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24901.

The rate constant for the decomposition of N_2O_5 in CCl_4 is 6.2 x 10^(-4) s^(-1) at 45°C. Calculate the rate constant at 100°C if the activation energy is 103 kJ mol^(-1) [Ant (2.49) = 309]

Answer»

`1.9 X 10^(-1) s^(-1)`
`3.5 x 10^(-2) s^(-1)`
`4.5 x 10^(-1) s^(-1)`
`1.75 x 10^(-2) s^(-1)`

ANSWER :A
Discussion
24902.

The rate constant for the first order decomposition of H_(2)O_(2) is given by the following equation : log K = 14.2 - (1.0 xx 10^(4))/(T) KCalculate E_a for this reaction and rate constant kif its half life period be 200 minutes. (Given : R = 8.314 JK^(-1) mol^(-1))

Answer»

Solution :log K = 14.2 - `(10^(4)K)/(T)`
Arrhenius equation : `k =Ae^(-Ea//RT)`
log k = log A - `(Ea)/(2.303 RT)`
Compare equation (i) and (II)
= `(cancel(-)Ea)/(2.303 RT) = (cancel(-)10^4)/(cancel(T))`
`Ea = 10^(4) xx 2.303 xx R`
`= 10^(4) xx 2.303 xx 8.314`
Ea = 191.47KJ/mol
`K = (0.693)/(t_(1//2))` [For 1ST order reaction]
`K = (0.693)/(200 xx 60) sec^(-1)`
`K = 5.78 xx 10^(-5)sec-1`
Discussion
24903.

The rate constant for the decomposition of N_(2)O_(5) at various temperatures is given below: Draw a graph beetween In k and 1//T and calculate the values of A and E_(a). Predict the rate constant at 30^(@) and 50^(@)C

Answer»

Solution :To draw the plot of log K versus `1//T`, we can re-write the given DATA as follows:

Draw the graph as shown on the next page, from the graph, we find that
Slope = `-2.4/(0.00074) = -E_(a)/(2.303 R)`
`therefore` Activation energy, `(E_(a)) = (2.4 xx 2.303 xx R)/(0.00047) = (2.4 xx 2.303 xx 8.314 J mol^(-1))/(0.00047)`
`=17, 689 J mol^(-1)= 17.689 kJ mol^(-1)`
As we know log k = log a -`(E_(a))/(2.303 RT)`
(Compare it with y = mx+c which is equation of line in intercept form)
`log k = (-E_(a))/(2.303RT)(1/T) + log A`

log A = VALUE of intercept on y-axis i.e., on the k-axis.
`=(-1+7.2) = 6.2``[y_(2)-y_(1) = -1-(-7.2)]`
Frequency factor, A= ANTILOG `6.2 = 1585000 = 1.585 xx 10^(6)` collisions `s^(-1)`
The value of rate constant k can be found from the graph as follows:
Discussion
24904.

The rate constant for the first order decompoistion of a certain reaction is described by the equation log k (s^(-1)) = 14.34 - (1.25 xx 10^(4)K)/(T) (a) What is the energy of activation for the reaction? (b) At what temperature will its half-life periof be 256 min?

Answer»

SOLUTION :`239.34 KJ "MOLE"^(-1) , 669K`
Discussion
24905.

The rate constant for the first-order decomposition at 45^@C of N_2O_5,dissolved in chloroform, is 6.2 xx 10^(-4) "min"^(-1). 2N_2O_5to4NO_2 +O_2(a) What is the rate of decomposition when [N_2O_5] = 0.40 M?(b) What are the rates of formation of NO_2 and of O_2 when [N_2O_5] = 0.40 M?

Answer»

Solution :`(a) 2.5 xx 10^(-4) (B) 5 xx 10^(-4) , 1.2 X 10^(-4)`
Discussion
24906.

