94420 + Interview Questions in Chemistry in Class 12 Page 494 InterviewSolution

24651.	The rate of chemical reaction depends on the nature of chemical reactions, because:
Answer» The threshold ENERGY level DIFFERS from one reaction to another Some of the reactant are solid at room temperature Some of the REACTANTS are coloured All Answer :A

Discussion

24652.	The rate of change in concentration of C in the reaction 2A+B to 2C +3D was 1.0 molL^(-1)S^(-1). Calculate the reaction rate as well as rate of change of concentration of A,B, and D
Answer» Solution :We have, `-(1)/(2)(d[A])/(dt)=(-d[B])/(dt)=(1)/(2)(d[C ])/(dt)=(1)/(3)(d[D])/(dt)=` rate of reaction `:. (d[C ])/(dt)=1.0 mol L^(-1)sec^(-1)` `:.-(d[A])/(dt)=(d[C])/(dt)=1.0molL^(-1)sec^(-1)` `-(d[B])/(dt)=(1)/(2)(d[C])/(dt)=(1)/(2)=0.5molL^(-1)sec^(-1)` `(d[D])/(dt)=(3)/(2)(d[C])/(dt)=(3)/(2)xx1=1.5mol L^(-1)sec^(-1)` Rate `=(1)/(2)(d[C])/(dt)=(1)/(2)xx1=0.5molL^(-1)sec^(-1)`

Discussion

24653.	The rate of change of concentration of (A) for reaction A to B is given by -(d[A])/(dt)=k[A]^(1//3) Derive expression for half-life period of the reaction.
Answer» Solution :`-(d[A])/(dt)=K[A]^(1//3):.-int(d[A])/([A]^(1//3))=k" dt or "-int[A]^(-1//3)d[A]=k" dt "-(3)/(2)[A]^(2//3)=KT+C[int X^(n)dx=(x^(n+1))/(n+1)]""...(i)` At `t=0,[A]=[A_(0)]:. -(3)/(2)[A_(0)]^(2//3)=C` Substituting the value of C in eqn. (i), `-(3)/(2)[A]^(2//3)=kt-(3)/(2)[A_(0)]^(2//3)" or "kt=(3)/(2)[A_(0)]^(2//3)-(3)/(2)[A]^(2//3)=(3)/(2){[A_(0)]^(2//3)-[A]^(2//3)}` At `t=t_(1//2),[A]=([A_(0)])/(2)` `:.kt_(1//2)=(3)/(2){[A_(0)]^(2//3)-[(A_(0))/(2)]^(2//3)}` or `t_(1//2)=(3)/(2k)[A_(0)]^(2//3)[1-(1)/((2)^(2//3))]=(3[A_(0)]^(2//3))/(2k)[((2)^(2//3)-1)/((2)^(2//3))]=(3[A_(0)]^(2//3))/(k)[((2)^(2//3)-1)/(2^(5//3))]`

Discussion

24654.	The rate of chemical reaction at constant temperature is proportional to
Answer» The amount of products formed The product of MASSES of the REACTANTS The product of the molar concentration of the reactants The MEAN free path of the reaction Solution :The rate of CHEMICAL reaction `prop` The product of the molar CONC., of the reactants (at constant T).

Discussion

24655.	The rate of a reaction that not involve gases is not dependent on
Answer» Pressure TEMPERATURE CONCENTRATION Catalyst SOLUTION :The rate of reaction depends upon conc., of REACTANT , surface area of reactant , temperature , presence of light and catalyst .

