94420 + Interview Questions in Chemistry in Class 12

1.	For a gaseous reaction 2A+B_(2)to2AB, the following rate data were obtained at 300 K {:("Rate of disappearance of "B_(2),,,"""Concentration",,,),("(""mol lit"^(-1)"min"^(-1)")",,,[A],,,[B_(2)]),((i)1.8xx10^(-3),,,0.015,,,0.15),((ii)1.08xx10^(-2),,,0.09,,,0.15),((iii)5.4xx10^(-3),,,0.015,,,0.45):} Calculate the rate constant for the reaction and rate of formation of AB when [A] is 0.02 and [B_(2)] is 0.04 "mol lit"^(-1)" at 300 K".
Answer» Solution :From (i) and (II), when `[B_(2)]` is KEPT constant and [A] is made 6 times, the rate also six times. Thous, Rate `prop[A].` Further, from (i) and (iii), when [A] is kept constant and `[B_(2)]` is made three times, rale also becomes three times. Thus, Rate `prop[B_(2)]`. Hence, rate `=k[A][B_(2)].` PUT the values of [A] and `[B_(2)]` and calculate k. It comes out to be 0.8 LITRE `"mol"^(-1)min^(-1)`. When [A] = 0.02 M and `B_(2)=0.04" M",` Rate of formation of `AB=2xx" Rate of disappearance of "B_(2)`

Discussion

2.	For a gas-solid system, classical adsorption isotherm is applicable. If mg of solid adsorb xg of gas at pressure p, then which of the following plots will be a straight line?
Answer» <P>`x/m vs.p` `P/(x//m)Vs.` `logx_m Vs.` `logx_M Vs.` SOLUTION :It is according to Freundlich adsorption ISOTHERM `log""x/3 g+1/n log `

Discussion

3.	For a gaseous equilibrium, A +2B hArr C + 3D the partial pressures of A, B, C and D are found to be 0.20, 0.10, 0.30 and 0.50 atm respectively. Predict the value of equilibrium constant.
Answer» 11.25 18.75 5 3.75 Answer :B

Discussion

4.	Fora gas - solidadsorbentsystem, theadsorptionisotherm (x)/(m) = kp^(1//m) is applicable , where n=5in the specifiedcase . Select the correctstatement.
Answer» `(x)/(m)` vs p plotis a linear graph with slope EQUAL to 5. `LOG (x)/(m)` vs logp plot is a straight line with slopeequal to 5. `log (x)/(m)` vs log p is a straightline withslopeequal to 0.2. None of these SOLUTION :Slope`= (1)/(N) = (1)/(5) = 0.2`

Discussion

5.	For a gas reaction ArarrP at T (K) the rate is given by : Rate=k^(') p_(A)^(2) atm/hr
Answer»

Discussion

6.	For a gas phasereaction , the unit of reaction rat is .......
Answer» `s^(-1)` `"ATM " s^(-1)` `MOL L^(-1)s^(-1)` `mol^(-1)L^(-1)s^(-1)` Answer :B

Discussion

7.	For a gas (R//C_(v)) = 0.67, the gas is made up of molecules which are
Answer» Polyatomic Mixture of GAS Diatomic Monatomic Answer :D

Discussion

8.	For a gas ((R)/(C_v)) = 0.67, the gas is made up of molecule which are :
Answer» Monoatomic Diatomic Polyatomic Mixture of gases Answer :A

Discussion

9.	For a gas at room temperature (289 K)and 1 atm, each molecules undergoes approximately ............ Per second.
Answer» SOLUTION :`10^9` COLLISIONS

