94420 + Interview Questions in Chemistry in Class 12 Page 491 InterviewSolution

24501.	The ratio of 'd' values in NaCl crystal is
Answer» `0.707 : 1 : 1.154` `1:0.707 : 1.154` `1 : 1.154 : 0.707` `0.707 : 1.154 : 1` ANSWER :B

Discussion

24502.	The ratio of Cu and Sn metals in bronze
Answer» Solution :In Brounze CU is 88% and Sn is 12% APPROXIMATLEY therefore the RATIO of Cu and Sn `(88xx63.5)/(12xx1.58)=3`

Discussion

24503.	The ratio of CN and OS of x in [x(NH_(3))_(5)SO_(4)]Cl is
Answer» Solution :In EDTA, two N ATOMS and FOUR O atoms are the donor atoms.

Discussion

24504.	The ratio of cis and trans-isomers of the complex [Ma_(2)bcde]^(n+-) is:
Answer» `5:3` `2:1` `7:3` `3:1` SOLUTION : No. off cis-isomers=6 They are: No. off trans-isomers=3

Discussion

24505.	The ratio of closed packed atoms to tetrahedral holes in cubic close packing is:
Answer» `1:1` `1:2` `1:3` `2:1` ANSWER :B

Discussion

24506.	The ratio of close packed atoms to tetahedral hole in cubic packing is ………….
Answer» `1:1` `1:2` `2:1` `1:4` Solution :Hint : If number of CLOSE packed = N , then, The number of TETRAHEDRAL holes formed = 2N Number of OCTAHEDRAL holes formed = N. THEREFORE `N:2N=1:2`

Discussion

24507.	The ratio of cations to anion in a octahedral close packing is :
Answer» 0.414 0.225 0.02 None Answer :A

Discussion

24508.	The ratio of cations to anion in a closed pack tetrahedral is-
Answer» 0.414 0.225 0.02 None Answer :B

Discussion

24509.	The ratio of cations to anion in a closed pack tetrahedral is :
Answer» 1 0.225 0.02 None Answer :B

Discussion

24510.	The ratio of cationic radius to anionic radius in an ionic crystal is greater than 0.732. its coordination number is
Answer» 6 8 1 4 Solution :If RADIUS RATIO `((r_(c)^(+))/(r_(a)^(-)))` is more than 0.732 its permitted coordination NUMBER is 8.

Discussion

24511.	The ratio of C^(14)//C^(12) in dead tissue is less than that in fresh tissue.
Answer» ANSWER :T

Discussion

24512.	The ratio of carbon, hydrogen and oxygen in 2- methyl benzoic acid is:
Answer» (4:4:2) (4:4:1) (4:2:2) (2:4:1) ANSWER :B

Discussion

24513.	The ratio of Boyle's temperature and critical temperature for a gas is :
Answer» 8/27 27/8 1/2 2/1 Answer :B

Discussion

24514.	The ratio of average speed of an oxygen molecular to the r.m.s. speed of a N_(2) molecular at the same temperature is :
Answer» `((3PI )/( 7) )^(1//2)` `(( 7)/( 3pi ))^(1//2)` `((3)/( 7pi ))^(1//2)` `((7pi)/( 3))^(1//2)` Solution :`u_(av)` of `O_(2) = sqrt((8RT)/( M)) = sqrt(( 8 RT)/( pi xx 32))` `u_("rms") ` or `N_(2) = sqrt(( 3RT)/( M)) = sqrt(( 3RT)/( 28))` `( u_(av))/( u_("rms")) = sqrt((8RT)/(pi xx 32))xxsqrt((28)/( 3RT)) = (7^(1/2))/( 3pi)`

Discussion

24515.	The ratio of average speed and most probable speed is :
Answer» `2 // SQRT( pi)` ` sqrt( 8 RT ) // (pi M )` `sqrt( 8 // 3PI )` ` pi // sqrt( 2)` ANSWER :A

Discussion

24516.	The ratio of area covered by second orbit to the first orbit in hydrogen atom is :-
Answer» 0.043055555555556 0.052777777777778 0.33402777777778 0.66736111111111 Answer :D

