94420 + Interview Questions in Chemistry in Class 12

1.	For a reaction the graph drawn between t_(1//2) and a gives a straight line passing through the origin with the slope 2xx10^(2)"mole"^(-1) lit min. lf the initial concentation of the reactant is 1M, then the half life period is _______________X10^(2) min.
Answer» Solution :`t_(1/2)=(C_(0))/(2xxK)impliest_(1/2)=(1/(2K))xxC_(0),t_(1/2)=(1xx4xx10^(2))/(2xx1),t_(1/2)=(4xx10^(2))/2=2xx10^(2)`

Discussion

2.	For a reaction the following mechanism has been proposed , 2A+BtoD+E A+BtoC+D (slow), A+CtoE(fast) The rate law a expression for the reaction is
Answer» `r=k[A]^(2)[B]` `r=k[A][B]` `r=k[A]^(2)` `r=k[A][C]` ANSWER :B

Discussion

3.	For a reaction, the energy of activation is zero. What is the value of rate constant at 300 K, if k=1.6xx10^(6)s^(-1)" at 280 K ? "[R=8.31" JK"^(-1)mol^(-1)]
Answer» SOLUTION :`E_(a)=0:.LOG""(k_(2))/(k_(1))=0" or "(k_(2))/(k_(1))=1" or "k_(2)=k_(1)`

Discussion

4.	For a reaction the activation energy is zero. What is the value of rate constant at 300K. ["Given"k_(280K)=1.6xx10^(6)s^(-1),R=8.314J "mol"^(-1)K^(-1)]
Answer» `2.08xx10^(-3)` `2.08xx10^(-2)` `6.93xx10^(-3)` `6.93xx10^(-2)` ANSWER :C

Discussion

5.	For a reaction taking place in three steps, the rate constants are k_(1),k_(2) and k_(3). The overall rate constant is k=(k_(1)k_(2))/(k_(3)). If the energy of activation values for the first, second and third stages are 40, 50 and 60" kJ mol"^(-1) respectively, then the overall energy of activation in kJ mol"^(-1) is
Answer» 30 150 50 60 Solution :`K=(k_1k_3)/k_3` `AE^(-E//RT)=(Ae^(-E_(1//RT))Ae^(-E_2//RT))/(Ae^(-E_3//RT))` or`e^(-E//RT)=e^((-E_(1)-E_(2)+E_(3))//RT)` or`-E/(RT)=(-E_1-E_2+E_3)/(RT)` or`E=E_1+E_2-E_2=40+50-60` `=30" kJ mol"^-1`

Discussion

6.	Fora reaction taking place in three steps , the rate constants are k_(1) , k_(2) and k_(3) . The overall rate constant k = (k_(1) k_(2))/(k_(3)) . If the energy of activation values for the first , second and third stages are respectively 40 , 50 , and 60 kJ mol^(-1), then the overall energy of activation in kJ mol^(-1) is
Answer» 30 40 60 50 Answer :a

Discussion

7.	For a reaction, temperature increases by 10^(@)C, the equilibrium will be attained faster
Answer» 2 times same `(1)/(2)` same 4 times Solution :Rate of REACTION becomes almost twice with an increase of `10^(@)C` in TEMPERATURE (from 298 K to 308 K).

Discussion

8.	For a reaction taking place is a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant K in terms of change in entropy is described by
Answer» With increase in TEMPERATURE, the VALUE of K for endothermic reaction INCREASES because unfavourable change in entropy of the surroundings decreases With increase in temperature, the value of K for exothermic reaction decreases because favourable change in entropy of the surroundings decreases With increase in temperature, the value of K for exothermic reaction decreases because the entropy change of the system is positive With increase in temperature, the value of K for endothermic reaction increases because the entropy change of the system is negative Answer :A::B::C

Discussion

9.	For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant K in terms of change in entropy is described by
Answer» With increase in TEMPERATURE, the value of K for endothermic REACTION increases because unfavourable CHANGE in ENTROPY of the surroundings decreases With increase in temperature, the value of K for exothermic reaction decreases because FAVOURABLE change in entropy of the surroundings decreases With increase in temperature, the value of K for endothermic reaction increases because the entropy change of the system is negative With increase in temperature, the value of K for exothermic reaction decreases because the entropy change of the system is positive Solution :`DeltaS_("Surr")=(-DeltaH)/(T_("Surr"))` For endothermic, if `T_("surr")` increases, `DeltaS_("surr")` will increases. For exothermic, if `T_("surr")` increases, `DeltaS_("surr")` will decreases.