Discussion

10.	For a fixes amount of real gas when a graph of z v//s was plotted than at very high pressure slope was observed to be 0.01 atm^(-1). At the same temperature if a graph is plotted b//w pv v//s P then for 2 moles of the gas 'Y' intercept is found to be 40 atm-litre. calculate excluded volume in litres for 20 moles of the real gas.
Answer» Solution :`:.` very high pressure `:.` NEGLECT (a) `:. Z = 1+(Pb)/(RT) ..(1)` comparing above equation with `y = mx +c` `:. M = (b)/(RT)` `rArr (b)/(RT) = 0.01` (GIVEN `m = 0.01)` `b = 0.01 RT ..(2)` `Z = ((PV)_(real))/((PV)_("ideal"))` ltbr. `Z = ((PV)_(real))/(nRT)` (given for `N = 2, PV = 40)` `Z = (40)/(2RT)` `Z (20)/(RT) ...(3)` as, `Z = 1+(Pb)/(RT)` from equation (3) `(20)/(RT) = 1+(Pb)/(RT) ..(4)` `10 = RT +Pb` `Pb = 20 - RT ...(5)` `(PV)_(real) = 40= ZnRT = (1+(Pb)/(2RT)) 2RT` `rArr 40 = (1+(20-RT)/(2RT)) 2RT` `40 = 2RT +20 - RT` `20 = RT...(6)` From (2) & (4) `b = 0.01 xx 20` `b = 0.2` excluded volume for 20 moles `nb = 20 xx 0.2` `nb = 4`

Discussion

11.	For a fixed amount of an ideal gas P us T plot is given as shown. Identify the correct option.
Answer» The change from A to B should be isochoric VOLUME first INCREASES reached MAXIMA and then decreases `PV = NRT` is not applicable None of the statements are correct Answer :D

Discussion

12.	For a fist order reaction the rate constant is 6.909 min^(-1) . The time taken for 75% conversion in minutes is :
Answer» `(3)/(2) "LOG"2` `(2)/(3)` log 3 `(2)/(3)` log 2 `(3)/(2)` log `(3)/(4)` Solution :(C ) K = 6.909 `min^(-1)` `t= (2.303)/(6.909)"log" (100)/(25)` `=(2.303)/(6.909)xx"log" 4` `= (2.303)/(6.909) xx2 "log" 2 = (2)/(3 ) "log" 2 `

Discussion

13.	For a first reaction, t_(0.75) is 138.6 seconds. Its specific rate constant (in sec^(-1)) is
Answer» `10^(-2)` `10^(-4)` `10^(-5)` `10^(-6)` Answer :C

Discussion

14.	For a first reaction , show that time required for 99% completion is twice the time required for the completion of 90% of reaction.
Answer» SOLUTION :For a FIRST REACTION, `t =k =(2.303)/t log.a/(a-x)` 99% completion means that `x =99% ` of a = 0.99a `:. t_(99%)=(2303)/k log. a/(a-0.99a)=(2.303)/klog 10^2=2xx(2303)/k` 90% completion means that `x=90% " of " a = 0.90a` `:. t_(99%)=(2303)/k log. a/(a-0.99a)=(2.303)/klog 10=2303/k` `(t_(99%))/(t_(90%))=((2xx2303)/k)//(2.303)/k=2 " or " t_90%=2xxt_(90%)`

Discussion

15.	For a first order reaction with rate constant k, which expression gives the half-life period ?
Answer» `kln2` `1//ka` `0.639//k` `3/(2ka^2)` ANSWER :C

Discussion

16.	For a first order reaction with rate constant 'k' and initial concentration 'a', the half-life period is given by
Answer» `(ln2)/(k)` `(1)/(KA)` `(0.693)/(k)` `(3)/(2ka^(2))` Answer :A::C

Discussion

17.	For a first order reaction with rate constant k and initial concentration a, the half-life period is given by
Answer» In `3//k` `1//ka` `0.693//k` `3//2ka^(2)` ANSWER :C

Discussion

18.	For a first order reaction with half-life of 150 seconds, the time taken for the, concentration of the reactant to fall from M/10 to M/100 will be approximately
Answer» 1500 s 500 s 900 s 600 s Solution :`k = 0.693/(t_(1//2))= 0.693/150 s^(-1)= 2.303/t log. (1//10)/(1//100)` `or t = (2.303xx 150)/(0.693) log 10 - 498 -=500S`

Discussion

19.	For a first order reaction we have K=100s^(-1). The time for completion of 50% reaction is
Answer» `10^(-2)s` `4XX10^(-5)s` `6.93xx10^(-3)s` `7XX10^(-5)s` ANSWER :C

