Explore topic-wise InterviewSolutions in Current Affairs.

This section includes 7 InterviewSolutions, each offering curated multiple-choice questions to sharpen your Current Affairs knowledge and support exam preparation. Choose a topic below to get started.

1.

ForthereactionbetweenA andBA+Btoproductthefollowingratedatawereobtained . the reactionis

Answer»

Zeroorder
firstorder
secondorder
thirdorder

SOLUTION :Rate `=K[A]^(x)[B]^(y)`
forexpt, (1)`1.2xx10^(-2)=k[0.1]^(x)[0.2]^(y) `
`(2)3.6xx10^(-2)=k [ 0.3 ]^(x)[0.2]^(y)`
`(3) 4.8xx10^(-2)=k[0.3]^(x)[0.2]^(y)`
DividingII by`I,3=[3]^(x)thereforex=1`
DividingIII by ,I ,4 `=[2]^(y) thereforey=2`
orderofreaction`x+y=1+2=3`
2.

For the reaction below, The structure of the product 'Q' is:

Answer»




ANSWER :B
3.

For the reaction below. the structure of the product Q is

Answer»




SOLUTION :
4.

For the reaction Aunderset(k_(2)"sec"^(-1))overset(K_(1)"sec"^(-1))hArr B following graph is given, k_(1) =4 xx 10^(-@)"sec"^(-1). Which is/are correct statement (s) (In 2=0.7 , In 8/7 =0.14)

Answer»

EQUILIBRIUM constant is 4.0
Time taken for the completion fo 50% fo equilibrium conc. Of B is 14 sec.
Time taken for the completion of 10% of INITIAL conc. Of A is 2.8 sec.
Rate constant of BACKWARD reaction is `10^(-2)"sec"^(-1)`

Answer :A::B::C::D
5.

Forthe reactionbelow , (##MOD_SPJ_CHE_XII_P2_C12_E07_083_Q01.png" width="80%"> The structureof theporduct Q is

Answer»




SOLUTION :
6.

For the reaction AtoB,DeltaH=+23kJ//mol and BtoC,DeltaH=-18kJ//mol, the decreasing order of enthalpy of A,B,C follows the order

Answer»

A,B,C
B,C,A
C,B,A
C,A,B

Solution :`A to B,DeltaH=+24" kJ "mol^(-1)`
`H_(B)-H_(A)=+24`. . ..(i)
`B to C, DeltaH=18" kJ "mol^(-1)`
`H_(C)-H_(B)=-18`. . (ii)
`H_(B)-H_(C)=+18`. . (iii)
`H_(C)-H_(A)=6`(from (i) and (ii))
`H_(B) GT H_(C) gt H_(A)`.
7.

For the reaction at 500 K NO_(2)(g)+CO(g) to CO_(2)(g)+NO(g) The proposed mechanism is as follows: (i) NO_(2)+NO_(2) to NO+NO_(3) (slow) (ii) NO_(3)+CO to CO_(2)+NO_(2) (fast) What is the rate law for the reaction?

Answer»

Solution :Rate LAW : `(dx)/(dt)=K[NO_(2)]^(2)`
Thus is because the SLOW step determines the rate of reaction and there are two `NO_(2)` reactant molecules in the slow step of the reaction.
8.

For the reaction Ato products the following dats is given for a particular run time (in): 0""5""15""35 1/([A])(M^(-1)): 1""2""4""8 Determine the order of the reaction.

Answer»


Solution :`((-DA)(DT)_(1))/((-dA//dt)_(2)) =((C_(1))/(C_(2)))^(4),((0.5//5))/((0.25//10))=(1/0.5)^(n),0.100/0.025=(10/5)^(n),100/25=(2)^(n),2^(2)=2^(n)impliesn=2`
9.

For the reaction, Ato products the rate of reaction is 7.5xx10^(-4)" mole. lit"^(-1)"sec"^(-1). If the concentration of A is 0.5" mole lit"^(-1), the rate constant is

Answer»

`1.5xx10^(-3)"sec"^(-1)`
`2.5xx10^(-5)"sec"^(-1)`
`3.75xx10^(-4)"sec"^(-1)`
`8XX10^(-4)"sec"^(-1)`

ANSWER :A
10.