The rate constant for the decomposition of N_(2)O_(5) at various temperature is given below : {:(T//^(@)C,0,20,40,60,80),(10^(5)xxk//s^(-1),0.0787,1.70,25.7,178,2140):} Draw a graph between In k & 1//T , calculate the values of A and E_(a) . Predict the rate constant at 30^(@)& 50^(@)C.

Answer»

Solution :From the GIVEN data, we have

`slope =(y_(2)-y_(1))/(x_(2)-x_(1))=(-14.055-(-3.84))/((3.66-2.83)xx10^(-3))=10.215xx10^(3)`
From Arrhenius equation , `-(E_(a))/(R)` = slope
or , `E_(a)=(-"slope"xxR)`
`=-(-10.215xx10^(3)xx8.314)-102.322"kJ.mol"^(-1)`
Also, In k `In A-(E_(a))/(RT) " or, ""InA"=Ink+(E_(a))/(RT)`
or, `A=ke^(-E_(a)//RT)`
At `40^(@)C, A = 25.7xx10^(-5)exp[(102322)/(8.314xx313)]=3.06xx10^(13)s^(-1)`
At `30^(@)C, T =303K therefore(1)/(T)=3.3xx10^(-3)K^(-1)`
From the plot , Ink`=-9.62 therefore k = 6.64xx10^(-5)s^(-1)`
At `50^(@)C, T =323K therefore(1)/(T)=3.1xx10^(-3)K^(-1)`
From the plot , Ink = -7.15`therefore k = 7.8xx10^(-4)s^(-1)`
Discussion
24907.

The rate constant for the decomposition of hydrocarbons is 2.418xx10^(-5)s^(-1) at 546 K.If the energy of activaton is 179.9 KJ/mol, what will be the value of pre-exponetial factor.

Answer»

Solution :According to Arrhenius EQUATION.
log k=log A-`(E_(a))/(2.303 RT)`
log A=log `K+(E_(a))/(2.303RT)`
Where ,T=546 k
k=`2.418xx10^(-5)s^(-1)`
log k=-log `2.418xx10^(-5)=-4.6165`
`E_(a)=179.9 KJ=179900 J MOL^(-1)`
A=(?)
`therefore log A=-4.6165+(179900)/(2.303xx8.314xx546)` `therefore` A=antilog 12.5917
`=3.9057xx10^(12)=3.91xx10^(12)s^(-1)`
Discussion
24908.

The rate constant for the decomposition of a certain substance is 2.80xx10^(-3)M^(-1)s^(-1) at 30^(@)C and 1.38xx10^(-1)m^(-1)s^(-1) at 50^(@)C. Evaluate the Arrheniusparameters of the reaction. (R=8.314xx10^(-3)kJ mol^(-1)K^(-1))

Answer»

Solution :Energy of activation `(E_(a))` and pre-exponential FACTOR A are Arrhenius PARAMETERS.
`k_(1)=2.80xx10^(-3)m^(-1)s` at `303k`
`k_(2)=1.38xx10^(-2)m^(-1)s^(-1)` at `323k`
As `log_(10).(k_(2))/(k_(1))=(E_(a))/(R )[(T_(2)-T_(1))/(T_(1)xxT_(2))]`
`:.E_(a)=(2.303RT_(1)T_(2))/(T(T_(2)-T_(1)))log_(10).(k_(2))/(k_(1))`
`=(2.303xx8.314xx10^(-3)xx303xx323)/((323-303)log_(10))((1.38xx10^(-2))/(2.80xx10^(-3)))=64.91 kJ mol^(-1)`
Also `k=Ae^(-Ea//RT)`
or `A=ke^(Ea//ET)=2.80xx10^(-3)e^(64.91//8.314)xx10^(-3)xx303=4.34xx10^(8)s^(-1)`
Discussion
24909.

The rate constant for the decomposition of hydrocarbons is 2.418xx10^(-5)s^(-1) at 546 K. If the energy of activation is 179.9 kJ/mol, what will be the value of pre-exponential factor?