Discussion

20.	For a first order reaction with half life of 150 seconds, the time taken for the concentration of the reactant to fall from M/10 to M/100 will be approximately
Answer» 1500 s 500 s 900 s 600 s Answer :B

Discussion

21.	For a first order reaction velocity constant , K = 10^(-3) s^(-1) . Two third life for it would be
Answer» 1100 s 2200 s 3300 s 4400 s Solution :We know that `k = (2.303)/(t)` log `(a)/(a-x)` `10^(-3) = (2.303)/(t)` log `(a)/((a - (2a)/(3))) , 10^(-3) = (2.303)/(t)` log 3 `10^(-3) = (2.303)/(t) xx 0.4771 , t = (2.303 xx 0.4771)/(10^(-3))` `= 1098.7` SEC = 1100 sec .

Discussion

22.	For a first order reaction , to obtain a positive slope , we need to plot {where [A] is the concentration of reactant A}
Answer» `- "log"_(10) [A]` VS t `- "log"_(E) [A]` vs t `log _(10) [A] vs ` log t [A] vs t Solution :For a FIRST order reaction to obtain a positive SLOPE , a graph is PLOTTED between `-"log"_(e)`[A] vs t.

Discussion

23.	For a first order reaction, time taken for half of the reaction to complete is t_(1) whereas that for 3//4th of the reaction to complete is t_(2). How are t_(1) and t_(2) related to each other ?
Answer» SOLUTION :`t_(2)=2t_(1).`This is because for `3//4th` of the reaction to COMPLETE, time required = two half-lives.

Discussion

24.	For a first order reaction, time taken for half of the reaction to complete ist_(1)and 3/4 of the reaction to complete is t_(2). How are t_(1)and t_(2) related?
Answer» Solution :`t_2=2t_1` because for 3/4th of the reaction to complete TIME REQUIRED is equal to TWO HALF LIVES.

Discussion

25.	For a first order reaction , the time taken to reduce the initial concentrations by a factor of 1/4 is 20 minutes . The time required to reduce initial concentration by a factor of 1/16 is
Answer» 20 MIN 10 min 80 min 40 min SOLUTION :`1 overset(1^(st))to1//2 overset(2^(nd))to1//4overset(3^(rd))to1//8 overset(4^(th))to1//16` 1/16 is 4 half-lives and 1/4 is 2 half-lives . It is given that the time taken for 2 half-lives is 20 MINUTES . So the time taken for 4 half-lives will be 40 minutes .

Discussion

26.	For a first order reaction, the time taken to reduce the initial concentration by a factor of (1)/(4) is 20 minutes. The time required to reduce initial concentration by a factor of 1//16 is
Answer» 20 min 10 min 80 min 40 min Solution :Time taken to REDUCE by factor of `(1)/(4) = 20` min. `therefore` Time taken to reduce by factor of `(1)/(2)` i.e., `t_(1//2) = 10 min` `aoverset(t_(1//2))to(a)/(2)overset(t_(1//2))to(a)/(4)overset(t_(1//2))to(a)/(8)overset(t_(1//2))to(a)/(16)` Toral time required ` = 4 xx t_(1//2) = 4 xx 10 min = 40 min`

Discussion

27.	For a first order reaction, the time required for 99.9% of the reaction to take place is nearly
Answer» 10 TIMES that REQUIRED for HALF of the reaction 100 times that required for two-thirds of the reaction 10 times that required for one-fourth of the reaction 20 times that required for half of the reaction Answer :A

Discussion

28.	For a first order reaction the rate constant at 500 K " is " 8xx10^(-4)s^(-1) . Calculate the frequency factor , if the energy of activation for the reaction is 190"kJ mol"^(-1).
Answer» Solution :`k= 8XX10^(-4) , T = 500 K E_a=190 " kJ MOL "^(-1)A = ? ` According to ARRHENIUS equation , `k=Ae^(-E_a//RT)` ln k - ln a - `E_a/(RT)` `logk=logA-E_a/(2.303RT)` `logA=logk+(E_a)/(2.303RT)` `log(8xx10^(-4))+(190)/(2.303xx8.314xx10^(-3)kJ //K^(-1)xx500)` `logA=-3.096+190/(9573.57xx10^(-3))` `logA=16.744` A = Antilog (16.744) `A = 5.546xx10^(16)s^(-1)`

