)O_(11)+H_(2)O overset(H^(+))rarrC_(6)H_(12)O_(6)+C_(6)H_(12)O_(6)`D. `NH_(4)NO_(2)overset(Delta)rarr2H_(2)O+N_(2)` |
|
Answer» Correct Answer - B The larger the area of contact, the faster the reaction. Thus finely divided catalysts are more effetive, liquid bromine reacts slowly as compared to bromine vapor and a lump of coal burns at a moderate rate in air while coal dust burns explosively. |
|
| 1270. |
For `2NO+O_(2)to 2NO_(2)` change if the volume of the reaction vessel is doubled , the rate of the reactionA. Will diminsh to 1/4 of initial valueB. Will diminsh to 1/8 of initial valueC. Will grow 4 timesD. will grow 8 times |
|
Answer» Correct Answer - B Volume is doubled therefore concentration become half . Reaction is 3d order ` therefore ((1)/(2))^(3) =(1)/(8)` |
|
| 1271. |
Rate law is an expression relating the rate of a reaction to theA. temperature of the reactionB. rate constant of the reactantsC. the concentration of the reactantsD. both (2) and (3) |
|
Answer» Correct Answer - D The rates of chemical reactions differ greately. Lonic reactions in an aqueous solution are over in a fraction of a second whereas involving molecules take much longer. If we add brium ion to an aqueous solution of sulfate ion, a preciptate of barium sulfate forms almost immediately. On the other hand, reactions such as rusting i=of iron, are impreceptibly slow. Reactions taht occur in a cement mixture as it hardens to concrete reuire years for completion. A molecules reaction usually involves breaking of covalent bonds in the reactions. Different amounts of energies are required for different reactants and hence rates are different. |
|
| 1272. |
A catalyst canA. shift the equilibrium of a reactionB. diminish the enthanlpy of a reactionC. diminish the activation energy of a reactionD. Increases the rate constant of the forward reaction without changing that of the reverse reaction |
|
Answer» Correct Answer - C In presence of catalyst , temperature required is less hence less energy of activation . |
|
| 1273. |
On increasing the temperature by 10K, the rate of reaction becomes double. Which of the following is the most appropriate reason?A. Collision frequency increasesB. activation energy decreases by increases in termperatureC. The fraction of molecules having energy equal to threshold enregy or more increaseD. the value of threshold energy decreases |
|
Answer» Correct Answer - C Rate of reaction increase with temperature . |
|
| 1274. |
The specific reaction rate (or rate constant) of a reaction depends upon (i) temperature of reaction (ii) concentration of reactants and products (iii) activation energy of reaction (iv) presence of absence of a catalystA. (i), (iii), (iv)B. (i), (ii), (iii), (iv)C. (i), (iii)D. (i), (ii) |
|
Answer» Correct Answer - A A rate law is an equation that relates the rate of a reaction to the rate constant and the concentrations of reactants (and catalst) raised to various power. For a general reaction of the type `aA+bBrarrcC+dD` the rate law takes the form Rate `=k[A]^(x)[B]^(y)` Thus, rate law is the expression in which reaction rate is given in terms of molar concentration of reatants with each term raised to same power , which may or may not be same as the stoichiometric coefficient of the reacting species in a balance chemical equation If we know the values of `k,x`, and `y` , we can ues the rate law to calculate the rate of the reaction, given the concentrations of `A` and `B` . Like `k,x` and `y` must be determined exprimentally. In short, we can say that rate laws are always determined experimentallty. |
|
| 1275. |
For the reaction ` 2SO_(3) to 2SO_(2) +O_(2)` rate is expressed as ,A. `-(1)/(2)(d[SO_(3)])/(dt)=-(1)/(2) (d[SO_(2)])/(dt)=(d[O_(2)])/(dt)`B. `-(2d[SO_(3)])/(dt)=(2d[SO_(2)])/(dt)=(d[O_(2)])/(dt)`C. `-(1)/(2)(d[SO_(3)])/(dt)=(1)/(2)(d[SO_(2)])/(dt)=(d[O_(2)])/(dt)`D. `(-d[SO_(3)])/(dt)=(d[SO_(2)])/(dt)=(2d[O_(2)])/(dt)` |
