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24551.

The rate of the process: CutoNi+""_(+1)^(0)e

Answer»

CHANGES with the CHANGE in PRESSURE
changes with the change in temperature
changes with the change in pressure and temperature
is INDEPENDENT of pressure and temperature

Answer :D
Discussion
24552.

The rate of the first order reaction X to Products is 7.5 xx 10^(-4) mol L^(-1) min^(-1) when the concentration of X is 0.5mol L^(-1). The rate constant is

Answer»

`3.75xx 10^(-4)s^(-1)`
`2.5 xx 10^(-5)s^(-1)`
`1.5 xx 10^(-3)s^(-1)`
`3.0 xx 10^(4)s^(-1)`

SOLUTION :RATE `=7.5 xx 10^(-4) mol L^(-1)mi n^(-1)`
`=(7.5 xx 10^(-4))/60 mol L^(-1)s^(-1)`
`= 1.25 xx 10^(-5) mol L^(-1) s^(-1)`
Rate = k [X].
HENCE, `k=(1.25xx 10^(-5))/0.5= 2.5 xx 10^(-5)s^(-1)`.
Discussion
24553.

Therateof thefirstreactionA toproducts is0.01 M//s, whenreactant conentrationconcentrationis 0.2M .Therateconstant for theractionwill be

Answer»

`0.05 s^(-1) `
`0.05 "MIN "^(-1) `
`0.1 S^(-1) `
`0.01 S^(-1) `

SOLUTION :RATE `=K[A],K=(0.01)/(0.2)=0.05S^(-1)`
Discussion
24554.

The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K. Calculate E_(a).

Answer»

Solution :Here we are GIVEN that
When `T_(1)=298K, k_(1)=k` (say)
When `T_(2)=308K, k_(2)=2k`
`"log"(k_(2))/(k_(1))=(E_(a))/(2.303R)((T_(2)-T_(1))/(T_(1)T_(2)))`
Substituting these values in the equation, we get
`"log"(2k)/(k)=(E_(a))/(2.303xx8.314J K^(-1)"mol"^(-1))xx(308K-298K)/(298K xx 308K)`
or `log 2=(E_(a))/(2.303xx8.314) xx (10)/(298xx308)`
or `0.3010=(E_(a)xx10)/(2.303xx8.314xx298xx308)`
or `E_(a)=(0.3010xx2.303xx8.314xx298xx308)/(10)`
or `E_(a)=52897.8J mol^(-1)=53.6"kJ mol"^(-1)`.
Discussion
24555.

The rate of the chemical reaction doubles for an increase of 10 K in absolute temperature from 298 K.Calculate E_(a). (R=8.314 J K^(1) mol^(-1))

Answer»

Solution :`log(k_(2))/(k_(1))=(E_(a))/(2.303 R)((1)/(T_(2))-(1)/(T_(2)))`
`(E_(a))/(2.303R)(T_(2)-T_(1))/(T_(1)T_(2))`
Where ,`T_(1)=298 K,T_(2)=(298+10)=308 K`
R=8.314 J`K^(-1) MOL^(-1)`
`(k_(2))/(k_(1))=2.0` (`because` Rate becomes double)
`therefore log 2.0=(E_(a)XX(308-298)K)/(2.303xx8.314 JK^(-1)mol^(-1)(308xx298))`
`therefore E_(a)=(0.3010xx8.314xx308xx298xx2.303)/(10)`
52898 J`mol^(-1)`
Discussion
24556.

The rate of the elementary reaction, 2NO+O_2rarr2NO_2 when the volume of the reaction vessel is doubled:

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Will GROW eight TIMES of its initial rate
REDUCE to one-eight of its initial rate
Will grow FOUR times of its initial rate
Reduce to one-fourth of its initial-rate

Answer :B
Discussion
24557.

The rate of the backward reaction in a reversible reaction

Answer»

positive
negative
Either (a) or (b)
NONE of the above

Solution :The rate of the bakwrad reactoin in a REVERSIBLE REACTION is positive. Rate of reaction can NEVER be negative.
Discussion
24558.