Discussion

10.	For a reaction taking place in a container inequilibrium with its surroundings the effect of temperature on its equilibrium constant K in terms of change in entrogy is described by
Answer» with increase in temperature, the value of K for endothermic reaction increases beacouse unfavourable change in entrogy of the SURROUNDINGS DECREASES With increase in temperature , the value of K for EXOTHERMIC REACTIONS decreause becacuse favourable change in entropy of the surrounding decreases With increase in temperature, the value of K endothermic rection increases becouse the entrogy change of the system is negative With increase in temperature, the value of K for endothemic reaction INCRESES because the entrogy change of the system is positive Answer :a,b

Discussion

11.	For a reaction t_(1//2)=[R]_(0)2k, reaction is of the
Answer» FIRST ORDER. SECOND order. third order. ZERO order. ANSWER :D

Discussion

12.	For a reaction represented by A rarr B, which of the following options do not given the value of order of the reaction. {:(A,ROR ="Rate of reaction",C="Concentration of reactant"),(,C_(0) ="Concentration of reactant",P="Concentration of product"),(B,t_(1//2) ="Half life of"A,):}
Answer» `(log(ROR)_(2) - log(ROR)_(1))/(log C_(2)- log C_(1))` `1+[(LOGT'_(1//2)-logt_(1//2))/(LOGC'_(0)-logC_(0))]` Slope of In ROR us ln C GRAPH y-intercept of log ROR us log C graph Answer :D

Discussion

13.	For a reaction Rate k=["acetone"]^(3//2) then unit of rate constant and rate of reaction respectively is ..........
Answer» `(mol L ^(-1) s^(-1) ) , (mol ^(-1//2) L^(1//2)s^(-1))` `(mol^(-1//2) L ^(1//2) s^(-1) ) , (MOLL^(-1)s^(-1))` `(mol L s^(-1)),(mol^(1//2)L^(1//2)s)` `(mol^(-1//2)s^(-1)),(molL^(-1)s^(-1))` SOLUTION :Rate `=k[A]^N` unit of rate `=(mol L ^(-1))/s = mol L ^(-1) s^(-1)` unit of rat CONSTANT =`((mol L^(-1)s^(-1)))/((mol L ^(-1))^n)` `= mol^(1-n) L^(n-1) s^(-1)` in this case, rate `= k ["Acetone"]^(3//2)` n = 3/2 `mol ^(1-(3//2)L^(3//2)s^(-1)` `mol^(-(1//2)L^(1//2)s^(-1)`

Discussion

14.	For a reaction Rate = "k [acetone]"^((3)/(2)) then unit of rate constant and rate of reaction respectively is
Answer» `("mol "L^(-1)s^(-1)), ("mol"^(-1//2)L^(1//2)s^(-1))` `("mol"^(-1//2)L^(1//2)s^(-1)), ("mol "L^(-1)s^(-1))` `("mol"^(1//2)L^(1//2)s^(-1)), ("mol "L^(-1)s^(-1))` `("mol "LS^(-1)), ("mol"^(1//2)L^(1//2)s)` Solution :Rate `=K[A]^(n)` Rate `=(-d[A])/(DT), " unit of rate"=("mol "L^(-1))/(s)="mol "L^(-1)s^(-1)` unit of rate constant `=(("mol "L^(-1)s^(-1)))/(("mol "L^(-1))^(n))="mol"^(1-n)L^(n-1)s^(-1)` In this CASE, rate `=k["Acetone"]^(3//2)` `n=3//2` `"mol"^(1-(3//2))L^((3//2)-1)s^(-1) rArr "mol"^(-(1//2))L^((1//2))s^(-1)`

Discussion

15.	For a reaction R to P, half-life (t_(1//2)) is observed to be independent of concentration of reactants. What is the order of reaction ?
Answer» SOLUTION :1 ST ORDER `(t_(1//2)=(0.693)/(K)).`

Discussion

16.	For a reaction of the order of 0.5, when the concentration of the reactant is doubled, the rate
Answer» doubles increases FOUR times decreases four times increases `SQRT2` times Answer :D