For the reaction at 500 K, "NO"_(2)(g)+"CO"(g)to"CO"_(2)(g)+"NO"(g), the proposed mechanism is as below : (i) "NO"_(2)+"NO"_(2)to"NO"+"NO"_(3)("slow")"""(ii) ""NO"_(3)+"CO"to"CO"_(2)+"NO"_(2)("fast") What is the rate law for the reaction ?

Answer»


SOLUTION :From the SLOW step,Rate `=K[NO_(2)]^(2).`
11.

For the reaction AtoB , the rate law expression is : Rate = k[A]. Which of the following statements is incorrect?

Answer»

The reaction is said to FOLLOW first ORDER kinetics
The half life of the reaction will depend on the initial concentration of the reactant
K’ is constant for the reaction at a constant temperature
The rate law provides a simple WAY of predicting the concentration of reactants and products at any time after the start of the reaction

Answer :B
12.

For the reaction at 800KN_(2)(g)+3H_(2)(g)hArr 2NH_(3)(g)the ratio of K_(p)and K_(c ) is:(R = 0.082 L atm mol^(-1)K^(-1))

Answer»

<P>`2.3xx10^(-4)`
`3.2xx10^(-6)`
`2.3xx10^(4)`
`3.2xx10^(6)`

Solution :`K_(p)=K_(c )(RT)^(DELTA N)`
or `K_(p)=K_(c )(RT)^(-2)`
`=(K_(c ))/((0.082xx800)^(2))`
or `(K_(p))/(K_(c ))=2.32xx10^(-4)`
13.

For the reaction at 298K, A_((g))+B_((g))rarrC_((g)) DeltaE=-5 cal and DeltaS =-10cal K^(-1)

Answer»

`DeltaG=+2612 CAL`
`DeltaG=-2612 cal`
`DeltaG=+261.2 cal`
`DeltaG=2379 cal`

SOLUTION :`DeltaH=DeltaE+DeltanRT=-5-1xx2.0xx298=-601`
`DeltaG=DeltaH-TDeltaS=-601-298xx(-10)=2379 cal`.
14.

For the reaction at 298 K : CO(g) + (1)/(2) O_(2)(g) rarr CO_(2)(g), Delta H^(@) = -282.8 kJ Standard entropies (in JK^(-1) mol^(-1)) : CO_(2)(g) = 213.6 CO(g) = 197.6 and O_(2)(g) = 205.0 Delta_(r) G^(@) for the reaction (kJ mol^(-1))

Answer»

`-306.02`
`-257.81`
306.02
`-157.03`

SOLUTION :`Delta S^(@) = S^(@) (CO_(2)) - [S^(@)(CO) + (1)/(2) S^(@)(O_(2))]`
`= 213.6 - [197.6 + (1)/(2)(205.0)]`
`= - 86.5 JK^(-1) MOL^(-1)`
`Delta G^(@) = -282.8 xx 10^(3) - 298(-86.5)`
`= -257.023 KJ mol^(-1)`
15.

For the reaction at 25^(@)C NH_(3)(g) to (1)/(2)N_(2)(g)+(3)/(2)H_(2)(g),DeltaH^(0)=11.04 kcal Calculate Deltau^(0) of the reaction at the given temperature.

Answer»


ANSWER :(10.44 KCAL)
16.

For the reaction at 298K, Ag^+(aq)+e^-=Ag(s),E^@=+0.80VSn^(2+)(aq) +2e^- =Sn(s),E^@=-0.14V what is the emf of the cell reprsented as Sn|Sn^(2+)||Ag^+| Ag,if each ion having unit concentration:

Answer»

0.66V
0.80V
0.94V
1.08V

Answer :C
17.

For the reaction Ararr2B , the rate of the reaction rate is .........

Answer»

`+(d[B])/(DT)=2-(d[A])/(dt)`
`+(d[A])/(dt)=1/2(d[B])/(dt)`
RATE `=1/2-(d[A])/(dt)`
Rate = `2 (d[B])/(dt)`

ANSWER :A
18.