Discussion

24656.	The rate of a reaction that not involve gases is not dependent on:
Answer» PRESSURE TEMPERATURE CONCENTRATION CATALYST ANSWER :A

Discussion

24657.	The rate of a reaction qudruples when the temperature changes from 293 K to 313 K.Calculation the energy of activation of the reaction assuming that it does not change with temperature.
Answer» SOLUTION :`T-(1)`=293 K rate=`r_(1)` means `k_(1)` and `E_(a)`=(?) `T_(2)`=313K rate =`r_(2)=4r_(1)` `k_(2)=4k_(1)` log `(K_(2))/(k_(1))`=log `(4k_(1))/(k_(1))` =log 4 log `(k_(2))/(k_(1))=(E_(a))/(2.303xx8.314)((313-293)/(313xx293))` `therefore 0.6021=(E_(a))/(2.303xx8.314)((20)/(313xx293))` `therefore E_(a)=(0.6021xx2.303xx8.314xx313xx293)/(20)`

Discussion

24658.	The rate of a reaction quadruples when the tamperature changes from 300 to 310k. The activation energy of this reaction is : (Assume activation energy and pre-exponential factor are independent of temperature , In2=0.693, R=8.314Jmol^(-1)K^(-1))
Answer» `107.2kJmol^(-1)` `53.6kJmol^(-1)` `26.8kJmol^(-1)` `214.4kJmol^(-1)` SOLUTION :`In(k_(1))/(k_(2))=(E_(a))/(R)[(1)/(T_(1))-(1)/(T_(2))]` `IN4=(E_(a))/(8,314)((310-300)/(310times300))` `2In2=(E_(a))/(8,314)((310-300)/(310times300))` `E_(a)=107.2kJ*mol^(-1)`

Discussion

24659.	The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction, assuming that it does not change with temperature.
Answer» SOLUTION :Using the relation `"log"(k_(2))/(k_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]`. In this CASE, `k_(2)=4k_(1)` `therefore "log"(4k_(1))/(k_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` Substituting the VALUES, we get `log 4=(E_(a))/(2.303x8.314)[(313-293)/(293xx313)] or 0.6021=(E_(a))/(19.147) xx (20)/(91709)` or `E_(a)=(0.6021xx19.147xx91709)/(20) or E_(a)=52862"J mol"^(-1)=52.862"kJ mol"^(-1)`.

Discussion

24660.	The rate of a reaction may be expressed by the following different ways : (1)/(2) (d[X])/(dt)=-(1)/(3) (d[Y])/(dt)=-(d[Z])/(dt) The reaction is :
Answer» `3 Y +Z rarr2X ` `2X+Yrarr Z` `3X+2Yrarr 6Z` `2Y +6Z rarr3X` ANSWER :A

Discussion

24661.	The rate of a reaction is expressed in different ways as follows +(1)/(2)(d[C])/(dt)=-(1)/(5)(d[D])/(dt)=+(1)/(3)(d[A])/(dt)=-(d[B])/(dt) The reaction is
Answer» `4A+B to 2C+3D` `B+5D to 3A+2C` `4A+2B to 2C` `B+(1)/(2)D to 4A+2C` ANSWER :B

Discussion

24662.	The rateof a reactionis givenas , rate= k[X]^(3//2) [Y]^(-1//2) Whatis the orderof thereaction ? If Y is takenlarge in excess , writethe rate equation.
Answer» Solution :Rate LAW of the reaction is given as, rate `=k[X]^(3//2)[Y]^(-1//2)`. ORDER with respect to X is 3/2 Order with respect to Y is -1/2 Over all order of the reaction `=3//2-1//2=1` If Y is taken large in excess, the rate is INDEPENDENT of its concentration, as [Y] is taken as constant during the course of reaction. The rate EQUATION where X is isolated is rate `=k[X]^(3//2)`.

Discussion

24663.	The rate of a reaction is expressed in the units
Answer» L `"mol"^(-1)t^(-1)` mol `DM^(-3)t_(-1)` Ms `M^(-1)s^(-1)` Answer :A::B::C::D

Discussion

24664.	The rate of a reaction is doubled when the concentration of reactant is doubled. What is the order of the reaction ?
Answer» two one half one and a half SOLUTION :For a FIRST order REACTION, rate `PROP` CONC. Of rectant.