Answer»

SOLUTION :The FOLLOWING data is provided:
`K=2.418xx10^(-5)s^(-1), E_(a)=179.9kJ" mol"^(-1), T=546K`
Applying Arrhenius equation,
`k=Ae^(-E_(a)//RT)" or In k"="In A"-(E_(a))/(RT)`
or `log k= LOGA -(E_(a))/(2.303RT) "or log A"=log k+(E_(a))/(2.303RT)`
Substituting the values, we get
`logA=log (2.418xx10^(-5)s^(-1))+(179.9"kJ mol"^(-1))/(2.303xx8.314xx10^(-3)"kJ K"^(-1)"mol"^(-1)xx546K)`
`=(-5+0.3834) s^(-1)+17.2081=12.5924s^(-1)`
or `A="Antilog"(12.5924)s^(-1)=3.912xx10^(12)s^(-1)`.
Discussion
24910.

The rate constant for the decomposition of a hydrocarbon is 2.418xx10^(-5)s^(-1) at 546 K. If the energy of activation is 179.9" kJ/mol", what will be the value of pre-exponential factor ?

Answer»

Solution :Here, `k=2.418xx10^(-5)s^(-1),E_(a)=179.9" kJ mol"^(-1),T=546" K".`
ACCORDING to Arrhenius equation,
`k="A E"^(-E_(a)//RT)" or ln "k="ln "A-(E_(a))/(RT)" or "logk=logA-(E_(a))/(2.303" RT")`
or `logA=logk+(E_(a))/(2.303" RT")=log(2.418xx10^(-5)s^(-1))+(179.9" kJ mol"^(-1))/(2.303xx8.314xx10^(-3)" kJ K"^(-1)" mol"^(-1)xx546" K")`
`=(-5+0.3834)s^(-1)+17.2081=12.5924" s"^(-1)`
or `A=" ANTILOG "(12.5924)s^(-1)=3.912xx10^(12)s^(-1).`
Discussion
24911.

The rate constant for reaction is 10.8 xx 10^(-5) mol dm^(-3) sec^(-1). The reaction obeys

Answer»

first ORDER
zero order
second order
HALF order

SOLUTION :The unit of RATE constant is to be compared with `n^(th)` order unit , i.e., `(mol dm^(3))^(1-n) sec^(-1) n = 0 therefore` Unit of k = `mol dm^(-3) sec^(-1)`
Discussion
24912.

The rate constant for an isomerisation reaction A to Bis 4.5 xx 10^(-3) "min"^(-1) . If theinitial concentration of A is 1 M, calculate the rate after 1 h.

Answer»

SOLUTION :`3.43 XX 10^(-3) `M / MIN
Discussion
24913.

The rate constant for forward and backward reactions of hydrolysis of ester are 1.1xx10^(-2) and 1.5xx10^(-3) per minute respectively. Equilibrium constant for the reaction is CH_(3)OOC_(2)H_(5)+H_(2)OhArrCH_(3)COOH+C_(2)H_(5)OH

Answer»

`4.33`
`5.33`
`6.33`
`7.33`

Solution :`K_(f)=1.1xx10^(-2),K_(B)=1.5xx10^(-3)`
`K_(C)=(K_(f))/(K_(b))=(1.1xx10^(-2))/(1.5xx10^(-3))=7.33.`
Discussion
24914.

The rate constant for an isomerisation reaction, AtoB is 4.5xx10^(-3)" min"^(-1). If the initial concentration of A is 1 M, calculate the rate of reaction after 1 h.