For the reaction ArarrB+C, DeltaH = +25 kJ mole The activation energy for the reaction is

Answer»

`25 kJ//mole`
LESS than `25 kJ//mole`
more than `25 kJ//mole`
NONE of these

Answer :C
19.

For the reaction A--->B,the rate law is, rate=K[A].Which of the following statement is incorrect ?

Answer»

The reaction follows FIRST order kinetics
The `t_1/2` of reaction depends upon initial concentration of reactant
K is CONSTANT for the reaction at at constant temperature
The rate law PROVIDES a simple way of predicting the concentration of reactants and PRODUCT at any time after the start of the reaction

Answer :B
20.

For the reaction [Ag(CN)_2]iffAg^(+) + 2CN^(-), the equilibrium constant at 25^@Cis4 xx 10^(-19) .Calculate the silver ion concentration in a solution which was originally 0.1 molar in KCN and 0.03 molar in AgNO_3 .

Answer»

SOLUTION :`7.5 XX 10^(-18) M`
21.

For the reaction, Ag_((aq.))^(+) + Cl_((aq.))^(-) hArr AgCl_((s)) the Deltag^(@) Values for Ag_(aq.)^(@), Cl_(aq.) and AgCl_(s) are +77, -129 and -109 kj mol^(-1), Write the cell representation of above reaction and calculate E^(@) at 298 K, also calculate thelog_(10) K_(sp) of AgCl at 298K. (b) If 6.539 xx 10^(-2)g of metallic zinc is added to 100 mL saturated solution of AgCl, find the value of log_(10).([Zn^(2+)])/([Ag^(+)]^(2)). How many moles of Ag will be precipitated in this reaction ? Given, E_(Zn^(2+)//Zn)^(@) = -0.76 V.

Answer»

Solution :`Ag_((s)) + (1)/(2)Cl_(2_(g))rarr AgCl_((s)), DeltaG^(@) = -109kJ` ....(1)
`Ag_((s)) rarr Ag_(kJ)^(+) + e, DeltaG^(@) = +77 kJ` ....(2)
`(1)/(2)Cl_(2(g))^(-) + e rarr Cl_((aq.))^(-), DeltaG^(@) = -129 kJ` ....(3)
By eq. `(1)` - `(2)` - `(3)`
`Ag_((aq.))^(+) + Cl_((aq.))^(-) rarr AgCl_((s)), DeltaG^(@) = -109 - 77 + 129 = -57 kJ`
`:' -DeltaG^(@) = nE^(@)F`
`57 xx 10^(3) = 1 xx E^(@)F`
`:. E^(@) = 0.59 V`
The CELL is `Ag|AgCl_((s))|Cl_((aq.))^(-)||Ag_((aq.))^(+)|Ag`
`E_(cell) = E_(OP_(Ag//AgCl//Cl^(-))) -0.059log.(1)/([Cl^(-)])+E_(RP_(Ag^(+)//Ag)) + 0.059log [Ag^(+)]`
at equilibrium
`E_(cell) = 0`, thus
`0 = E_(Ag//AgCl//Cl^(-))^(@) + E_(RP_(Ag//AgCl//Cl^(-)))^(@) + 0.059 log [Ag^(+)][Cl^(-)]`
or `0 = E_(cell)^(@) + 0.059 log K_(SP)`
or `E_(cell)^(@) = -0.059 log K_(SP)`
or `logK_(SP) = -(E_(cell)^(@))/(0.059) = - (0.59)/(0.059) = -10`
`:. K_(SP)AgCl = 1 xx 10^(-10) M^(2)`
(b) Let solubility of `AgCl` be `S`, then
`S = sqrt(K_(SP)) = sqrt(10^(-10)) = 10^(-5) M`
Total milli-mole of `AgCl` in its SATURATED solution of `10^(-5)M` in `100 ML = 10^(-5) xx 10^(2) = 10^(-3)`
`:.` Mole of `AgCl` in `100 mL` solution
`= 10^(-3) xx 10^(-3) = 10^(-6)`
Also mole of `Zn` added `= (6.539 xx 10^(-2))/(65.39) = 10^(-3)`
`DeltaG^(@)` for `Ag rarr Ag^(+) + e` is `77 kJ`
`DeltaG^(@) = nE^(@)F`
`:. E_(Ag//Ag^(+))^(@) = (-77 xx 10^(3))/(1 xx 96500) = - 0.80 V`
For the redox change on ADDITION of `Zn`
`{:(ZnrarrZn^(2+)+2e,E^(@)=+0.76),(2Ag^(+)+2erarr2Ag,E^(@)=+0.80):}/(Zn+2Ag^(+)rarrZn^(2+)+2Ag,E_(cell)^(@)=0.76+0.80=1.56V)`
`E_(cell) = E_(OP_(Zn))^(@) + E_(RP_(Ag))^(@) + (0.059)/(2)log.([Ag^(+)]^(2))/([Zn^(2+)])`
At equilibrium, `E_(cell) = 0`
`:.= E_(cell)^(@)+(0.059)/(2)log.([Ag^(+)]^(2))/([Zn^(2+)])`
`:. log.([Zn^(2+)])/([Ag^(+)]^(@)) = (E_(cell)^(@) xx 2)/(0.059) = (1.56 xx 2)/(0.059) = 52.88`
`K_(c)=([Zn^(2+)])/([Ag^(+)]^(2)) = 7.6 xx 10^(52)`
Since `K_(c)` is appreciably HIGH and thus nearly whole of the `Ag^(+)` is converted to`Ag`. Thus moles of `Ag` formed `=` moles of `Ag^(+)` in `100 mL` saturated solution
`= (10^(-3))/(10^(3)) = 10^(-6)`
22.