Discussion

24665.	The rate of a reaction is doubled for every 10^(@)C rise in temperature . The increase in reaction rate as a result of temperature rise from 10^(@) C to 100^(@)C is
Answer» 112 512 400 614 Solution :`(k_(t) + 10)/(k_(t)) = (r_(t) + 10)/(r_(t)) = 2`. For an increase to `90^(@)C` i.e. 9 times , the rate increases by `2^(9)` times i.e. 512.

Discussion

24666.	The rate of a reaction is doubled for every 10^@C rise in temperature. The increase in rate as a result of increase in temperature from 10^@C to 100^@C is:
Answer» 112 512 400 256 Answer :B

Discussion

24667.	The rate of a reaction ((dx)/(dt)) varies with nature, physical state and concentration of reactants, temperature, exposure to light and catalyst, whereas rate constant (K) varies with temperature and catalyst only. The rate constant K is given as K=Ae^(pmE_(a)//RT)where A is Arrhenius parameter or pre-exponential factor and E_(a) is energy of activation. The minimum energy required for a reaction is called threshold energy and the additional energy required by reactant molecules to attain threshold energy level is called energy of activation. At what conditions exponential factor is 1 for a reaction:
Answer» Infinite temperature FREE RADICAL combination `E_(a) =0` All of these Answer :D

Discussion

24668.	The rate of a reaction gets doubled when the temperature change from 7^(@)C to 17^(@)C. By what factor will it change for the the temperature change from 17^(@)C to 27^(@)C ?
Answer» `1.81` `1.71` `1.91` `1.76` ANSWER :C

Discussion

24669.	The rate of a reaction increases to four times when the prevailing temperature is raised from 300 K to 320 K. Calculate the energy of activation of this reaction assuming that it does not change with temperature.""[R=8.314"J mol"^(-1)K^(-1)]
Answer» Solution :Arrhenius EQUATION relating the RATE CONSTANT and temperature is given below: `"log"(k_(2))/(k_(1))=(E_(a))/(2.303R) xx ((1)/(T_(1)) - (1)/(T_(2)))` SUBSTITUTING the values, we get `log 4= (E_(a))/(2.303xx8.314)xx ((1)/(300)-(1)/(320)) or log 4= (E_(a))/(19.147) ((20)/(300xx320))` `or E_(a)=(19.147 xx 32xx3xx0.6021)/(20)=55.336"kJ mol"^(-1)`.

Discussion

24670.	The rate of a reaction doubles when its temperature changes from 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)
Answer» 53.6 KJ `MOL^(-1)` 48.6 KJ `mol^(-1)` 58.5 KJ `mol^(-1)` 60.5 KJ `mol^(-1)` ANSWER :A

Discussion

24671.	The rate of a reaction increased from 2 units to 54 units due to change in temperature from 27^(@)C to 57^(@)C. What is the approximate temperature coefficient of the reaction.
Answer» SOLUTION :`(r_(2))/(r_(1))=mu^(x),(54X)/(2R)=mu^(3),27=mu^(3),3^(3),mu=3`

Discussion

24672.	The rate of a reaction doubles when its temperature changes from 300 K to 310 K . Activation energy of such a reaction will be : (R = 8.314 JK^(-1) mol^(-1) and log2 = 0.301)
Answer» `53.6 kJ mol^(-1)` 48.6 kJ `mol^(-1)` 58.5 kJ `mol^(-1)` 60.5 kJ `mol^(-1)` SOLUTION :log `(K_(2))/(K_(1)) = (-E_(a))/(2.030R) ((1)/(T_(2)) - (1)/(T_(1)))` `(K_(2))/(K_(1)) = 2 , T_(2) = 310 K"" T_(1) = 300` K `implies ` log `2 = (-E_(a))/(2.303 XX 8.314 ) ((1)/(310) - (1)/(300))` `implies E_(a) = 53598.6` J/mol = 53.6 KJ/mol