|
Answer» Correct Answer - C `-(1)/(2)(d[So_(3)])/(dt)=(1)/(2) (d[SO_(2)])/(dt) =(1)/(2) (d[SO_(2)])/(dt) =(d[O_(2)])/(dt)` |
|
| 1276. |
The rate of a chemical reaction depends onA. Atomic MassB. Equivalent MassC. Molecular MassD. Active Mass |
|
Answer» Correct Answer - D Factors affecting rate of reaction . |
|
| 1277. |
The units of rate constant and rate of a reaction are idential for:A. third - order reactionB. second - order rectionC. first - order reactionD. zero - order reaction |
|
Answer» Correct Answer - D Be careful not to confuse the rate of a reaction and the rate constant. The rate depends on concentrations, whereas the rate constant does not (it is a constant). Hiher the activation energy, smaller the rate constant (Arrhenius equation: `k=Ae^(-E_(0)//RT)` . Only a temperature change or the introduction of a catalst can chamge the value of `k` . |
|
| 1278. |
The units of rate constant and rate of a reaction are idential for:A. zero-order reactionB. first-order reactionC. second-order reactionD. third-order reaction |
|
Answer» Correct Answer - A Rate `= k[A]^(n)[B]^(m)(Aoverset(k)rarrB)` `rArr` Units of Rate and `k` are same if `m = 0`. |
|
| 1279. |
Which of the following statement is (are) correct?A. (a) It is possible to change the specific rate constant by changing the temperatureB. (b) The specific rate constant is independent of the concentrations of the reacting speciesC. (c ) In step-wise reaction the rate determining step is the slowest oneD. (d) The rate of a catalysed reaction is always independent of the concentration of the catalyst |
| Answer» Correct Answer - a, b, c | |
| 1280. |
Assertion : For a chemical reaction with rise in temperature by `10^(@)` the rate constant is nearly doubled. Reason : At t + 10, the fraction of molecules having energy equal to or greater than activation energy gets doubled.A. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
| Answer» Correct Answer - A | |
| 1281. |
In a zero order reaction:A. (a) The rate is independent of the temperature of the reactionB. (b) The rate is independent of the concentration of the reactantsC. (c ) The half-life depends on the concentration on the reactantsD. (d) The rate constant has the unit mol `litre^(-1) s^(-1)` |
| Answer» Correct Answer - b, c, d | |
| 1282. |
In a reaction `2X+Y rarrX_(2),Y` the reactant `X` will disappear atA. Half the rats as that of disappearance of `Y`B. The same rate as that of disappearance of `Y`C. Twice the rate as that of appearance of `X_(2)Y`D. Twice the rate as that of disappearance of `Y` |
|
Answer» Correct Answer - C::D `2X +Y rar X_(2) Y` Rate `= (1)/(2) (-d[X])/(dt) = (1)/(1)(-d[Y])/(dt) = +(d[X_(2)Y])/(dt)` |
|
| 1283. |
In the reaction `2A+ B to A_(2)B ` the rate of consumption of reactant A isA. Half of the consumption rate of BB. Equal to the consumption rate of BC. Twice to the consumption rate BD. equal to the rate of formation of `A_(2 )` B |
|
Answer» Correct Answer - C `-(1)/(2)(d[A])/(dt) =-(d[B])/(dt)` |
|
| 1284. |
Assertion : The rate of a reaction sometimes does not depend on concentration. Resaon : Lower the activation energy, faster is the reactionA. If both assertion and reason are true and reason is the correct explanation of assertion.B. If both assertion and reason are true but reason is not the correct explanation of assertion.C. If assertion is true but reason is false.D. If both assertion and reason are false. |
| Answer» Correct Answer - B | |
| 1285. |
In a reaction `2A + B rarr A_(2)B`, the reactant `A` will disappear atA. (a) The same rate at which `B` will decreaseB. (b) Twice the rate at which `B` will decreaseC. (c ) Half the rate at which `B` will decreaseD. (d) Twice the rate at which `A_(2)B` will form |