Discussion

29.	For a first- order reaction, the time required for 99.9%of the reaction to take place is nearly
Answer» 10 times that required for HALF the REACTION 100 times that required for two-third of the reaction 10 times that required for one- fourth of the reaction 20 times that required for half of the reaction . ANSWER :A

Discussion

30.	For a first order reaction, the rate constant is 6.909 "min"^(-1), the time taken for 75% conversion in minutes is
Answer» `((3)/(2))log2` `((2)/(3))log2` `((3)/(2))LOG((3)/(4))` `((2)/(3))log((4)/(3))` SOLUTION :`k=((2.303)/(t))log(([A_(0)])/([A]))` `[A_(0)]=100,[A]=25` `6.909=((2.303)/(t))log((100)/(25))` `t=((2.303)/(6.909))log(4)impliest=((1)/(3))log^(2)` `t=((2)/(3))log^(2)`

Discussion

31.	For a first order reaction , the rate constant is 6.909 "min"^(-1) . The time taken for 75% conversion in minutes is ........
Answer» `(3/2)LOG2` `(2/3)log2` `(3/2)log(3/4)` `(2/3)log(4/3)` Solution :`k=((2.303)/t)log.(([A_0])/([A]))` `[A_0]=100,[A]=25` `6.909=((2.303)/t)log(100/25)` `t=(2.303)/(6.909)log(4)impliest=(1/3)log2^2` `t = (2/3)log2`

Discussion

32.	For a first order reaction, the plot of log[A]_(t)vs t is linear with a
Answer» POSITIVE SLOPE and ZERO intercept positive slope and NON zero intercept negative slope and zero intercept negative slope and non zero intercept Answer :A

Discussion

33.	For a first order reaction, the half life is 50 sec. Identify the correct statement form the following.
Answer» the reaction is almost COMPLET in 500 sec the same QUANTITY of reactant is CONSUMED for every 50 sec of the reaction quantity of reactant remaining after 100 sec is half of what remains after 50 sec All the above three Solution :`100overset(50"sec")(to)50 overset(50"sec")(to)25`

Discussion

34.	For a first order reaction temperature coefficient is 2. If the value of K at 310K is 2 xx 10^(-2) "min''^(-1),t_(1//2) of reaction at 300K will be (in min)
Answer» 69.3 23.03 46.06 69.1 Answer :A

Discussion

35.	For a first order reaction ,the half-life period is independent of…….
Answer» INITIAL concentration cuber ROOT of initial concentration first POWER of FINAL concentration square root of final concentration Answer :D

Discussion

36.	For a first order reaction, t_(av) (average life time), t_(50%) and t_(75%) are in the order :
Answer» `t_(50)lt t_(av) lt t_(75),` `t_(50)lt t_(75) lt t_(av),` `t_(av)lt t_(50) lt t_(75),` `t_(av)=t_(50) lt t_(75),` Solution :`t_(av) = (1)/(k) = (1)/(0693//t_(1//2)) = (t_(1//2))/(0693)= 1*44 t_(50%)` ALSO, we KNOW that `t_(75%) = 2 XX t_(50%)` Hence, `t_(50%) lt t_(av) lt t_(75%)`

Discussion

37.	For first order reaction t_(0.75) is 138.6 sec. Its specific rate constant is (ins^(-1))
Answer» `10^(-2)` `10^(-4)` `10^(-5)` `10^(-6)` Answer :A

Discussion

38.	For a first order reaction t_(0.75) is 1368 seconds, therefore, the specific rate constant in sec^(-1) is
Answer» `10^(-3)` `10^(-2)` `10^(-9)` `10^(-5)` Solution :`k=(2.303)/(1386)"LOG"(100)/(100-75)` on solving we GET, `k=10^(-3)`