Discussion

24673.	The rate of a reaction depends upon the temperature and is quantitatively expressed as K=Ae^(-E_(a)//RT) (i) If a graph between log K and 1/t write the expression for the slope of the reaction ? (ii) If at under different condition E_(a1)and E_(a2) are the activation energy of two reactions . If E_(a1)=40j/mol and E_(a_(2))=80j/mol. Which of the two has larger value of the rate constasnt ?
Answer» Solution : In `k=-(E_(a))/(RT) + In A` This equation is of the tyep y = MX+ C SLOPE` = -(E_(a))/( R)` (ii) Smaller the ACTIVATION a=energy activation energy , GREATER is the value of rate cosntant . `therefor E _(1)> K_(2)`

Discussion

24674.	The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be
Answer» `60.5" KJ MOL"^(-1)` `53.6" kJ mol"^(-1)` `48.6" kJ mol"^(-1)` `58.5" kJ mol"^(-1)` ANSWER :B

Discussion

24675.	The rate of a reaction doubles for energy 10^(@)C rise of temperature. If the temperature is raised by 50^(@)C, the rate of reaction increases by about
Answer» 10 TIMES 24 times 32 times 64 times Answer :C

Discussion

24676.	The rate of a reaction depends upon the
Answer» Volume pressure force concentration of reactant Answer :D

Discussion

24677.	The rate of a reaction depends upon
Answer» CONCENTRATION of reactant temperature catalyst all of these Answer :D

Discussion

24678.	The rate of a reaction depends on all factors except :
Answer» reaction temperature concentration of reactants magnitude of EQUILIBRIUM CONSTANT specific rate constant. Solution :(C ) Rate of a reaction does not DEPEND upon the magnitude of equilibrium constant .

Discussion

24679.	The rate of a reaction can be increased in general by all the factors except:
Answer» USING a catalyst Increasing the temperature Increasing the ACTIVATION energy Increasing the CONCENTRATION of reactants Answer :C

Discussion

24680.	The rate of a reaction can be increased in general by all the factors except
Answer» USING a catalyst increasing the TEMPERATURE invreasing the ACTIVATION energy increasing the conc. of REACTANTS Solution :Energy of activation increase, RATE of reaction will decrease.

Discussion

24681.	The rate of a reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (E_(a)) of the reaction aassuming that it does not change with temperature.
Answer» Solution :Given if rate at 293 K is R thus at 313 K rate BECOMES = 4R `LOG.(k_(2))/(k_(1)) = (E_(a))/(2.303 R)[(T_(2)-T_(1))/(T_(1) xx T_(2))]` `log.(4R)/(R ) = (E_(a))/(2.303 xx 8.314)[(313 - 298)/(293xx 313)]` `log4 = (E_(a))/(19.1471)[(20)/(91709)]` `0.6021 = (E_(a))/(19.1471)[(20)/(91709)]` `(0.6021 xx 19.1471 xx 91709)/(20) = E_(a)` `E_(a) = 52863.2177 J mol^(-1)` or `= 52.863 kJ mol^(-1)`

Discussion

24682.	The rate of a reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (E_(a)) of the reaction assuming that it does not change with temperature.""[R=8.314JK^(-1)"mol"^(-1), log 4=0.6021]
Answer» Solution :`"log"(k_(2))/(k_(1))=(E_(a))/(2.303R)[(T_(2)-T_(1))/(T_(1)T_(2))]` Substituting the values, we have `log 4=(E_(a))/(2.303xx8.314JK^(-1)"mol"^(-1))[(313K-293K)/(293K XX 313K)]` or `0.6021=(E_(a))/(19.147JK^(-1)"mol"^(-1))xx(20)/(91709K)` or `E_(a)=(0.6021xx19.147xx91709)/(20)"J mol"^(-1)=52863"J mol"^(-1)=52.86"kJ mol"^(-1)`

Discussion

24683.	The rate of a reaction at a particular temperature is r and at another temperatureis 2r. What could be approximate difference betweenthe two temperatures?
Answer» SOLUTION :`10^(@)`