Discussion

17.	For a reaction, Ox+ne^(-)tored, the neerst equation has a form of:
Answer» `E=E^(@)+(RT)/(nF)ln([RED])/([OX])` `E=E^(@)-(RT)/(nF)ln([red])/([Ox])` `E=E^(@)-2.303(RT)/(nF)LOG([red])/([Ox])` `E=E^(@)+(RT)/(nF)log([red])/([Ox])` Answer :B::C

Discussion

18.	For a reaction pA+qBto products the rate law expression is r=k[A]^(1)[B]^(m) then
Answer» `(p+q)=(1+m)` `(p+q)GT(1+m)` `(p+q)` MAY or may not be equal to `(1+m)` `(p+q)NE(1+m)` ANSWER :C

Discussion

19.	For a reaction of second order , t_(75%)=xt_(50%). The value of X is:
Answer» Solution :For the second order reaction: `t=1/k[1/[[A]]-1/[A]_(0)]` For `t_(1//2)` or `t_(50%)` `t_(50%) = 1/k[(1/[A]_(0))/2-1/[A]_(0)]=1/k[2/[A]_(0)-1/[A]_(0)]` `=1/k xx 1/[A]_(0)` For `t_(3//4)` or `t_(75%)` `t_(75%) = 1/k[(1/[A]_(0))/4-1/[A]_(0)]=1/k[4/[A]_(0)-1/[A]_(0)]=1/k xx 3/[A]_(0)` `t_(75%) = X t_(50%)` `t_(75%)/(t_(50%))=(1/k xx 3/[A]_(0))/(1/k xx 1/[A]_(0))=3` `THEREFORE x=3`

Discussion

20.	For a reaction of II order kinetics, t_(1//_2) is:
Answer» `PROP a` `prop a^-3` `prop a^2` `prop a^-1` ANSWER :D

Discussion

21.	For a reaction nArarrB, following plots are given: The order of reaction for above plots are:
Answer» `1,0,2` `0,1,2` `1,1,2` `0,2,1` SOLUTION :`(d[A])/(dt)=K.[A],` for FIRST ORDER reaction `[A]=[A]_(0)-kt,` for zero order reaction `1/([A]_(t))=1/([A]_(0))+kt` for SECOND order reaction

Discussion

22.	For a reaction NO_(g) +O_(2) (g)rarr 2NO_(2)(g) Rate=k[NO^(2)] [O_(2)] if the volume of the reaction vessel is doubled the rate of the reaction :
Answer» will diminish to 1/4 of initial value will diminish to 1/8 of initial value will grow 4 times will grow 4times will grow 8 times SOLUTION :(B) On increasing the volume to twice the value the concentration of each species is REDUCED by a factor of 2 . Therefore RATE = `k[NO]^(2) [O_(2)]` `"Rate"_(2)= k(([NO])/(2))^(2)(([O_(2)])/(2))` `("Rate"_(2))/("Rate"_(1)) = (1)/(8)`

Discussion

23.	For a reaction, N_(2)(g)+3H_(2)(g)rarr 2NH_(3)g),identify dihydrogen (H_(2)) as a limiting reagent in the following reaction mixtures.
Answer» `"56 G of "N_(2)+"10 g of "H_(2)` `"35 g of "N_(2)+"8 g of "H_(2)` `"28 g of "N_(2)+"6 g of "H_(2)` `"14 g of "N_(2)+"4 g of "H_(2)` Solution :According to the stoichiometry of balanced equation `28g N_(2)` react with `6gH_(2)` `underset("28g")underset("1 mole")(N_(2))+underset("6g")underset("3 mole")(3H_(2))RARR 2NH_(3)` `therefore"For 56 g of "N_(2), 12 g " of " H_(2)` is required.

Discussion

24.	For a reaction, N_(2)(g) + 3H_(2)(g) to 2NH_(3)(g) , identify dihtdrogen (H_(2)) as a limiting reagent in the following reaction mixtures.
Answer» `14G of N_(2) + 4G of H_(2)` `28g of N_(2) + 6G of H_(2)` `56g of N_(2) + 10g of H_(2)` `35G of N_(2) + 8g of H_(2)` Answer :C

Discussion

25.	For a reaction, K=2xx10^(13)e^(-30000//RT). When log K (y-axis) is ploted against 1/T (x-axis) slope of line be. . . ..Cal
Answer» `(30000)/(4.6)` `(-30000)/(4.6)` `(-30000)/(2.3030)` `(-30000)/(4.6)` ANSWER :D