The rate of SN^(1)reaction is fastest with

Answer»




ANSWER :A
Discussion
24559.

The rate of SN^2 will be negligible in

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SOLUTION :bridgeheaded HALOGENS doe.t INVOLVED EITHER`SN^1 & SN^2` - RXns
Discussion
24560.

The rate of S_N^1 reaction is fastest in the hydrolysis of which of the following halides?

Answer»

`C_(6)H_(5)CH_(2)Br`
`CH_(3)Br`
`(CH_(3))_(2)CHBr`
`(CH_(3))_(2)CBR`

Solution :Because of the formation of most stable carboniumion `C_(6)H_(5)-overset(o+)(C)H_(2)`.
Discussion
24561.

The rate of second order reaction depends on ……….

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only concentration of SUBSTRATE
concentration of two REACTANTS
concentration of PRODUCT
concentration of neucleophile

Answer :B
Discussion
24562.

The rate of reaction X to Y becomes 8 times when the concentration of the reactant X is doubled. Write the rate law of the reaction.

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SOLUTION :`(DX)/(dt)=K[X]^(3)`.
Discussion
24563.

The rate of reactions exhibiting negative activation energy

Answer»

Decreases with INCREASING temperature
INCREASES with the increasing temperature
Does not DEPEND on temperature
DEPENDS on the height of the potential barrier

Answer :a
Discussion
24564.

The rate of reaction ............with the increase is the concentration of the reactants.

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SOLUTION :INCREASES
Discussion
24565.

The rate of reaction which does not involve gases, does not depend upon

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TEMPERATURE
concentration
PRESSURE
CATALYST

Answer :C
Discussion
24566.

The rate of reaction starting with initial concentration of 2xx10^(-3) M "and"10^(-3) M are equal to 2.4 xx10^(-4) "mol dm"^(-3) sec^(-1) "and" 0.6 xx 10^(-4) "mol dm"^(3) sec^(-1) respectively. Find rate constant of reaction in units [mol^(-1) dm^(3n-3) sec^(-1)] where n=order of reaction .

Answer»


ANSWER :60
Discussion
24567.

The rate of reaction that not involve gases is not dependent on…..

Answer»

Pressure
Temperature
Concentration
Cayalyst

Answer :A
Discussion
24568.

The rate of reaction is doubled when the temperature changes from 27^@C to 37^@C. Calculate the energy of activation.

Answer»

SOLUTION :We KNOW that` "log" K_2/K_1=E_a/(2.303R)[1/T_1-1/T_2]`Substituting the VALUES,
`"LOG2"=E_a/(2.303xx8.314)[1/300-1/310]`
`E_a=(0.3010xx2.303xx8.314xx300xx310)/10=535986.5=53.5kj=535986.5=53.5kj`
Discussion
24569.

The rate of reaction increases with temperature due to

Answer»

DECREASE in activation ENERGY
Increase in activation energy
Increase in collision frequency
Increase in CONCENTRATION

Solution :When heat energy is supplied , kinetic energy of reactant molecules increases. This will increase the number of COLLISIONS and ultimately rate of reaction will be ENHANCED.
Discussion
24570.

The rate of reaction increases with increase of temperature because a

Answer»

ACTIVATION energy BARRIER is LOWERED
the average energy of the products increases
THRESHOLD energy changes
activation energy is lowered

Solution :INCREASE of temperature lowers the activation energy as average energy of the reactants increases.
Discussion
24571.

The rate of reaction increases with rise in temperature because of

Answer»

INCREASE in number of activated molecules
increase in energy of activation
decrease in energy of activation
increase in the number of EFFECTIVE COLLISIONS

Solution :There are two effects of increasing temperature: greater COLLISION intensity i.e., no. of activated molecules and more FREQUENT collisions.
Discussion
24572.