Discussion

39.	For a first order reaction, show that the time required for 99% completion of a first order reaction is twice the time required for the completion of 90%.
Answer» SOLUTION :For FIRST ORDER reaction, `t=(2.303)/(k)log""(a)/(a-x)` `99%` completion means that `x=99%" of a=0.99" a"` `t_(99%)=(2.303)/(k)log""(a)/(a-0.99a)=(2.303)/(k)log10^(2)=2XX(2.303)/(k)` `90%` completion means that `x=90%" of "a=0.90" a"` `:.""t_(90%)=(2.303)/(k)log""(a)/(a-0.99a)=(2.303)/(k)log10=(2.303)/(k)` `:.""(t_(99%))/(t_(90%))=((2xx2.303)/(k))//((2.303)/(k))=2" or "t_(99%)=2xxt_(90%).`

Discussion

40.	For a first order reaction, show thattime required for 99% completion is twice the time required for the completion of 90% of reaction.
Answer» Solution :For first order REACTION, `t=(2.303)/(k)"log"(a)/(a-x)` At 99% COMPLETION`x=99%` of a i.e., 0.99a `therefore t(99%)=(2.303)/(k)"log"(a)/(a-0.99a)=(2.303)/(k) log 10^(2)= 2XX(2.303)/(k)` At 90% completion MEANS that `x=90%` of a = 0.90 a `therefore t(90%) =(2.303)/(k)"log"(1)/(a-0.90a)=(2.303)/(k)log 10=(2.303)/(k)` `therefore (t(99%))/(t(90%))=((2xx2.303)/(k))//((2.303)/(k))=2.`

Discussion

41.	For a first order reaction R rarrP, the rateconstantis k. if the initialconcentration of R is [R_(0)], theconcentration of R at any time't'is given by the expression-
Answer» `[R_(0)]E^(KT)` `[R_(0)](1-e^(-kt))` `[R_(0)]e^(-kt)` `[R_(0)](1-e^(kt))` SOLUTION :N//A

Discussion

42.	For a first order reaction, rate constant is 0.6932 "hr"^(-1) , then half-life for the reaction is
Answer» `0.01` hr 1HR 2HR 10 hr Solution :`t_(1//2) = (0.693)/(k)= (0.693)/(0.6932 hr^(-1)) = 1`hr .

Discussion

43.	For a first-order reaction of the type nA to product where initial concentration of A is 'a' moles/litre. The correct expression for the rate constant or half-life is
Answer» `k=2.303/(NT)"LOG"a/((a-x))` `k=2.303/(t)"log"a/(a-x)` `t_(1/2)=0.6932/k` `t_(1/2)=0.6932/(NK)` ANSWER :A::D

Discussion

44.	For a first order reaction : R rarr P t_(1//2) is proportional to :
Answer» `[A]^(1//2)` `[A]^(0)` [A] `1//[A]` ANSWER :B

Discussion

45.	For a first order reaction, it takes 5 minutes for the initial concentration of 0.6" mol L"^(-1) to become 0.4" mol L"^(-1). How long in all will it take for the initial concentration to become 0.3" mol L"^(-1) ?
Answer» SOLUTION :`a=0.6" MOL L"^(-1),(a-x)=0.4" mol L"^(-1),t=5" min",""K=(2.303)/(5" min")log""(0.6)/(0.4)=0.0811" min"^(-1),` `0.6" mol L"^(-1)to0.3" mol L"^(-1)"means "t_(1//2).` Hence, `t_(1//2)=(0.693)/(0.0811" min"^(-1))=8.54" min".`

Discussion

46.	For a first order reaction , nAtoB whose concentration vs time curve is as shown in the figure. If half -life for this reaction is 24 minutes. Find out the value of n .
Answer» 1 2 3 4 Solution :At t=48 min, [A]=[B] Also, `[A]=([A]_0)/4=[B]` `underset([A]_0-nx)(nA)toundersetx(B)` `X=([A]_0)/4implies [A]_0-n([A]_0)/4=([A]_0)/4 implies n=3`

Discussion

47.	For a first order reaction, it takes 16 min to complete 50% reaction. How much time does it take to complete 75% reaction?
Answer» Solution :`K=0.693/t_(0.5)=0.693/16=0.0433min^-1` Time required to COMPLETE` 75%`REACTION, `t_(0.75)=2.303/K"log"100/(100-75)=2.303/0.0433"log"100/25=53.187xx0.06021=32`MINUTES.