Discussion

26.	For a reaction(K_((1+10)))/(K_((1)))=x. When temperature is increased from 10^(@)C to 100^(@)C rate constnat (K) increased by a factor of 512. Then value of x is
Answer» `1.5` `2.5` `3` `2` ANSWER :D

Discussion

27.	For a reaction involving 1 mol of Zn and 1 mol of H_(2)SO_(4) in a bomb calorimeter-
Answer» `DELTAU GT0, W gt0` `DeltaU gt0,w=0` `DeltaU LT0,w lt0` `DeltaU lt0,w=0` ANSWER :D

Discussion

28.	For a reaction k is 2.5 xx 10^(-3) 1/(mol xx sec). The order of reaction is
Answer» half one two three Solution :UNIT of 'k' for the REACTION is L `mol^(-1) SEC^(-1)` which represents a second order reaction.

Discussion

29.	For a reaction K=2xx10^-2 mole lit^-1 sec^-1. If the concentration of the reactant is IM, the half life period of the reaction is
Answer» 20 sec 25 sec 34.6 sec 50 sec Answer :B

Discussion

30.	Fora reaction in gaseous state to reach an equilibrium state the reaction should be carried out in
Answer» An OPEN vessel Closed vessel Glass vessel Iron vessel Answer :B

Discussion

31.	For a reaction in a galvanic cell the value of -DeltaG^(@) at certain temperatureis not necessarily equal to
Answer» `nFE^(@)` `RT LN K ` ` T. DeltaS^(@) - DELTAH^(@)` Zero Solution :`DeltaG = -2.303RT log K , DeltaG^(0) = 0` at EQUILIBRIUM

Discussion

32.	For a reaction if k_(p)gtk_(c) the forward reaction is favoured by
Answer» LOW PRESSURE hifh pressure HIGH temperature Low temperature Answer :A

Discussion

33.	For a reaction having rate law expression : Rate = k[A]^(3//2) [B]^(-1//2) If the concentration of both A and B become four times the rate of reaction
Answer» become FOUR times becomes 16 times DECREASES four times reamains same . Solution :(A) `r_(1)=k (a)^(3//2)(b)^(-1//2)` `r_(2) = k (4a)^(3//2)(4b)^(-1//2)` `(r_(2))/(r_(1))=(4)^(3//2)(4)^(-1//2)=4 ` .

Discussion

34.	For a reaction: HX(aq) + H_2O
Answer» `CLO^(-)` `F^(-)` `Cl^(-)` `NO_(2)^(-)` ANSWER :C

Discussion

35.	For a reaction if K_(p) gt K_(c), the forward reaction is favoured by
Answer» A) LOW PRESSURE B) HIGH pressure C) high temperature D) low temperature Solution :Low pressure.

Discussion

36.	For a reaction, H_2O_2rarrH_2O_2+1//2O_2 ({d [H_2O_2])}/{dt}=K_1[H_2O_2], (d[H_2O_2])/dt=K_2[H_2O],(d[O_2])dt=K_3[H_2O_2] Therelationbetween K_1, K_2 and K_3is
Answer» (a) `K_1=K_2=K_3` (B) `K_1=K_2=2K_3` (C) `2K_1=4K_2=K_3` (d) `2K_1=2K_2=K_3` ANSWER :B

Discussion

37.	For a reaction : H_(2)+Cl_(2) underset("Rate"=k)(overset(hv)to 2HCl) (i) Write the order and molecularity of this reaction. (ii) Write the unit of k.
Answer» Solution :(i) When rate of the reaction is equal to k, that means it is reaction of ZERO ORDER. THUS Order = 0 (zero) Molecularity = 2 (ii) As k = rate and the UNITS of rate are moles `L^(-1)s^(-1)`. Therefore units of `k="mol L"^(-1)s^(-1)`.

Discussion

38.	For a reaction DeltaH = 9.08 kJ mol^(-1) and DeltaS=35.7 JK^(-1)mol^(-1) Which of the following statements is correct for the reaction
Answer» REVERSIBLE and Isothermal Reversible and Exothermic Spontaneous and Endothermic Spontaneous and Exothermic Solution :`DELTAH` and `DeltaS` both are +ve for spontaneous CHANGE, and `DeltaH=+ve` for endothermic REACTION.