Answer»

Solution :Rate CONSTANT in `min^(-1)` shows that it is a REACTION of 1st order. Hence, `k=(2.303)/(t)log""([A]_(0))/([A])`
or `4.5xx10^(-5)min^(-1)=(2.303)/(60" min")log""(1)/([A])" or "log[A]=-0.1172=1.8828" or "[A]=" Antilog "1.8828`
`:.[A]=0.7635" mol L"^(-1).` This is the concentration after 1 h.
Rate after `1 h=k[A]=4.5xx10^(-3)xx0.7635" mol L"^(-1)min^(-1)=3.44xx10^(-3)" mol L"^(-1)mil^(-14).`
Discussion
24915.

The rate constant for first order reaction is 1.54 xx10^(-3) s^(-1). Calculate its half life time .

Answer»

SOLUTION :We KNOWN that `,t_(1//2) = 0.693//k`
`t_(1//2) = 0.693//1.54xx10^(-3)s^(-1)=450S`
Discussion
24916.

The rate constant for a zero order reaction is 2xx10^(-2) mol L^(-1)sec^(-1). If the concentration of the reactant after 25 sec is 0.5M, the initial concentration must have been

Answer»

`0.5M`
`1.25M`
`12.5M`
`1.0M`

ANSWER :A
Discussion
24917.

The rate constant for a second order reaction is8 xx10^(-5) M^-1 min^-1. How long will it take a IM solution to be reduced to 0.5 M:

Answer»

`8.665 X 10^3` minute
`8 x IO^(-5)` minute
`1.25 x 10^4` minute
`4 x 10^(-5)`minute

Answer :C
Discussion
24918.

The rate constant for a reaction of zero order in A is 0.0030 "mol L"^(-1) s^(-1). How long will it take for the initial concentration of A to fall from 0.10 M to 0.075 M ?

Answer»

SOLUTION :`[R]t = -K t + [R]_(0)`
`0.075 M = -(0.0030 mol L^(-1) s^(-1))t + 0.10 M`
`-0.025 M = -(0.003 mol L^(-1) s^(-1))t`
`= 8.33 s`
Discussion
24919.

The rate constant for a reaction is 2.05xx10^(-5) mole "lit"^(-1)."sec"^(-1). The rection obeys ……order

Answer»

FIRST
SECOND
Zero
Half

Answer :C
Discussion
24920.

The rate constant for a reaction is 10.8 xx 10^-5 mole litre^-1sec^-1. The reaction obeys:

Answer»

FIRST order
Zero order
Second order
Half order

Answer :B
Discussion
24921.

The rate constantfor areactionat 500^(@) C " is" 1.6 xx 10^(3) M^(-1)s^(-1). Whatis thefrequencyfactorof thereactionif itsenergyof activationis 56 kJ mol^(-1)

Answer»


SOLUTION :Given : `k = 1.6xx 10^(3)M ^(-1) s^(-1), T = 273+ 500= 773K`
Energyof activation= `E_(a)= 56 kJ mol^(-1) =56000 J mol^(-1)`
Frequencyfactor = A= ?
ByArrheniusequation
`k= AXX e^(-E_(a) //RT) :. " In" k= " In" A-(E_(a))/(RT)`
`2.303 log_(10)k = 2.303log_(10)A - (E_(a))/(RT):. log_(10) k =log_(10)A - (E_(a))/(2.303 RT)`
`:. log_(10) A= log_(10) k + (E_(a))/(2.303 RT)`
`= log_(10)1.6 xx 10^(3) + (56000)/(2.303 xx 8.314 xx 773) = 3.204 + 3.784= 6.988`
`:. A = AL 6.988= 9.727 xx 10^(6) s^(-1)`
Discussion
24922.

The rate constant for a gaseous reaction, X(g) rightarrow2Y(g) is 0 2 mol L^(-1) s^(-1). If Initial concentration of the reactant, X is 0.8 mol L^(-1)which of the following graph is correct?

Answer»




ANSWER :D
Discussion
24923.

The rate constant for a reaction is 10.8 xx 10^(-5) mole "litre"^(-1) "sec"^(-1) . The reaction obeys

Answer»

HALF ORDER
zero order
SECOND order
First order

Answer :B
Discussion
24924.