The rateof reaction for N_(2)+3H_(2)to2NH_(3) may be represented as

Answer»

`R=-(d[N_(2)])/(dt)=-1/3(d[H_(2)])/(dt)=+1/2(d[NH_(3)])/(dt)`
`r=-(d[N_(2)])/(dt)=1/3(d[H_(2)])/(dt)=+1/2(d[NH_(3)])/(dt)`
`r=-(d[N_(2)])/(dt)=3(d[H_(2)])/(dt)=+1/2(d[NH_(3)])/(dt)`
`r=-(d[N_(2)])/(dt)=-1/3(d[H_(2)])/(dt)=+2(d[NH_(3)])/(dt)`

ANSWER :A
Discussion
24573.

The rate of reaction for certain reaction is expressed as :1/3 (d[A])/(dt) = -1/2(d[B])/(dt) = -(d[C])/(dt) The reaction is:

Answer»

`3A to 2B + C`
`2B to 3A + C`
`2B + C to 3A`
`3A + 2B to C`

ANSWER :A
Discussion
24574.

The rate ofreaction for certain reaction is expressed as : 1/3 (d[A])/(dt)=-(1)/(2)(d[B])/(dt)=-(d[C])/(dt) The reaction is

Answer»

`3A RARR 2B+C`
`2Brarr 3A+C`
`2B+Crarr3A`
`3A+2BrarrC`

ANSWER :A::B::C
Discussion
24575.

The rate of reaction depends on the concentration of the reacting species, i.e., reactants in rate equations or rate law, consider the gaseous reactions: aA bB to "Products" The rate laws is r=k[AP][Y]^(y) Here (x+y) is the order with respect to that reactant will be taken as zero. In the reaction A+B to "Products", if the concentration of B is kept fixed and the concentration of A is increased 3 times, the reaction rate increases 27 times. If the concentration of both A and B are doubled, then the increases 8 times. The order w.r.t. A is:

Answer»

3
2
1
zero

Solution :a) in the first experiment, Rate `propto [A]^(X)`
`[3]^(3)=27 propto [3]^(x)`
x=3
Discussion
24576.

The rate of reaction depends on the concentration of the reacting species, i.e., reactants in rate equations or rate law, consider the gaseous reactions: aA bB to "Products" The rate laws is r=k[AP][Y]^(y) Here (x+y) is the order with respect to that reactant will be taken as zero. 12 A to "Products". If concentration of A increases four times then its rate increases two times. The order w.r.t A will be:

Answer»

2
1
`1/2`
Can't predicted.

Solution :c) Rate `propto [A]^(x), 2 propto [4]^(x)`
`[4]^(1/2) propto [4]^(x)` or `x=1//2`.
Discussion
24577.

The rate of reactionCl_3"CC"HO+NOrarrCHCl_3+NO+COis given by equation,Rate=k[Cl_3"CC"HO][NO]If concentration is expressed in moles/litre, the units of k are

Answer»

`LITRE^(2)"mole"^(-2)SEC^(-1)`
mole `litre^(-1)sec^(-1)`
litre `"mole"^(-1)sec^(-1)`
`sec^(-1)`

SOLUTION :As `R=[Cl_3"CC"HO][NO]`
`thereforek=r//[Cl_3"CC"HO][NO]`
=`("mole "litre^(1)sec^(-1))/("mole"^(2)" "litre^(-2))=mol^(-1)litre^(1)sec^(-1)`
Discussion
24578.

The rate of reaction between two reactants A andB decreases by a factor of 4 if the concentration of reactant B is doubled . The order of this reaction with respect to reactant B is

Answer»

-1
-2
1
2

Solution :R = `K[B]^(n) , (1)/(4) R = k [2B]^(n) , 4 = ((1)/(2))^(n) , 4 = 2^(-n) , n = -2`.
Discussion
24579.