Discussion

39.	For a reaction, H_2+I_2hArr 2HI at 721 K , the value of equilibrium constant is 50. If 0.5 moles each of H_2 and I_2 is added to the system the value of equilibrium constant will be :
Answer» 40 60 50 30 Solution :The EQULIBRIUM constant does not CHANGE when concentration of reactant is CHANGED as the concentration of PRODUCT also get changed accordingly.

Discussion

40.	For a reaction for which the activation energies of forward and reverse reactions are equal:
Answer» `DeltaH=O` `DeltaS=O` The ORDER is zero There is no catalyst Answer :A

Discussion

41.	For a reaction DeltaH=(+3kJ), DeltaS=(+10J//K)beyond which temperature this reaction will be spontaneous
Answer» 300 K 200 K 273 K 373 K Solution :`DeltaG=DeltaH-TDeltaS`. For the SPONTANEOUS REACTION the `DeltaG` MUST be NEGATIVE. `DeltaH=3kJ=+3000 J` `DeltaS=+10 J//K` If `T=300 K implies DeltaG=3000-300xx10=0` If `T=200 K implies DeltaG=3000-200xx10=1000 J` If `T=273 K implies DeltaG=3000-273xx10=270 J` If T = 373 K `DeltaG=3000-373xx10=-730 J` Hence, beyond 373 temperature the reaction will be spontaneous.

Discussion

42.	For a reaction C(s)+CO_(2)(g)hArr2CO(g)
Answer» The equlibrium shifts in backward direction if more of CO is added The equilibrium shifts in backward direction if more of CO is added The equlibrium shifts in forward direction if more of corbon is added On adding carbon, the equlibrium is not disturbed SOLUTION :`Q_(p)=(P_(CO)^(2))/(P_(CO_(2))),Q_(C)=([CO]^(2))/([CO_(2)])` When more of `CO_(2)` is added `Q_(p)` DECREASES, hence in order to reach equlibrium, SYSTEM moves in forward direction, hence choice (a) is correct. When more of CO is added, `Q_(p)` increases and therefore to reach equlibrium, system moves in backward DIRECTIN, so choice (b) is correct. Since active mass of colids is 1, hence changing the concentrations of carbon does not change Q. Hence adding carbon does not shift the equlibrium, (c) is incorrect. Whereas (d) is correct.

Discussion

43.	For a reaction [CaCO_3(s) implies CaO(s) + CO_2(g)] (K_c = 0.5 moll^-1), the maximum moles of CO_2(g) formed at equilibrium in 5 L container is
Answer» 5 2.5 1.24 10 Answer :4

Discussion

44.	For the raction A+Brarr products, it is found that order of A is 2 and the order of B is 3. In the rate expression when the concentration of both A and B are doubled the rate will increases by a factor
Answer» 12 16 32 10 Solution :Given `"Rate"_(1st) = K[A]^(2) [B]^(3) "" … (i)` `"Rate"_(2nd) = K[2A]^(2) [2B]^(3) "" …. (ii)` On DIVIDING equation (i) and (ii) we get `("Rate"_(1st) )/("Rate"_(2nd)) = (K[A]^(2)[B]^(3))/(K4[A]^(2)8[B]^(3)) = (1)/(32)` `therefore "Rate"_(2nd) = 32 XX "Rate"_(1st)`

Discussion

45.	For a reaction AtoB, the rate of reaction can be denoted by -(dA)/(dt)" or "+(dB)/(dt). State the significance of plus and minus signs in this case.
Answer» Solution :Rate of reaction is always positive. Minus SIGN INDICATES decrease in the concentration of the reactant, i.e., `d[A]` is -ve so that `(-)xx(-)=+"ve"`. PLUS sign indicates INCREASE in the concentration of the product with time during the reaction, i.e., `d[B]` is + ve.

Discussion

46.	For a reaction AtoB, E_(a) = 10kJ mol^(-1) and DeltaH = 5 kJmol^(-1). Which is correct potential energy profile for the reaction?
Answer» Solution :II is the CORRECT POTENTIAL ENERGY profile for the reaction which is of endothermic nature.