For the reaction: [Ag(CN)_2]^- hArrAg^+ + 2CN^- the equilibrium constant K_c at 25^@C is 4.0 xx 10^(-19) then the silver ion concentration in a solution which was originally 0.1 molar in KCN and 0.03 molar inAgNO_3

Answer»

`7.5xx10^(18)`
`7.5xx10^(-18)`
`7.5xx10^(19)`
`7.5xx10^(-19)`

ANSWER :B
23.

For the reaction, Ag_2O(s)to2Ag(s)+1/2O_2(g),/_\H,/_\S and T are 40kJ, 100J and 380K respectively Hence /_\Gin KJ is :

Answer»


SOLUTION :`DeltaG=DeltaH-TDeltaS =40-(380xx100)/(1000)=2kJ`
24.

For the reaction : A(g) rarr B(g) + C(g) Concentration of 'A' varies with time (in sec) as : [A]_(T)=[10=0.2T^(2)]M The time at which rate of formation of 'B' is 1 M/sec.

Answer»

2 SEC
3 sec.
4 sec.
`2.5` sec.

Answer :D
25.

For the reaction A(g) to 2B(g) (d[B])/(dt) = 6xx10^(-4) M sec^(-1) "when" [A] = 0.1 M (d[B])/(dt) = 2.4xx10^(-3) M sec^(-1) when [A] =0.4 M Find the time taken (in seconds) for concentration of A to change from 0.6 M to 0.15 M.

Answer»


ANSWER :462
26.

For the reaction A_((g)) + B_((g))hArr C_((g)) + D_((g)), the degree of dissociation alpha would be –

Answer»

`(sqrtK)/(sqrtK+1)`
`sqrtK+1`
`sqrtKpm1`
`sqrtK-1`

ANSWER :A
27.

For the reaction A(g)+2B(g)rarr2C(g)+3D(g),the value of DeltaE at 27^(@)C is 19.0 kcal. The value of DeltaH for the reaction would be (R=2.0 cal K^(-1) mol^(-1))

Answer»

20.8 KCAL
19.8 kcal
18.8 kcal
20.2 kcal

Solution :USE `DeltaH=DeltaE+DeltanRT`
`DeltaH=19+2xx2xx10^(-3)xx300=20.2 kcal , Deltan=2`.
28.