Discussion

48.	For a first order reaction involving decomposition of N_(2) O_(5) the following information is available : 2N_(2)O_(5)(g) rarr 4NO_(2) (g) +O_(2)(g) Rate =k [N_(2)O_(5)] N_(2)O_(5)(g) rarr 2NO_(2)(g) +1//2 O_(2) (g) Rate =k [N_(2)O_(5)) Which of the following expressions is true ?
Answer» k=k k=2k k=1/2k k `gt k` Solution :(B) `2N_(2)O_(5)(G) rarr4NO_(2) +O_(2)(g) ` Rate = `=-(1)/(2)([N_(2)O_(5)])/(DT)=2k[N_(2)O_(5)]` or Rate=` -(d[N_(2)O_(5)])/(dt)=2k[N_(2)O_(5)]` For `N_(2)O_(5)rarr2NO_(2)+(1)/(2)O_(2)` `-(d[N_(2)O_(5)])/(dt)=k[N_(2)O_(5)]` or Rate `= - (d[N_(2)O_(5)])/(dt)= k[N_(2)O_(5)]` Since rate must be same k = 2k

Discussion

49.	For a first order reaction, if the time taken for completion of 50% of the reactionis t second, the time required for completion of 99.9% of the reaction is:
Answer» 10t 5t 100t 2t Solution :`(Kxxt_(1))/(Kxxt_(2))=("LOG"100/50)/("log"100/0.1)impliest/(t_(2))=(log_(2))/(3log_(10)^(10)),t_(2)=(txx3)/0.3010=10xxt`

Discussion

50.	For a first order reaction half life period t(1/2)is independent of initial concentration of its reacting species. By deriving the equation for t(1/2)of first order reaction, prove that t_(1/2)is independent of initial concentration of its reacting species.
Answer» SOLUTION :For a FIRST order reaction,`k=2.303/tlog[[R_0]]/[[R]]`When t=`t_(1/2)[R]=[[R_0]]/2,` `t_(1/2)=2.303/k LOG([[R_0]]/[[R_0]])/2 ` `2.303/klog2=(2.303xx0.3010)/k thereforet^(1/2)=0.693/k`

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

For a gas-solid system, classical adsorption isotherm is applicable. If mg of solid adsorb xg of gas at pressure p, then which of the following plots will be a straight line?

For a gaseous equilibrium, A +2B hArr C + 3D the partial pressures of A, B, C and D are found to be 0.20, 0.10, 0.30 and 0.50 atm respectively. Predict the value of equilibrium constant.

Fora gas - solidadsorbentsystem, theadsorptionisotherm (x)/(m) = kp^(1//m) is applicable , where n=5in the specifiedcase . Select the correctstatement.

For a gas reaction ArarrP at T (K) the rate is given by : Rate=k^(') p_(A)^(2) atm/hr

For a gas phasereaction , the unit of reaction rat is .......

For a gas (R//C_(v)) = 0.67, the gas is made up of molecules which are

For a gas ((R)/(C_v)) = 0.67, the gas is made up of molecule which are :

For a gas at room temperature (289 K)and 1 atm, each molecules undergoes approximately ............ Per second.

For a fixed amount of an ideal gas P us T plot is given as shown. Identify the correct option.

For a fist order reaction the rate constant is 6.909 min^(-1) . The time taken for 75% conversion in minutes is :

For a first reaction, t_(0.75) is 138.6 seconds. Its specific rate constant (in sec^(-1)) is

For a first reaction , show that time required for 99% completion is twice the time required for the completion of 90% of reaction.

For a first order reaction with rate constant k, which expression gives the half-life period ?