Therateconstantfor areactionat 500^(@)C is 1.6 xx 10^(3)M ^(-1)s^(-1)Whatis thefrequencyfactorof thereactionif itsenergyof activationis 56 kJ "mol"^(-1)

Answer»

SOLUTION :FREQUENCYFACTOR =A=9.727 `xx 10^(6) S^(-1)`
Discussion
24925.

The rate constant for a fist order reaction is k = 7xx10^(-4)s^(-1) .The time taken for the reactant to be reduced to 1/2 of the initial concentration is :

Answer»

990 s
1980s
445 s
2970 s

SOLUTION :(B) `t=(2.303)/(k)"log"(1a)/(1//4a)`
`=(2.303)/(7XX10^(-4))"log"4 `
`t=(2.303)/(7xx10^(-4))xx0.0602`
= 1980 s
Discussion
24926.

The rate constant for a first order reaction is 60s^(-1) . How much time will it take to reduce the initial concentration of the reactant to its 1//16^(th) value?

Answer»

SOLUTION :`t = (2.303)/klog.x/(x//16)=(2.303)/(60s^(-1))LOG 16 = 4.62xx10^(-2)s`
Discussion
24927.

The rate constant for a first order reaction is 6.93 xx 10^(-2) mi n^(-1). How long will it take a1M solution to be reduced to 0.5 M?

Answer»

600 MIN
6 hr
600 sec
10 sec

Solution :For 1 order, `t= 2.303/k log.(a)/(a-x)`
`= 2.303/(6.93 xx 10^(-2)mi N^(-1))xx log.(1)/0.5`
`= (2.303 xx 0.6990)/(6.93 xx 10^(-2)min`
= 10 min = 600 sec.
Discussion
24928.

The rate constant for a first order reaction is 60s^(-1). How much time will it take to reduce the initial concentration of the reactant to its 1//16^(th) value.

Answer»

Solution :Data : `K = 60S^(-1), [R]_(0) = 1, [R] = (1)/(16) , t = ? `
formlula, `k = (2.303)/(t) log ""([R]_(0))([R])`
Substitution : `t = (2.303)/(60) log ""((1)/(1))/(16) rArr t = (2.303)/(60) log 16`
`t = (2.303)/(60) xx 1.204 rArr t = 4.6 xx 10^(-2)` seconds.
Discussion
24929.

The rate constant for a first order reaction is 60s^(-1). How much time will it take to reduce the concentration of the reactant to 1//10th of its initial value ?

Answer»


SOLUTION :`t=(2.303)/(K)LOG""(a)/(a-x)=(2.303)/(60S^(-1))log""(a)/(a//10)=(2.303)/(60s^(-1))log10=0.0384s`
Discussion
24930.

The rate constant for a first order reaction is 60 s^(-1) .How much time will it take to reduce the initial concentration of the reactant to its (1)/(16)^(th) value?

Answer»

SOLUTION :Order of REACTANT =1
rate constant of reaction k=`60s^(-1)`
Suppose ,initial concentration =`[R]_(0) MOL L^(-1)`
Concentration after t time `=(1)/(6)` initial concentration
`therefore[R]_(t)=(1)/(6) mol L^(-1)`
For first order reaction ,
`t=(2.303)/(k)` log `([R]_(0)6)/([R]_(0))`
`=(2.303)/(60 s^(-1))xxlog6(2.303)/(60 s^(-1))xx0.7782`
`=2.9868xx10^(-2)s`
Note:If the concentration is `(1)/(16)` after t time,
`t=(2.303)/(k)` log `([R]_(0)xx16)/([R]_(0))`
`=(2.303)/(k)` log 16 =`(2.303)/(60 s^(-1))xx1.2041`
Note ::Here ,the CALCULATION of `(1)/(16)` is according to english textbook .
Discussion
24931.