For the reaction, A(g)+2B(g) to C(g)+D(g) (dx)/(dt)=k[A][B]^(2) Initial pressure of A and B are respectively 0.60 atm and 0.80 atm. At a time when pressure of C is 0.20 atm, rate of the reaction, relative to the initial value is

Answer»

`(1)/(6)`
`(1)/(48)`
`(1)/(4)`
`(1)/(24)`

ANSWER :A
29.

For the reaction AB_((g))hArrA_((g))+B_((g)), Ab is 33% dossociated at a total pressure of P. Therefore, P is related to K_(P) by one of the following options

Answer»

<P>`P=K_(p)`
`+p3K_(p)`
`P=6K_(p)`
`P=8K_(p)`

Solution :`AB_((g))hArrA_((g))+B_((g))`
`{:("Initial MOLES",1,0,0),("Moles at EQM.",1-0.33,0.33,0.33):}`
`P_(A)((0.33)/(1.33))P, P_(B)=((0.33)/(1.33))P`
`P_(AB)=((0.66)/(1.33))P`
`P_(AB)=(P_(A)xxP_(B))/(P_(AB))`
`K_(p)=(0.33xx0.33xxP)/(0.66xx1.33)`
`P/K=8impliesP=8K_(p),`
30.

For the reaction aC(s) + CO_2(g) hArr 2CO(g), the partial pressure of CO_2 and CO are 2.0 and4.0 atm respectively at equilibrium. TheK_p for reaction is:

Answer»

0.5
4
8
32

Answer :C
31.

For the reaction, AB(g)hArr A(g)+B(g), AB" is "33% dissociated at a total pressure of 'p' Therefore, 'p' is related to K_(p) by one of the following options

Answer»

<P>`P = 3K_p`
`p = K_p`
`P = 8K_p`
`P = 4K_p`

ANSWER :C
32.

For the reaction AB_(2)(g) rarr AB(g) + B(g), the initial pressure of AB_(2)(g)was 6 atm and at equilibrium, total pressure was found to be 8 atm at 300 K. The equilibrium constant of the reaction at 300 K is……

Answer»

Solution :`AB_(2)(G) rarr AB(g) + B(g)`
Given P = 6 atm
P + P' = 8 atm
`K_(p) = 2xx2/4 = 1`
33.

For the reaction :A(aq) to 2B(aq) +C(aq), the rate is 2.4xx10^(-3)"mol"L^(-1)s^(-1) on 20% reaction of A and the rate is 9.6xx10^(-5)"mol" L^(-1)s^(-1) on 84% reaction of A the order of reaction is :

Answer»


ANSWER :2
34.

For the reaction : aA+bB to Product (dx)/(dt)=k[A]^(a)[B]^(b) If concentration of A is doubled, rate becomes four times. If concentration of B is made four times, rate becomes doubled . Hence

Answer»

`-(d[A])/(DT)=-(d[B])/(dt)`
`-(d[A])/(dt)=-4(d[B])/(dt)`
`-4(d[A])/(dt)=-(d[B])/(dt)`
NONE of these

ANSWER :B
35.

For the reactionA to B. The rate of reaction becomes three times when the concentration of A is increased by nine times. What is the order of the reaction?

Answer»

SOLUTION :ORDER of the REACTION`=(1)/(2)`.
36.

For the reaction AtoB, the rate law expression is : rate = K[A]. Which of the following statements is incorrect ?

Answer»

The REACTION FOLLOWS first ORDER kinetics
The `t_(1//2)` of reaction depends on initial concentration of reactants
K is constant for the reaction at a constant temperature
The rate law provides a simple WAY of PREDICTING the concentration.

Answer :B
37.

For the reaction , A to B , k_1=10^8 e^(-6000/"8.34 T") and P to Q , k_2=10^10 e^(-8000/"8.34T") The temperature at which k_1=k_2 is

Answer»

386 K
221 K
26 K
52 K

Solution :`k_1=k_2`
`10^8 e^(-6000/"8.34 T")=10^10 e^(-8000/"8.34 T")`
`10^8/10^10 = e^((-2000)/(8.34 T))`
ln `10^(-2) = (-2000)/(8.34 T) RARR -ln 100=(-2000)/(8.34 T)`
`rArr` 2.303 log 100= `2000/(8.34T) rArr T= 2000/(2.303xx2xx8.34)`= 52K
38.