| Answer» Correct Answer - b, d | |
| 1286. |
The Arrhenius relationship of two different reactions is shown below. Which reaction is faster at a lower temperature and which is more sensitive to changes of temperature ? A. `B` faster, `A` more sensitiveB. `B` in both casesC. `A` in both casesD. `A` faster, `B` more sensitive |
|
Answer» Correct Answer - A `K = Ae-(E_(a))/(RT)" "log k = log A + ((E_(a))/(2.303R)) xx (1)/(T)` `(-E_(a))/(2.303R) =` slope Slope of `A gt` Slope of `B` `rArr (E_(a))_(A) gt (E_(a))_(B)` |
|
| 1287. |
Which of the following statement is (are) true for the given reaction, `4Ararr B hArr 2C+2D`?A. (a) The rate of disappearance of `B` is one-forth the rate of disappearance of `A`B. (b) The rate of appearance of `C` is half the rate of disappearance of `B`C. (c ) The rates of formation of `C` and `D` are equalD. (d) The rate of formation of `D` is half the rate of disappearance of `A` |
| Answer» Correct Answer - a, c, d | |
| 1288. |
The first-order rate constant `k` is related to temperature as `log k = 15.0 - (10^(6)//T)`. Which of the following pair of value is correct ?A. `A = 10^(15)` and `E = 1.9 xx 10^(4) kJ`B. `A = 10^(-15)` and `E = 40 kJ`C. `A = 10^(15)` and `E = 40 kJ`D. `A = 10^(-15)` and `E = 1.9 xx 10^(4) kJ` |
|
Answer» Correct Answer - A `log k = 15.0 - (10^(6))/(T)` compare this relation with `log k = log A - (E)/(2.303 R)` we find `A = 10^(15)" "E = 1.9 xx 10^(4)J` |
|
| 1289. |
The rate of a chemical reaction doubles for every `10^(@)C` rise of temperature. If the temperature is raised by `50^(@)C`, the rate of the reaction increases by aboutA. 24 timesB. 32 timesC. 64 timesD. 10 times |
|
Answer» Correct Answer - B Temperature coefficient `mu =2 ` `""_(mu ) (DeltaT)/(10)=(k_(2))/(k_(1)),` `""_(2)(50)/(10)=2^(5)=32=(k_(2))/(k_(1))` therefore `32 k_(1)=k_(2)` |
|
| 1290. |
In general, it is observed that the rate of a chemical reaction doubles with every `10^(@)` rise in temperature. If the generalisation holds for a reaction in the temperature range 295 K to 305 K, what would be the value of activation energy for the reaction? |
|
Answer» According to Arrheneius equation: `logk_(2)/k_(1) = E_(a)/(2.303R)[1/T_(1)-1/T_(2)]` `k_(2)//k_(1)=2, T_(1)=295 K, T_(2)= 305K, R=8.314J K^(-1)mol^(-1)` `log2= E_(a)/(2.303 xx (8.314 JK^(-1)mol^(-1)))[1/295K -1/305k]` or `0.3010 = E_(a)/(2.303 xx (8.314Jmol^(-1))) xx (10)/(295 xx 305)` `E_(a) = (0.310 xx 2.303 xx 8.314 xx 295 xx 305)/(10) (J mol^(-1))= 51855 J mol^(-1)` `=51.855 kJ mol^(-1)` |
|
| 1291. |
Which of the following statement is (are) correct?A. (a) The order of a reaction can be zeroB. (b) The order of an elementary reaction is equal to its molecularityC. (c ) The order of the inversion of sucrose is `2`D. (d) The oder of a reaction may change if the experimental conditions are changed |
| Answer» Correct Answer - a, b, d | |
| 1292. |
The rate of a chemical reaction doubles for every `10^(@)C` rise of temperature. If the temperature is raised by `50^(@)C`, the rate of the reaction increases by aboutA. `10` timesB. `24` timesC. `32` timesD. `64` times |
|
Answer» Correct Answer - C For every `10^(@)C` rise of temperature, rate is doubled. Thus, temperature coefficient of the reaction `= 2` When temperature is increased by `50^(@)`, rate becomes`= 2^((50//10)) = 2^(5) "times" = 32` times |
|
| 1293. |
A container of `2` litre contain `4` moles of `N_(2)O_(5)`. On heating to `100^(@)C, N_(2)O_(5)` undergoes complete dissociation to `NO_(2)` and `O_(2)`. If rate constant for decomposition of `N_(2)O_(5)` is `6.2xx10^(-4) sec^(-1)`, select the correct statements:A. (a) The mole ratio before and after dissociation is `4:2`B. (b) The time required to complete `40%` of reaction is `824 sec`C. (c ) `t_(1//2)` of `N_(2)O_(5)` is `1117.7` sec and it is independent of temperatureD. (d) If volume of container is doubled, the rate of decomposition becomes half of the initial rate |