Discussion

24684.	Therate of areactionat 600 Kis 7.5 xx 10^(5)timesthe rate ofthe samereactionat 400 K. Calculatethe energyof activationfor the reaction.
Answer» SOLUTION :Given : `(R_(2))/(R_(1)) = 7.5 xx 10^(5)` Fromthe hint `,(R_(2))/(R_(1))= (K_(2))/(K_(1)) = 7.5 xx 10^(5)` Initialtemperature= `T_(1)= 400 K` Finaltemperature = `T_(2) = 600 K` ENERGYOF activation`=E_(a) = ?` `log_(10).(k_(2))/(k_(1))= (E_(a) (T_(2)-T_(1)))/(2.303 R xx T_(1) xx T_(2))` `= (2.303 xx 8.314 xx 400 xx 600)/((600 -400)) log_(10)7.5 xx 10^(5)` `=(2.303 xx 8.314 xx 400 xx 600xx 5.875)/(200)` `= 135000 J mol ^(-1)=135 kJ mol^(-1)`

Discussion

24685.	The rate of a reaction 2X + Y toProducts is given by (d[Y])/(dt) = k[X]^2[Y] if X is present in large excess, the order of the reaction is
Answer» zero two ONE three Solution :The RATE does not depend UPON the reactant present in excess `THEREFORE` Rate = k [ Y] Hence , order of REACTION = 1.

Discussion

24686.	The rate of a reaction Ararr product, increases by a factor of 100, when cone, of ‘A’ is increased 10 fold. The order of the reaction is
Answer» 3 2 1 0.5 Answer :B

Discussion

24687.	The rate of a particular reaction quadruples when the temperature changes from 293 K to 313 K. Calculate activation energy in KJ/mol.
Answer» Solution :`K_2//K_1`=4 `T_1`=293 K , `T_2`=313 K `logK_2/K_1=-(E_a)/(2.303R)[1/T_1-1/T_2]` Thus, on calculating and substituting values, we GET : `E_a=52.86 KJ "MOL"^(-1)`

Discussion

24688.	The rate of a particular reaction quadruples when the temperature changes from 293K to 313 K. Calculate the energy of activation for such a reaction assuming that it does not change with temperature.
Answer» SOLUTION :`k_(2)=4k_(1)" or "k_(2)//k_(1)=4.`

Discussion

24689.	The rate of a particular reaction doubles when temperaturechanges from 27^(@) C to 37^(@)C. Calculate the energy of activation of such a reaction.
Answer» Solution :Here, we are given that When `T_(1)=27^(@)C=300" K, "k_(1)=k("say"). "When "T_(2)=37^(@)C=310" K",k_(2)=2k` Substituting these VALUES in the equation : `log""(k_(2))/(k_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2))),` we get, `log""(2k)/(k)=(E_(a))/(2.303xx8.314)XX(10)/(300xx310("J MOL"^(-1)))" or "log2=(E_(a))/(2.303xx8.314)xx(10)/(300xx310("J mol"^(-1)))` This on solving gives `E_(a)=53598.6" J mol"^(-1)=53.6" kJ mol"^(-1)`

Discussion

24690.	The rate of a heterogeneous reaction such as iron (solid) and any gas (oxygen) does not depend on
Answer» CONCENTRATION of reactants surface AREA of reactants pressure of reactant gases potential energy of reactant Solution :RATE not DEPENDS on P.E. of reactant

Discussion

24691.	The rate of a gaseous reaction is given by the is expression k[A][B].If the volume of the reaction vessel is suddenly reduced to 1//4 of the initial volume, the reaction rate relating to original rate will be
Answer» `1//10` `1//8` 8 16 Solution :SINCE active mass = CONCENTRATION in moles/litre Thus when the volume of the reaction vessel reduces to `1/4^(th)` of its initial volume the molar concentration of each REACTANT increases four times. Thus r=K[4A][4B]-16 k [A][B] i.e. 16 times.