For the reaction A to B, following curves represent reaction The correct curves are

Answer»

`1, 2` only
`2, 3` only
`1, 4` only
`3, 4` only

Answer :C
39.

For the reaction : A rarrProduct, the order of reaction is equal to :

Answer»

`("LN"(r_(0))_(1)+"ln"(r_(0))_(2))/("ln"[A_(0)]_(1)+"ln"[A_(0)]_(2))`
`1-("ln"(t_(1//2))_(1)-"ln"(t_(1//2))_(2))/("ln"[A_(0)]_(1)-"ln"[A_(0)]_(2))`
`("ln"(t_(1//2))_(1)-"ln"(t_(1//2))_(2))/("ln"[A_(0)]_(1)-"ln"[A_(0)]_(2))`
`1+("ln"(t_(1//2))_(1)-"ln"(t_(1//2))_(2))/("ln"[A_(0)]_(1)-"ln"[A_(0)]_(2))`

ANSWER :B
40.

For the reaction A rarr C, it is found that the rate of the reaction quadruples when the concentration of A is doubled. The rate for the reaction is Rate = [A]^(n). The value of n is _________

Answer»

1
2
0
3

Answer :B
41.

For the reaction , A overset(k)(to)B+C Following data are obtained What will be the rate of reaction (in mol L^(-1) min^(-1), when conc of 'A' is 3M.

Answer»

`2.079xx10^(-3)`
`6.75xx10^(-1)`
`7.5xx10^(-2)`
none

Answer :D
42.

For the reaction, A hArr B ,deltaH for the reaction is -33.0 kJ/mol. Calculate (i) Equilibrium constant K_c for the reaction at 300 K (ii) If E_a (f) and E_a(r ) in the ratio 20:31 , calculate E_a(f) at 300 K . Assuming pre-exponential factor same for forward and reverse reaction.

Answer»


Answer :(i) `5.572 XX 10^5 "at" 300 K`
`E_(a(R ))=+93Kj`
43.

For the reaction A overset(K_(1))underset(K_(2))(hArr)B, if 'a' is the initial concentration of A and n is the number of moles of A reacting if initially no B was present then :

Answer»

`(DX)/(DT)=[K_(1)(a-x)-K_(2)x]` initially
`K_(1)(a-x_(e ))-K_(2)x_(e )=0` at equilibrium
`K_(1)+K_(2)=K_(1).(a)/(xe)`
`(K_(1)+K_(2))=(1)/(t) ln.(x_(e ))/(x_(e)-x)`

Solution :Initially, `(dx)/(dt)=K_(1)(a-x)-K_(2)x`………`(1)`
Hence choice `(A)` is correct.
If `x_(e)=` equilibrium concentration of B, where net rate `=0`
`K_(1)(a-x_(e ))-K_(2)x_(e)=0`………..`(2)`
Hence choice (B) is correct.
From EQUATION `(2)` , `K_(2)=(K_(1)(a-x_(e )))/(x_(e ))`
Substituting the value of `K_(2)` in equation `(1)`, we get
`(dx)/(dt)=K_(1)(a-x)-K_(1)((a-x_(e ))x)/(x_(e ))`
`=K_(1)((x_(e )-x)a)/(x_(e ))`
`IMPLIES(dx)/((x_(e )-x))=(a)/(x_(e ))K_(1)dt`
Hence by integrating and further calculation, we get
`(K_(1)+K_(2))=K_(1)(a)/(x_(e ))=(1)/(t)ln.(x_(e ))/((x_(e )-x))`
44.

For the reaction , A hArrB : K_c =2,B hArrC : K_c =4,C hArrD : K_c =6 K_c for the reaction A hArr D is :

Answer»

`(2div4div6)`
`(2xx4)/6`
`(4xx6)/2`
`2xx4xx6`

ANSWER :D
45.