Discussion

47.	For a reaction AtoB+C. It was found that at the end of 10 minutes from the start.The total optical rotation of the system was 50^(@) and when the reaction is complete it qws 100^(@). Assume that only B and C are optically active and dextro rotatory, the rate constant of this first order reaction wouldbe
Answer» `6.9"min"^(-1)` `0.069"min"^(-1)` `0.69"min"^(-1)` `6.9xx10^(-2)"min"^(-2)` Solution :`[C_(0)-X[x""x` `0""C_(0)""C_(0)` `C_(0)alpha50,xoo25.K=2.303/10xxlog(50/25)=(2.30xx0.3010)/10=0.693/10=0.0693"min"^(-1)`

Discussion

48.	For a reaction Ato products, the graph drawn between rate of reaction Vs (a-x) gives a straight line with the slope 0.0693 "min"^(-1).Initial concentration of the reactant is 0.4M . After 20 minutes the rate of reaction is calculated as 6.93xx10^(-x). Report the value of x?
Answer» Solution :`r=K.(a-x)^(N)impliesn=1,r=K(a-x),0.0693xx20=2.303xx"log"(0.4/(Ct))` `2xx0.3010=log(0.4/(Ct)),4=0.4/(C_(t))impliesC_(t)=0.1,r=K.(C_(t))^(1)=0.693xx(0.1)^(1)=6.93xx10^(-3)`

Discussion

49.	For a reaction Ato products, the rate of reaction is increased by 8 times when the concentration of A is doubled then
Answer» order of the reaction is 4 order of the reaction is 3 units of rate constant are `"mole"^(-2)"LIT"^(2)"TIME"^(-1)` rate `=K[A]^(3)` Solution :`r=K[A]^(3),M/t=Kxx(M)^(3)impliesK=M^(-2)"time"^(-1)`

Discussion

50.	For a reaction Ato products, half life is 40 min, when the initial concentration of A is 4 molar. Half life of same reaction is 80 mm when the initial concentration orf A is 2 molar. Then
Answer» order is 1 order of the REACTION is 2 units of rate constant is `"time"^(-1)` unit of rate of reaction MOLE `"LIT"^(-1)"time"^(-1)` Solution :`[A]_(0)""4""2` `t_(1/2)""40""80` `((t_(1/2))_(1))/((t_(1/2))_(2))=((C_(2))/(C_(1)))^((n-1)),(40/80)=(2/4)^((n-1)),(1/2)^(1)=(1/2)^(n-1),1=n-1,n=2`

Discussion

Explore topic-wise InterviewSolutions in Current Affairs.

For a reaction the graph drawn between t_(1//2) and a gives a straight line passing through the origin with the slope 2xx10^(2)"mole"^(-1) lit min. lf the initial concentation of the reactant is 1M, then the half life period is _______________X10^(2) min.

For a reaction the following mechanism has been proposed , 2A+BtoD+E A+BtoC+D (slow), A+CtoE(fast) The rate law a expression for the reaction is

For a reaction, the energy of activation is zero. What is the value of rate constant at 300 K, if k=1.6xx10^(6)s^(-1)" at 280 K ? "[R=8.31" JK"^(-1)mol^(-1)]

For a reaction the activation energy is zero. What is the value of rate constant at 300K. ["Given"k_(280K)=1.6xx10^(6)s^(-1),R=8.314J "mol"^(-1)K^(-1)]

For a reaction, temperature increases by 10^(@)C, the equilibrium will be attained faster

For a reaction taking place is a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant K in terms of change in entropy is described by

For a reaction taking place in a container in equilibrium with its surroundings, the effect of temperature on its equilibrium constant K in terms of change in entropy is described by

For a reaction taking place in a container inequilibrium with its surroundings the effect of temperature on its equilibrium constant K in terms of change in entrogy is described by

For a reaction t_(1//2)=[R]_(0)2k, reaction is of the

For a reaction represented by A rarr B, which of the following options do not given the value of order of the reaction. {:(A,ROR ="Rate of reaction",C="Concentration of reactant"),(,C_(0) ="Concentration of reactant",P="Concentration of product"),(B,t_(1//2) ="Half life of"A,):}

For a reaction Rate k=["acetone"]^(3//2) then unit of rate constant and rate of reaction respectively is ..........

For a reaction Rate = "k [acetone]"^((3)/(2)) then unit of rate constant and rate of reaction respectively is

For a reaction R to P, half-life (t_(1//2)) is observed to be independent of concentration of reactants. What is the order of reaction ?