The rate constant for a first order reaction is 60 s^(-1). How much time will it take to reduce the initial concentration of the reactant to its 1/16th value ?

Answer»

Solution :`t=(2.303)/(K)"log"(a)/(a//16)=(2.303)/(60S^(-1))log 16=4.62xx10^(-2)s`.
Discussion
24932.

The rate constant for a first order reaction is "60 s"^(-1). How much time will it take to reduce the initial concentration of the reactant to its (1)/(16) th value ?

Answer»

SOLUTION :If we carry out HALF - change FOUR times, the CONCENTRATION will reduce to `(1)/(16)`.
`t_(1//2)=(0.693)/(K)=(0.693)/("60 s"^(-1))=0.01155s`
Total time taken `=4xxt_(1//2)=4xx0.01155s=0.0462s`
Discussion
24933.

The rate constant for a first order reaction is 60 s^(-1) . The time required to reduce the concentration of the reactant to 1/ 10th of its initial value is

Answer»

380 s
`7.6xx10^(-2)` s
`3.8xx10^(-2) ` s
760

Solution :(C ) `t_(1//2)` for a FIRST order reaction is
`t_(t//2) = (0.693)/(k) "or " = (IN2)/(k)`
Discussion
24934.

The rate cnstant k, for thereaction N_(2)O_(5)(g) rarr 2NO_(2)(g) + (1)/(2)O_(2)(g) is 23 xx 10^(-2) s^(-1). Which equation given below decribes the change of [N_(2)O_(5)] with time? [N_(2)O_(5)]_(0) and [N_(2)O_(5)]_(t) correspond to concentration of N_(2)O_(5) initially and at time, t

Answer»

`[N_(2)O_(5)] = [N_(2)O_(5)]_(0) + kt`
`[N_(2)O_(5)] = [N_(2)O_(5)]_(t)E^(kt)`
`log_(10) [N_(2)O_(5)]_(t) =log_(10) [N_(2)O_(5)]_(0) - kt`
`"ln"([N_(2)O_(5)]_(0))/([N_(2)O_(5)]t) = kt`

SOLUTION :Rate constant `= 2.3 xx 10^(-2) SEC^(-1)`
It means it is a first order reaction (because unit of rate constant is `sec^(-1)`)
For first order reaction`K = (1)/(t)ln(a)/(a-x)`
`kt = ln(a)/(a-x) = ln[N_(2)O_(5)]_(0)/([N_(2)O_(5)]_(t)`
Discussion
24935.

The rate constant for a first order reaction is 60 s^(-1). How much time will it take to reduce 1 g of the reactant to 0.0625 g?

Answer»

Solution :The EQUATION for a first order reaction is :
`t=(2.303)/(k)"LOG"([R]_(0))/([R])`
`[R]_(0)=1 g, [R]=0.0625 g, k=60s^(-1)`
Substituting the values in the first order equation, we have
`t=(2.303)/(60) "log"(1)/(0.0625)=(2.303)/(60)log 16=(2.303)/(60)xx1.204=0.0462s`
Discussion
24936.

The rate constant for a first order reaction is 100 S^(-1). The time required for completion of 50% of reaction is-

Answer»

0.0693 MILLISECONDS
0.693 milliseconds
6.93 milliseconds
69.3 milliseconds

ANSWER :C
Discussion
24937.

The rate constant for a first order reaction is 2.3 xx10^(-4)s^(-1). If the initial concentration of the reactant is 0.01M. What concentration will remain after 1 hour?

Answer»

Solution :Rate constant of a first order REACTION `K = 2.3 xx10^(-4)s^(-1)`
Initial concentration of the reactant `[A_0]=0.01M`
Concentration will REMAIN after 1 hour [A] = ?
`k=(2.303)/TLOG.([A_0])/([A])`
`2.3xx10^(-4)=(2.303)/(1"hour")log.([0.01])/([A])`
`(2.3xx10^(-4)xx1)/(2.303)=log[0.01]0log[A]`
`9.986xx10^(-5)=-2-log[A]`
`11.986xx10^(-5)=-log[A]`
`[A]="Antilog"(-11.986xx10^(-5))`
`[A]=0.9997M`
Discussion
24938.