The rate of reaction:CH_3C(S) NH_2(aq) + H_2O to H_2S (aq) + CH_3C(O)NH_2(aq)is given by the rate law:Rate = k[H_3O^+] [CH_3C(S) NH_2] Consider 1 litre of solution that is 0.20 M in CH_3C(S)NH_2and 0.15 M in HCl at 25^@C.(a) For each of the following changes, state whether the rate of reaction increases, decreases or remains the same.(i) A 4 g sample of NaOH is added to the solution(ii) 500 mL of water is added to the solution(iii) The 0.15 M HCl solution is replaced by a 0.15 M acetic acid solution.(b) State whether the value of k will increase, decrease or remain the same.(i) A catalyst is added to the solution (ii) The reaction is carried out at 15^@C instead of 25^@C

Answer»

SOLUTION : (i), (ii) and (iii) decreases (B) (i) INCREASES (ii) decreases
Discussion
24580.

The rate of reaction can be expressed by Arrhenius equation R=Ae^(-K//RT) . In this equation. E represents

Answer»

the energy below which all the colliding molecules will react
the energy below which colliding molecules will not react
The TOTAL energy of the REACTING molecules at a TEMPERATURE T.
The fraction of the molecules with energy energy greater than the activation energy of the REACTANTS.

Solution :b) It is the correct answer.
Discussion
24581.

The rate of reaction between two reactants A and B decreases by a factor of 4 if the concentration of reactant B is doubled .The order of this reaction with respect to reactant B is…

Answer»

`-1`
`-2`
1
2

Answer :B
Discussion
24582.

The rate of reaction between two reactants A and B decreases by a factor of 4 if the concentration of reactant B is doubled. The order of this reaction with respect to reactant B is

Answer»

`-1`
`-2`
`1`
`2`

ANSWER :B
Discussion
24583.

The rate of reaction between A and B increases by a factor of 100 when the concentration of A is changed from 0.1 mol L^(-1) to 1 mol L^(-1) . The order of reaction with respect to A is :

Answer»

10
1
3
2

Solution :Concentration has been increased 10 TIMES `10^(2) = 100` . HENCE, order = 2 .
Discussion
24584.

The rate of reaction between two reactants A and B is expressed as rate = K [A] [B]^2. On doubling the concentration of both the reactants A and B, the reaction rate increases by

Answer»

3 times
4 times
6 times
8 times

Answer :C
Discussion
24585.

The rate of reaction between A and B increases by a factor of 100 , when the concentration of A is increased 10 folds . The order of reaction with respect to A is

Answer»

10
1
4
2

Solution :`r = K[A]^(n) , 100 r = k [10A]^(n)`
THUS `(1)/(100) = ((1)/(10))^(n) ` or n = 2.
Discussion
24586.

The rate of reaction becomes 2 times for every 10^@C rise in temperature. How the rate of reaction will increase when temperature is increased from 30^@C" to" 80^@C ?

Answer»

16
32
64
128

Solution :`(k_(t + 10))/k_t=(r_(t+ 10))/r_1=2`,
For an increase of temperature to `50^@C`, i.e., 5 times, the rate increases by `2^5` times, i.e., 32 times.
Discussion
24587.

The rate of reaction becomes 2 times for every 10^@C rise in temperature. How the rate of reaction will increases when temperature is increased from 30^@C to 80^@C

Answer»

16
32
64
128

Answer :B
Discussion
24588.

The rate of reaction becomes two times for every 10^(@)C rise in temperature. If the rate of reaction increases by 32 times when the temperature is increased from 30^(@)C to (10x)^(2)C. Then X=

Answer»

`16`
`32`
`64`
`28`

ANSWER :B
Discussion
24589.

The rate of reaction becomes 2 times for evergy 10^(@)C rise in temperature. How many times the rate of reaction will increase when temperature is increased from 30^(@)C to 80^(@)C

Answer»

16
32
64
28

Answer :B
Discussion
24590.

The rate of reaction for Ato products is 10 mol. "lit"^(-1)."min"^(-1) at time t_(1)=2 minutes. What will be the rate ("in mol. lit"^(-1)"min"^(-1)) at time t_(2)=12 minutes ?