For a reaction of the order of 0.5, when the concentration of the reactant is doubled, the rate

For a reaction, Ox+ne^(-)tored, the neerst equation has a form of:

For a reaction pA+qBto products the rate law expression is r=k[A]^(1)[B]^(m) then

For a reaction of second order , t_(75%)=xt_(50%). The value of X is:

For a reaction of II order kinetics, t_(1//_2) is:

For a reaction nArarrB, following plots are given: The order of reaction for above plots are:

For a reaction NO_(g) +O_(2) (g)rarr 2NO_(2)(g) Rate=k[NO^(2)] [O_(2)] if the volume of the reaction vessel is doubled the rate of the reaction :

For a reaction, N_(2)(g)+3H_(2)(g)rarr 2NH_(3)g),identify dihydrogen (H_(2)) as a limiting reagent in the following reaction mixtures.

For a reaction, N_(2)(g) + 3H_(2)(g) to 2NH_(3)(g) , identify dihtdrogen (H_(2)) as a limiting reagent in the following reaction mixtures.

For a reaction, K=2xx10^(13)e^(-30000//RT). When log K (y-axis) is ploted against 1/T (x-axis) slope of line be. . . ..Cal

For a reaction(K_((1+10)))/(K_((1)))=x. When temperature is increased from 10^(@)C to 100^(@)C rate constnat (K) increased by a factor of 512. Then value of x is

For a reaction involving 1 mol of Zn and 1 mol of H_(2)SO_(4) in a bomb calorimeter-

For a reaction k is 2.5 xx 10^(-3) 1/(mol xx sec). The order of reaction is

For a reaction K=2xx10^-2 mole lit^-1 sec^-1. If the concentration of the reactant is IM, the half life period of the reaction is

Fora reaction in gaseous state to reach an equilibrium state the reaction should be carried out in

For a reaction in a galvanic cell the value of -DeltaG^(@) at certain temperatureis not necessarily equal to

For a reaction if k_(p)gtk_(c) the forward reaction is favoured by

For a reaction having rate law expression : Rate = k[A]^(3//2) [B]^(-1//2) If the concentration of both A and B become four times the rate of reaction

For a reaction: HX(aq) + H_2O

For a reaction if K_(p) gt K_(c), the forward reaction is favoured by

For a reaction, H_2O_2rarrH_2O_2+1//2O_2 ({d [H_2O_2])}/{dt}=K_1[H_2O_2], (d[H_2O_2])/dt=K_2[H_2O],(d[O_2])dt=K_3[H_2O_2] Therelationbetween K_1, K_2 and K_3is

For a reaction : H_(2)+Cl_(2) underset("Rate"=k)(overset(hv)to 2HCl) (i) Write the order and molecularity of this reaction. (ii) Write the unit of k.

For a reaction DeltaH = 9.08 kJ mol^(-1) and DeltaS=35.7 JK^(-1)mol^(-1) Which of the following statements is correct for the reaction

For a reaction, H_2+I_2hArr 2HI at 721 K , the value of equilibrium constant is 50. If 0.5 moles each of H_2 and I_2 is added to the system the value of equilibrium constant will be :

For a reaction for which the activation energies of forward and reverse reactions are equal:

For a reaction DeltaH=(+3kJ), DeltaS=(+10J//K)beyond which temperature this reaction will be spontaneous

For a reaction C(s)+CO_(2)(g)hArr2CO(g)

For a reaction [CaCO_3(s) implies CaO(s) + CO_2(g)] (K_c = 0.5 moll^-1), the maximum moles of CO_2(g) formed at equilibrium in 5 L container is

For the raction A+Brarr products, it is found that order of A is 2 and the order of B is 3. In the rate expression when the concentration of both A and B are doubled the rate will increases by a factor

For a reaction AtoB, the rate of reaction can be denoted by -(dA)/(dt)" or "+(dB)/(dt). State the significance of plus and minus signs in this case.

For a reaction AtoB, E_(a) = 10kJ mol^(-1) and DeltaH = 5 kJmol^(-1). Which is correct potential energy profile for the reaction?

For a reaction Ato products, the graph drawn between rate of reaction Vs (a-x) gives a straight line with the slope 0.0693 "min"^(-1).Initial concentration of the reactant is 0.4M . After 20 minutes the rate of reaction is calculated as 6.93xx10^(-x). Report the value of x?

For a reaction Ato products, the rate of reaction is increased by 8 times when the concentration of A is doubled then

For a reaction Ato products, half life is 40 min, when the initial concentration of A is 4 molar. Half life of same reaction is 80 mm when the initial concentration orf A is 2 molar. Then