For the reaction A+Bto Products, the following initial rates were obtained at various given initial concentrations. ltrbgt Write rate law and find the rate constant for the above reaction.

Answer»

Solution :The RATE law equation may be expressed as:
Rate = `k[A]^(p)[B]^(q)` ltrbgt COMPARING experiments, 1 and 2,
`(Rate)_(1)=k[0.1]^(p)[0.1]^(q)=0.05`
`(Rate)_(2) = k[0.2]^(p)[0.1]^(q)=0.10`
Dividing eqn. (ii) by eqn. (i),
`(Rate)_(2)/(Rate)_(1) = (k[0.2]^(p)[0.1]^(q))/(k[0.1]^(p)[0.1]^(q)) = 0.10/0.05 = 2:[2]^(p) = [2]^(1)` or p=1
Comparing experiments 1 and 3,
`(Rate_(1)) = k[0.1]^(p)[0.1]^(q)=0.05`
`(Rate)_(3) = k[0.1]^(p)[0.2]^(q)=0.05`
Dividing eqn. (iii) by eqn. (i),
`(Rate_(3))/(Rate_(1)) = (k[0.1]^(p)[0.2]^(q))/(k[0.1]^(p)[0.1]^(q)) = 0.05/0.05 = 1 , [2]^(1) = 1=[2]^(0) or q=0`
Rate law for the reaction, rate = `k[A]^(1)[B]^(0)`
Rate constant (k) for the reaction may be CALCULATED as follows:
`k[0.1]^(p)[0.1]^(q)=0.05 , k[0.1]^(1)[0.1]^(0)=0.05` , `k=(0.05)/0.1=0.5`
46.

For the reaction A+Bto products, it is found that order of A is 1 and order of B is 1//2. When concentrations of both A and B are increased four times the rate will increase by a factor

Answer»

6
8
4
16

Answer :B
47.

For the reaction A+Bto product ,the rate law is given by ,r=k[A]^((1)/(2))[B]^(2).What is the order of the reaction ?

Answer»

SOLUTION :(overall ORDER of REACTION )=(with respect to all reaction )
`=(1)/(2)+2=2.5`
48.

For the reaction, A+Brarr"product" If concentration of A is doubled, rate increases 4 times. If concentrations of A and B both are doubled, rate increases 8 times . The differential rate equation of the reaction will be

Answer»

`(DC)/(dt)-kC_(A)xxC_(B)`
`(dC)/(dt)="k "C_(A)^(2)xxC_(B)^(3)`
`(dC)/(dt)=kC_(A)^(2)xxC_(B)`
`(dC)/(dt)=kC_(A)^(2)xxC_(B)^(2)`

Solution :Let the ORDER with repect to A and B are x and y, respectively . Hence , Rate , `r=[A]^xx[B]^Y`
On doubling the concentration of A, rate increases 4 times , `4r=[2A]^xx[B]^Y`
From Eqs. (i) and (ii),
`(1)/(4)=((1)/(2))^(x)`
`:.""x=2`
`:.` Order with respect to A is 2.
If concentration of A and B both are doubled,
`8R=[2A]^x[2B]^y`
From Eqs. (i) and (ii), we get
`(1)/(8)=(1)/(2)^(x).(1)/(2)^(y)""[becausex=2]`
`(1)/(8)=(1)/(4xx2^(y))rArr2^(y)=2`
`:.""y=1`
Hence, the DIFFERENTIAL rate equation is
`rprop[A]^(2)[B]^(1)""or(dc)/(dt)=kC_(A)^(2)xxC_(B)`
`[Where, C_(A)andC_(B)=concentrations " of " A andB]`
49.

For the reaction a+brArrc+d, initially concentrations of a and b are equal and at equilibrium the concentration of a . What will be the equilibrium constant for the reaction ?

Answer»

2
9
4
3

Answer :3
50.

For the reaction ,A+BrarrC+D. The variation of the concentration of the products is given by the curve:

Answer»

X
Y
Z
W

Answer :B