Discussion

24692.	The rate of a gaseous reaction is given by the expression K[A][B]. If the volume of the reaction vessel is suddenly reduced to 1/4 th of the initial volume, the reaction rate relating to original rate will be
Answer» `1//10` `1//8` 8 16 Solution :When volume is REDUCED to `(1)/(4)` , CONCENTRATION become FOUR times and reaction RATE become 16 times because it is 2 order reaction .

Discussion

24693.	The rate of a gaseous reaction is given by k[A] [B]. If the volume occupied by the reacting gases is suddenly reduced to half the original volume, the rate of the reaction relation relative to the initial rate will be
Answer» `1//8` `1//4` `8//1` 4 Solution :Given `r_(1) = K [A]^(x) [B]^(y) … (i)` On reducing the ORIGINAL volume to HALF , the concentration of reactants A and B will be doubled . LET the new rate be `r_(2)` . Then , `r_(2) =k [2]^(x) [ 2 B] ^(y) or r_(2) = k xx 4 [A]^(x) [B]^(y) ""... (ii)` or `r_(2) = 4 r_(1)`The rate will become four times relative to the initial rate .

Discussion

24694.	The rate of a gaseous reaction is halved when the volume of the vessel is doubled. What is the order of reaction ?
Answer» SOLUTION :Suppose, order of reaction is n and the reaction is A (G) `to`Products RATE=`k[A]^n`…(i) When volume is doubled, molar conc. becomes half and rate of reaction gets halved. `"Rate"/2=k(A/2)^n`…(ii) DIVIDING equation (i) by equation (ii), `(2)^1 =(2)^n` n=1

Discussion

24695.	The rate of a gaseous reaction A + Bto C + D is equal to k [A][B]. The volume of the reaction vessel containing these gases is suddenly reduced to one-fourth of the initial volume . The rate of reaction would be
Answer» `1//16` `16//1` `1//8` `8//1` Solution :INITIALLY, `"rate"_1 = k[A]^x[B]^y` When VOLUME of the vessel is REDUCED to one-fourth, concentrations become 4 times. Hence, NEW rate i.e. `"rate"_2= k[4A]^x[4B]^y` `:."rate"_2= 16 xx "rate"_1`

Discussion

24696.	The rate of a first order reaction : R rarr P is 7.5xx10^(-4) molL^(-1) "s"^(-1) . The rate concentration of R is 0.5 mol L^(-1) . The rate constant is :
Answer» `2.5xx10^(-5) s^(-1)` `1.5xx10^(-3)s^(-1)` `4.16xx10^(-7) s^(-1)` `1.50xx10^(5)s^(-1)` Solution :(A) Rate `=7.5xx10^(-4) "mol L"^(-1) "MIN"^(-1)` `(7.5xx10^(-4))/(60) mol L^(-1) s^(-1)` Rate = k[R] k = `("Rate")/([R])` `= (7.4xx10^(-4))/(60xx0.5)` `= 2.5 xx10^(-5)`

Discussion

24697.	The rate of a first-order reaction is 1.5xx10^(-2)"mol"L^(-1) at 0.5 M concentration of the reactant. The half-life of the reaction is
Answer» 0.383 min 23.1 min 8.73 min 7.53 min Answer :B

Discussion

24698.	The rate of a first order reaction is 1.8xx10^(-3) "mol L"^(-1) when the initial concentration is 0.3 mol L^(-1) . The rate constant in the units of second is :
Answer» `1xx10^(-2) s^(-1)` `1xx10^(-4) s^(-1)` `6xx10^(-2) s^(-1)` `6xx10^(2) s^(-1)` Solution :[B] RATE = K( C) `1.8 xx10^(-3) = k(0.3)` or `k=(1.8xx10^(-3))/(0.3)` `= 6xx10^(-3) min^(-1)` `=(6xx10^(-3))/(60)=1xx10^(-4)s^(-1)`