|
Answer» Correct Answer - b, c, d `{:(,N_(2)O_(5),rarr,2NO_(2),+,1/2O_(2)),("Initial mole",4,,0,,0),("After diss. mole",0,,8,,2):}` `:.` Mole ratio`=4/10=2:5` `t_(1//2)=0.693/K=0.693/(6.2xx10^(-4))=1117.7 sec` But it depends upon temp. as `K` depends upon `T`. `t_(40%)=2.303/(6.2xx10^(-4))"log"100/((100-40))` `=824 sec` Rate `r_(1)=K [N_(2)O_(5)]`, If `V` is double the concentration becomes half. `:. r_(2)=Kxx1/2[N_(2)O_(5)]` `:. r_(1)/r_(2)=2/1` |
|
| 1294. |
The half-life period of any first order reaction:A. (a) is independent of the initial concentration of the reactantB. (b) is inversely proportional to the rate constantC. (c ) is always the same whatever the reactionD. (d) is half the specific rate constant |
| Answer» Correct Answer - a, b | |
| 1295. |
For the reaction, `A+2B rarr C`, the differential from of the rate law is:A. (a) `(d[C])/(dt)=-(d[A])/(dt)`B. (b) `(3d[C])/(dt)=(-d[B])/(dt)`C. (c ) `(-3d[B])/(dt)=(-d[C])/(dt)`D. (d) `(d[A])/(dt)=(d[C])/(dt)` |
| Answer» Correct Answer - a, b | |
| 1296. |
For a first order reaction,A. (a) The degree of dissociation isequal to `(1-e^(-Kt))`B. (b) A plot of reciprocal concentration of the reactant vs. time gives a straight lineC. (c ) The time taken for the completion of `75%` reaction is thrice the `t_(1//2)` of the reactionD. (d) The pre-exponential factor in the Arrhenius equation has the dimension of time, `(T^(-1))` |
|
Answer» Correct Answer - a, d For first order reaction, if degree of dissociation is `alpha`, Then `K=2.303/t"log"1/(a-alpha)` or `K. t="log"1/(1-alpha)=-log(1-alpha)` or `e^(-K.t)=1-alpha` `alpha=1-e^(-Kt)` According to Arrhenius equation, `K=A.e^(-E_(a)//RT)` Thus, dimension of `A` in the above expression is the same as that of rate constant, and for the I order reaction of K is `time^(-1)`. |
|
| 1297. |
Which of the following graphs is correct for the following reaction? `CH_(3)-CH_(2)CH=CH_(2)underset(300^(@)C)overset(H_(2)//Ni)rarrCH_(3)-CH_(2)-CH_(2)-CH_(3)`A. B. C. D. |
|
Answer» Correct Answer - A Reaction involves adsorption on the surface of Ni catalyst, hecne it is a zero order reaction. `a=t_(1//2)k " "` (for zero order) `log a =log t_(1//2)+log2k` Y=MX+C Here, slope M=1 i.e.,`tan theta=1` `:. theta=45^(@)` Intercept C= log 2 k |
|
| 1298. |
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. `53.6 JK^(-1) "mol"^(-1)`B. `48.6 JK^(-1) "mol"^(-1)`C. `58.5 JK^(-1) "mol"^(-1)`D. `60.5 JK^(-1) "mol"^(-1)` |
|
Answer» Correct Answer - A Form Arrhenius equation, `log` `(k_(2))/(k_(1)) = (-E_(a))/(2.303R)((1)/(T_(2)) - (1)/(T_(1)))` Given, `(k_(2))/(k_(1)) = 2, T_(2) = 310 K` `T_(1) = 300 K` On putting values, `rArr log 2 = (-E_(a))/(2.303 xx 8.314) ((1)/(310) - (1)/(310))` `rArr E_(a) = 53603.93 J//mol = 53.6 kJ//mol` |
|
| 1299. |
The rate of reaction:A. (a) Decreases with timeB. (b) Decreases with decrease in conc. Of reactantC. (c ) Decreases with increase in time and cecrease in conc. Of reactantD. (d) None of these |
| Answer» Correct Answer - a, b, c | |
| 1300. |
In the reaction, `3BrO^(-)rarrBrO_(3)^(-)+2Br^(-)` (aqueous alkaline medium at `80^(@)C`) the vlaue of the rate constant in the rate law in terms of `--(d)/(dt)[BrO^(-)]` is `0.056 L "mol"^(-1)s^(-1)`. What will be the rate constant when the law is stated in terms of `(d)/(dt)[BrO^(-)]`?A. `18.7xx10^(-3) L"mol"^(-1)s^(-1)`B. `37.4xx10^(-3) L"mol"^(-1)s^(-1)`C. `0.0187 L"mol"^(-1)s^(-1)`D. `0.0187xx10^(-2) L"mol"^(-1)s^(-1)` |
|
Answer» Correct Answer - C `-(1)/(3)(d[BrO^(-)])/(dt)=(d[BrO_(3)^(-)])/(dt)=+(1)/(3)(d[Br^(-)])/(dt)` Required rate constant `=(1)/(3)xx0.056 L "mol"^(-1)s^(-1)` `=0.0187 L "mol"^(-1)s^(-1)` |
|