The rate constant for a first order reaction is 2.3xx10^(-4)s^(-1). If the initial concentration of the reactant is 0.01m. What concentration will remain after 1 hour ?

Answer»

Solution :`k=2.3xx10^(-4)s^(-1)`
T = 60 min = 3600 s
`[A]_(0)=0.01 M`
`k=(2.303)/(t)"log"([A]_(0))/([A])`
`rArr 2.3xx10^(-4)s^(-1)=(2.303)/(3600s)"log"(0.01)/([A])`
`rArr 2.3xx10^(-4)s^(-1)=(2.303)/(3600s)(-log[A])`
`rArr - log[A]=(2.3xx10^(-4)xx3600)/(2.303)`
[A] = ANTILOG `[-(2.3xx10^(-4)xx3600)/(2.303)]`
`= 0.142 "MOL L"^(-1)` (APPROXIMATELY)
Discussion
24939.

The rate at which a substance reacts depends on its…..

Answer»

ATOMIC WEIGHT
equivalent weight
molecular weight
ACTIVE mass

Answer :D
Discussion
24940.

The rate at which a substrance reacts is proportional to its active mass. This statement is :

Answer»

Le-Chatelier's principle
Faraday's LAW
Law of MULTIPLE PROPORTION
Law of MASS acction

Answer :D
Discussion
24941.

The rate at which a substance reacts, depends on its:

Answer»

ATOMIC weight
EQUIVALENT weight
Molecular weight
Active MASS

Solution :ACCORDING to law of mass ACTION .
Discussion
24942.

The rate at which a substance reacts depends upon its

Answer»

ATOMIC masss
Equivalet MASS
Molecular mass
ACTIVE mass

Solution :RATE `PROP` Active mass
Discussion
24943.

The rate and mechanism of chemical reactions are studied in chemical kinetics. The elementary reactions are single step reactions having no mechanism. The order of reaction and molecularity are same for elementary reactions. The rate of forward reaction aA + bB rarr cC + dD is given as : rate = =((dx)/(dt))=-(1)/(a) (d[A])/(dt) =-(1)/(b) (d[B])/(dt) =(1)/(c)(d[C])/(dt)=(1)/(d) (d[D])/(dt) or rate =K [A]^(a) [B]^(b). In case of reversible reactions net rate expression can be written as : rate = K_(1)[A]^(a)[B]^(b) – K_(2)[C]^(c)[D]^(d). At equilibrium, rate= 0. The constants K, K_(1), K_(2), are rate constants of respective reaction. In case of reactions governed by two or more steps reactions mechanism, the rate is given by the solwest step of mechanism. For a reaction, 2ND_(3) rarr N_(2) +3D_(2) : (-d[ND_(3)])/(dt) =K_(1)[ND_(3)], (d[N_(2)])/(dt) =K_(2) [ND_(3)], (d[D_(2)])/(dt) =K_(3) [ND_(3)], then :

Answer»

`K_(1)=K_(2)=K_(3)`
`3K_(1)=6K_(2)=2K_(3)`
`K_(1)=2K_(2)=K_(3)`
`K_(1)=K_(2)=2K_(3)`

Answer :B
Discussion
24944.