For a first order reaction with rate constant 'k' and initial concentration 'a', the half-life period is given by

For a first order reaction with rate constant k and initial concentration a, the half-life period is given by

For a first order reaction with half-life of 150 seconds, the time taken for the, concentration of the reactant to fall from M/10 to M/100 will be approximately

For a first order reaction we have K=100s^(-1). The time for completion of 50% reaction is

For a first order reaction with half life of 150 seconds, the time taken for the concentration of the reactant to fall from M/10 to M/100 will be approximately

For a first order reaction velocity constant , K = 10^(-3) s^(-1) . Two third life for it would be

For a first order reaction , to obtain a positive slope , we need to plot {where [A] is the concentration of reactant A}

For a first order reaction, time taken for half of the reaction to complete is t_(1) whereas that for 3//4th of the reaction to complete is t_(2). How are t_(1) and t_(2) related to each other ?

For a first order reaction, time taken for half of the reaction to complete ist_(1)and 3/4 of the reaction to complete is t_(2). How are t_(1)and t_(2) related?

For a first order reaction , the time taken to reduce the initial concentrations by a factor of 1/4 is 20 minutes . The time required to reduce initial concentration by a factor of 1/16 is

For a first order reaction, the time taken to reduce the initial concentration by a factor of (1)/(4) is 20 minutes. The time required to reduce initial concentration by a factor of 1//16 is

For a first order reaction, the time required for 99.9% of the reaction to take place is nearly

For a first order reaction the rate constant at 500 K " is " 8xx10^(-4)s^(-1) . Calculate the frequency factor , if the energy of activation for the reaction is 190"kJ mol"^(-1).

For a first- order reaction, the time required for 99.9%of the reaction to take place is nearly

For a first order reaction, the rate constant is 6.909 "min"^(-1), the time taken for 75% conversion in minutes is

For a first order reaction , the rate constant is 6.909 "min"^(-1) . The time taken for 75% conversion in minutes is ........

For a first order reaction, the plot of log[A]_(t)vs t is linear with a

For a first order reaction, the half life is 50 sec. Identify the correct statement form the following.

For a first order reaction temperature coefficient is 2. If the value of K at 310K is 2 xx 10^(-2) "min''^(-1),t_(1//2) of reaction at 300K will be (in min)

For a first order reaction ,the half-life period is independent of…….

For a first order reaction, t_(av) (average life time), t_(50%) and t_(75%) are in the order :

For first order reaction t_(0.75) is 138.6 sec. Its specific rate constant is (ins^(-1))

For a first order reaction t_(0.75) is 1368 seconds, therefore, the specific rate constant in sec^(-1) is

For a first order reaction, show that the time required for 99% completion of a first order reaction is twice the time required for the completion of 90%.

For a first order reaction, show thattime required for 99% completion is twice the time required for the completion of 90% of reaction.

For a first order reaction R rarrP, the rateconstantis k. if the initialconcentration of R is [R_(0)], theconcentration of R at any time't'is given by the expression-

For a first order reaction, rate constant is 0.6932 "hr"^(-1) , then half-life for the reaction is

For a first-order reaction of the type nA to product where initial concentration of A is 'a' moles/litre. The correct expression for the rate constant or half-life is

For a first order reaction : R rarr P t_(1//2) is proportional to :

For a first order reaction, it takes 5 minutes for the initial concentration of 0.6" mol L"^(-1) to become 0.4" mol L"^(-1). How long in all will it take for the initial concentration to become 0.3" mol L"^(-1) ?

For a first order reaction , nAtoB whose concentration vs time curve is as shown in the figure. If half -life for this reaction is 24 minutes. Find out the value of n .

For a first order reaction, it takes 16 min to complete 50% reaction. How much time does it take to complete 75% reaction?

For a first order reaction involving decomposition of N_(2) O_(5) the following information is available : 2N_(2)O_(5)(g) rarr 4NO_(2) (g) +O_(2)(g) Rate =k [N_(2)O_(5)] N_(2)O_(5)(g) rarr 2NO_(2)(g) +1//2 O_(2) (g) Rate =k [N_(2)O_(5)) Which of the following expressions is true ?

For a first order reaction, if the time taken for completion of 50% of the reactionis t second, the time required for completion of 99.9% of the reaction is:

For a first order reaction half life period t(1/2)is independent of initial concentration of its reacting species. By deriving the equation for t(1/2)of first order reaction, prove that t_(1/2)is independent of initial concentration of its reacting species.