Discussion

24699.	The rate of a first order reaction is 1.5xx10^(-2) mol L^(-1) "min"^(-1) at 0.5 M concentration of the reactant . The half life of the reaction is :
Answer» `23.1 min` `8.73 ` min `7.53 ` min `0.383 `min SOLUTION :( B) RATE =K [C] `k= ("Rate")/([C])` `=(1.5xx10^(-2))/(0.5) ` `3xx10^(-2)` `t_(1//2)= (0.693)/(3xx10^(-2))` = 23.1 min

Discussion

24700.	The rate of a first order reaction is 0.04mol L^(-1) sec^(-1) at 10 minute and 0.03mol L^(-1) at 20 minute after initiation. Find the half life of the reaction.
Answer» Solution :RATE `=K[A]` `0.04=k[A]_(10)` and `0.03=k[A]_(20)` `([A]_(10))/([A]_(20))=(0.04)/(0.03)=(4)/(3)` `t=(2.303)/(k)log.([A]_(10))/([A]_(20))` when `t=10^(1)`min. `10=(2.303)/(k)log.(4)/(3)` `k=(2.303)/(10)log.(4)/(3)=0.0288min^(-1)` `t_((1)/(2))=(0.693)/(k)=(0.693)/(0.0288)=24.06min`.

Discussion

Explore topic-wise InterviewSolutions in .

The rate of chemical reaction depends on the nature of chemical reactions, because:

The rate of change in concentration of C in the reaction 2A+B to 2C +3D was 1.0 molL^(-1)S^(-1). Calculate the reaction rate as well as rate of change of concentration of A,B, and D

The rate of change of concentration of (A) for reaction A to B is given by -(d[A])/(dt)=k[A]^(1//3) Derive expression for half-life period of the reaction.

The rate of chemical reaction at constant temperature is proportional to

The rate of a reaction that not involve gases is not dependent on

The rate of a reaction that not involve gases is not dependent on:

The rate of a reaction qudruples when the temperature changes from 293 K to 313 K.Calculation the energy of activation of the reaction assuming that it does not change with temperature.

The rate of a reaction quadruples when the tamperature changes from 300 to 310k. The activation energy of this reaction is : (Assume activation energy and pre-exponential factor are independent of temperature , In2=0.693, R=8.314J*mol^(-1)*K^(-1))

The rate of a reaction quadruples when the temperature changes from 293 K to 313 K. Calculate the energy of activation of the reaction, assuming that it does not change with temperature.

The rate of a reaction may be expressed by the following different ways : (1)/(2) (d[X])/(dt)=-(1)/(3) (d[Y])/(dt)=-(d[Z])/(dt) The reaction is :

The rate of a reaction is expressed in different ways as follows +(1)/(2)(d[C])/(dt)=-(1)/(5)(d[D])/(dt)=+(1)/(3)(d[A])/(dt)=-(d[B])/(dt) The reaction is

The rateof a reactionis givenas , rate= k[X]^(3//2) [Y]^(-1//2) Whatis the orderof thereaction ? If Y is takenlarge in excess , writethe rate equation.

The rate of a reaction is expressed in the units

The rate of a reaction is doubled when the concentration of reactant is doubled. What is the order of the reaction ?

The rate of a reaction is doubled for every 10^(@)C rise in temperature . The increase in reaction rate as a result of temperature rise from 10^(@) C to 100^(@)C is

The rate of a reaction is doubled for every 10^@C rise in temperature. The increase in rate as a result of increase in temperature from 10^@C to 100^@C is:

The rate of a reaction gets doubled when the temperature change from 7^(@)C to 17^(@)C. By what factor will it change for the the temperature change from 17^(@)C to 27^(@)C ?

The rate of a reaction increases to four times when the prevailing temperature is raised from 300 K to 320 K. Calculate the energy of activation of this reaction assuming that it does not change with temperature.""[R=8.314"J mol"^(-1)K^(-1)]

The rate of a reaction doubles when its temperature changes from 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)

The rate of a reaction increased from 2 units to 54 units due to change in temperature from 27^(@)C to 57^(@)C. What is the approximate temperature coefficient of the reaction.