Answer»

more than 10
10
less than 10
20

Answer :C
Discussion
24591.

The rate of reaction Ato products is 10mole/lit/min at time (t_(1))=2"min". What will be the rate in mole/lit/min at time (t_(2))-12 min

Answer»

more than `10`
`10`
LESS than `10`
`20`

ANSWER :C
Discussion
24592.

The rate of reaction, A+B rarr product, is proportional to the first power of concentration of A and second power of concentration B. The overall order of the reaction is :

Answer»

1
2
3
Zero

Answer :C
Discussion
24593.

The rate of reaction A + B + C to Products is given Rate = K[A]^(1/2)[B]^(1/3)[C]. The order of the reaction is

Answer»

1
3
`5/6`
`11/6`

SOLUTION :`1/2+1/3+1=11/6`
Discussion
24594.

The rate of reaction, A + B + C rarr P is given by: r=-(d[A])/dt=K[A]^(1//2)[B]^(1//2)[C]^(1//4) . The order of the reaction is:

Answer»

1
2
`1//2`
`5//4`

ANSWER :D
Discussion
24595.

The rate of reaction A+2Bto Products is given by -(d[A])/(dt)=k[A][B]^(2).If B is present in large excess, the order of reaction is

Answer»

`3`
`2`
`1`
zero

Answer :C
Discussion
24596.

The rate of reaction, A + 2B rarr products is given by the following equation :-(d[A])/(dt)=k[A][B]^(2)If B is present in large excess, the order of the reaction is

Answer»

zero
first
SECOND
third.

Solution : From the EXPRESSION
`-(d[A])/(dt)=k[A][B]^(2)`
when `beta` is present in large excess, rate will be independent upon the change in CONC. of B therefore order of REACTION w.r.t. A will be one.
Discussion
24597.

The rate of reaction:

Answer»

DECREASES with TIME
Decreases with DECREASE in CONCENTRATION of reactant
Decreases, with increase in time and decrease in concentration of reactant
None

Answer :C
Discussion
24598.

The rate of reaction 2NO+Cl_(2)to 2NOCl is doubled when concentration of Cl_(2) is doubled and it becomes eight times when concentration of both NO and Cl_(2) are doubled. Deduce the order of the reaction.

Answer»

Solution :The ORDER wrt `Cl_(2)` is one because the rate is doubled when the concentration is doubled. However the order wrt NO is 2 because the rate becomes 8 times when the concentrationof both NO and `Cl_(2)` are doubled. The rate becomes 4 times when the concentration of NO is doubled.
Order of the REACTION `=2+1=3`.
Discussion
24599.

The rate of reaction 2NO + Cl_(2) to 2NOCl is doubled when the concentration of Cl_(2) is doubled and becomes eight times when the concentration of both NO and Cl_(2) are doubled. Predict the order of reaction.

Answer»


Solution :Rate ( r) = `K[NO]^(x)[Cl_(2)]^(y)`…………..(i)
Rate (2r)`= k[NO]^(x)[2Cl_(2)]^(y)`………..(II)
Rate (8r) = `k[2NO]^(x)[2Cl_(2)]^(y)`…………(iii)
`(8r)/(2r) = (k[2NO]^(x)[2Cl_(2)]^(y))/(k[NO]^(x)[2Cl_(2)]^(y))`
`4= [2]^(x) or x=2`
`(2r)/(r)=(k[NO]^(x)[2Cl_(2)]^(y))/(k[NO]^(x)[Cl_(2)]^(y))`
`2= [2]^(y)` or y=1
Order of REACTION= 2+1=3.
Discussion
24600.

The rate of reaction : 2NO+Cl_(2) to 2NOCl is given by the rate equation "rate"=k[NO]^(2)[Cl_(2)]. The value of the rate constant can be increased by

Answer»

INCREASING the temperature
increasing the CONCENTRATION of NO
increasing the concentration of the `Cl_(2)`
doing all of these

Answer :A
Discussion