The rate and mechanism of chemical reactions are studied in chemical kinetics. The elementary reactions are single step reactions having no mechanism. The order of reaction and molecularity are same for elementary reactions. The rate of forward reaction aA + bB rarr cC + dD is given as : rate = =((dx)/(dt))=-(1)/(a) (d[A])/(dt) =-(1)/(b) (d[B])/(dt) =(1)/(c)(d[C])/(dt)=(1)/(d) (d[D])/(dt) or rate =K [A]^(a) [B]^(b). In case of reversible reactions net rate expression can be written as : rate = K_(1)[A]^(a)[B]^(b) – K_(2)[C]^(c)[D]^(d). At equilibrium, rate= 0. The constants K, K_(1), K_(2), are rate constants of respective reaction. In case of reactions governed by two or more steps reactions mechanism, the rate is given by the solwest step of mechanism. For the reaction : aA rarr bB, log [(-dA)/(dt) ]=log [(dB)/(dt)]+0.6," then "a:b is :

Answer»

`3.98`
`2.18`
`1.48`
0

Answer :A
Discussion
24945.

The rate at which a substance reacts depends on its:

Answer»

Atomic WEIGHT
Equivalent weight
MOLECULAR weight
ACTIVE mass

ANSWER :D
Discussion
24946.

The rate at which a substance reacts depends on its

Answer»

atomic weight
atomic number
molecular weight
active mass.

Solution :According to law of mass ACTION, "at a given temperature, the rate of a reaction at a particular instant is proportional to the PRODUCT of the active masses of the reactants at that instant raised to powers which are numerically EQUAL to the numbers of their respective molecules in the stoichiometric EQUATION DESCRIBING the reaction"
Discussion
24947.

The rate and mechanism of chemical reactions are studied in chemical kinetics. The elementary reactions are single step reactions having no mechanism. The order of reaction and molecularity are same for elementary reactions. The rate of forward reaction aA + bB rarr cC + dD is given as : rate = =((dx)/(dt))=-(1)/(a) (d[A])/(dt) =-(1)/(b) (d[B])/(dt) =(1)/(c)(d[C])/(dt)=(1)/(d) (d[D])/(dt) or rate =K [A]^(a) [B]^(b). In case of reversible reactions net rate expression can be written as : rate = K_(1)[A]^(a)[B]^(b) – K_(2)[C]^(c)[D]^(d). At equilibrium, rate= 0. The constants K, K_(1), K_(2), are rate constants of respective reaction. In case of reactions governed by two or more steps reactions mechanism, the rate is given by the solwest step of mechanism. The rate of formation of SO_(3) in the following reaction : 2SO_(2) +O_(2) rarr 2SO_(3)" is "10 g sec^(-1). The rate of disapperance of O_(2) will be :

Answer»

`5" g SEC"^(-1)`
`100" g sec"^(-1)`
`20" g sec"`
`2" g sec"^(-1)`

ANSWER :D
Discussion
24948.

The rat eof S_(N)2 reaction is maximum when the solvent is

Answer»

`CH_(3)OH`
`H_(2)O`
`DMSO`
benzene

Solution :POLAR aprotic SOLVENTS such as DMSO increase the rate of `S_(N` 2 reactions.
Discussion
24949.

The rare gases are

Answer»

Monoatomic
Diatomic
Triatomic
Polyatomic

Answer :A
Discussion
24950.

The rapiddepletion of fossil fuels has inspiredextensive research in the area of alternative and renewale energy sources. Of these, hydroden is the most Contemplated fuel of the future . Howevercost effective production and hazard free storage are major issuses is using H_(2) (Note : use the data in table-1 given at the end of partA,whenever necessary.) If the miximum theoretical work (calculate in 4.3 (i))0 is used to run an electric motor of 1 watt, under standard potential conditons. (i) for how many months the motor will be run? (ii) What is the value of the current produced by this motor? (Assume 30 days in all months.) Table-1: {:(,H_(2)(g),O_(2)(g),H_(2)O(I)),(S_(298)^(@) K//"JMolK"^(-1),131,205,70),(,H_(2)O(l),CO_(2)(g),),(DeltaH_(f)^(@)//"KMol"^(-1), -266 KJ "mol"^(-1),-394 KJ "mol"^(-1),):}

Answer»


SOLUTION :N//A
Discussion