The rate of a reaction doubles when its temperature changes from 300 K to 310 K . Activation energy of such a reaction will be : (R = 8.314 JK^(-1) mol^(-1) and log2 = 0.301)

The rate of a reaction doubles when its temperature changes from 300 K to 310 K. Activation energy of such a reaction will be

The rate of a reaction doubles for energy 10^(@)C rise of temperature. If the temperature is raised by 50^(@)C, the rate of reaction increases by about

The rate of a reaction depends upon the

The rate of a reaction depends upon

The rate of a reaction depends on all factors except :

The rate of a reaction can be increased in general by all the factors except:

The rate of a reaction can be increased in general by all the factors except

The rate of a reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (E_(a)) of the reaction aassuming that it does not change with temperature.

The rate of a reaction becomes four times when the temperature changes from 293 K to 313 K. Calculate the energy of activation (E_(a)) of the reaction assuming that it does not change with temperature.""[R=8.314JK^(-1)"mol"^(-1), log 4=0.6021]

The rate of a reaction at a particular temperature is r and at another temperatureis 2r. What could be approximate difference betweenthe two temperatures?

Therate of areactionat 600 Kis 7.5 xx 10^(5)timesthe rate ofthe samereactionat 400 K. Calculatethe energyof activationfor the reaction.

The rate of a reaction 2X + Y toProducts is given by (d[Y])/(dt) = k[X]^2[Y] if X is present in large excess, the order of the reaction is

The rate of a reaction Ararr product, increases by a factor of 100, when cone, of ‘A’ is increased 10 fold. The order of the reaction is

The rate of a particular reaction quadruples when the temperature changes from 293 K to 313 K. Calculate activation energy in KJ/mol.

The rate of a particular reaction quadruples when the temperature changes from 293K to 313 K. Calculate the energy of activation for such a reaction assuming that it does not change with temperature.

The rate of a particular reaction doubles when temperaturechanges from 27^(@) C to 37^(@)C. Calculate the energy of activation of such a reaction.

The rate of a heterogeneous reaction such as iron (solid) and any gas (oxygen) does not depend on

The rate of a gaseous reaction is given by the is expression k[A][B].If the volume of the reaction vessel is suddenly reduced to 1//4 of the initial volume, the reaction rate relating to original rate will be

The rate of a gaseous reaction is given by the expression K[A][B]. If the volume of the reaction vessel is suddenly reduced to 1/4 th of the initial volume, the reaction rate relating to original rate will be

The rate of a gaseous reaction is given by k[A] [B]. If the volume occupied by the reacting gases is suddenly reduced to half the original volume, the rate of the reaction relation relative to the initial rate will be

The rate of a gaseous reaction is halved when the volume of the vessel is doubled. What is the order of reaction ?

The rate of a gaseous reaction A + Bto C + D is equal to k [A][B]. The volume of the reaction vessel containing these gases is suddenly reduced to one-fourth of the initial volume . The rate of reaction would be

The rate of a first order reaction : R rarr P is 7.5xx10^(-4) molL^(-1) "s"^(-1) . The rate concentration of R is 0.5 mol L^(-1) . The rate constant is :

The rate of a first-order reaction is 1.5xx10^(-2)"mol"L^(-1) at 0.5 M concentration of the reactant. The half-life of the reaction is

The rate of a first order reaction is 1.8xx10^(-3) "mol L"^(-1) when the initial concentration is 0.3 mol L^(-1) . The rate constant in the units of second is :

The rate of a first order reaction is 1.5xx10^(-2) mol L^(-1) "min"^(-1) at 0.5 M concentration of the reactant . The half life of the reaction is :

The rate of a first order reaction is 0.04mol L^(-1) sec^(-1) at 10 minute and 0.03mol L^(-1) at 20 minute after initiation. Find the half life of the reaction.

The rate of a reaction quadruples when the tamperature changes from 300 to 310k. The activation energy of this reaction is : (Assume activation energy and pre-exponential factor are independent of temperature , In2=0.693, R=8.314Jmol^(-1)K^(-1))