Discussion

24517.	The ratio of any colligative property for KCl solution to that of sugar solution of same molality is
Answer» 1 0.5 2 3 Answer :C

Discussion

24518.	The ratio of activities of two radionuclides X and Y in a mixture at time t=0 was found to be 8:1 after two hour's the ratio of activities become 1:1. If the t_(1//2) of radionuclide X is 20 min find the t_(1//2)[in minutes] of radionuclide Y.
Answer» ANSWER :40

Discussion

24519.	Ratio of the amounts of H_(2)S needed to precipitate all the metal ions from 100 ml of 1M AgNO_(3) and 100 ml of 1M CuSO_(4) will be :
Answer» `1:2` `2:1` zero infinite ANSWER :a

Discussion

24520.	The ratio of amounts of H_(2)S needed to precipitate all the metal ions from 100 ml of 1 M AgNO_(3) and 100 ml of 1 M CuSO_(4) will be
Answer» `1:2` `2:1` Zero Infinity Solution :In ACIDIC medium 2 MOLECULES of `KMnO_(4)` gives 5 ATOMS of OXYGEN `2KMnO_(4)+3H_(2)SO_(4) to K_(2)SO_(4)+2MnSO_(4)+3H_(2)O+5(O)` `2xx158=(316xx8)/(80)=31.6` So, equivalent wt. of `KMnO_(4)` in acidic medium is `=31.6`gm

Discussion

24521.	The ratio fo fraction of molecules present in the velocity range 450 m/sec to (450+0.01) m/sec for O_(2) at 100K and SO_(2) present at 200 K is :
Answer» 2 `(1)/(2)` 4 1 Solution :Since `(T)/(M)` RATIO for both `O_(2)` at 100K and `SO_(2)` at 200K is same, curve will be same for both `&` ratio=1.

Discussion

24522.	The ratio gamma((C_(P))/(C_(V))) for iner gases is
Answer» 1.33 1.66 2.13 1.99 Solution :`gamma =(C_(p))/(C_(V))=(5)/(3) = 1.66` ( For MONOATOMIC asHe,Ne,AR)

Discussion

24523.	The ratio C_p//C_vfor noble gases is
Answer» 1.67 1.33 1.42 1.84 Answer :A

Discussion

24524.	The ratio between the root mean square velocity of H_(2) at 50K and that of O_(2) at 800K is
Answer» 4 2 1 `1//4` SOLUTION :`m(H_(2)) = sqrt(( 3R xx 500)/( 2))` `m ( O_(2)) = sqrt(( 3R xx 800)/( 32))` `( mu ( H_(2)))/( mu ( O_(2))) = sqrt(( 3R xx 500)/( 2) ) xx sqrt( ( 32 ) /( 3R xx 800))` `= sqrt(( 500 xx 32)/( 2 xx 800)) = 1 `

Discussion

24525.	The ratio between the root mean square velocity of H_(2) at 50 K and that of O_(2) at 800 K is
Answer» 4 2 1 `1//4` ANSWER :C

Discussion

24526.	The ratio between the root mean square speed of H_2 at 50k and that of O_2 at 800k is :
Answer» 4 2 1 1/4 Answer :C

Discussion

24527.	The ratio between the maximum tolerated dose of a drug and a minimum curative dose is called
Answer» ISO electric point THERAPEUTIC INDEX critical point iso thermal point Solution :therapeutic index

Discussion

24528.	The van der Walls' equation explains the behaviour of :
Answer» ATM litre `mol^(-1)` `atm dm^3 mol^(-1)` DYNE CM `mol^(-1)` all Answer :D

Discussion

24529.	The rates of reaction invrease with increase of temperature because
Answer» activation ENERGY of the REACTION molecules decreases kinetic eriergy of the product molecules increases the fraction of the REACTING molecules possessing an· energy equal to the activation energy the collisions between molecules decrease SOLUTION :With increase in temperature, tehe factor `._eE_a//RT`which gives fraction of molecules with energy increases and hence `E_a` increases.

Discussion

24530.	The rates of most the reactions double when the temperature is raised from 298 K to 308 K Calculate their activation energy.
Answer» Solution :From ARRHENIUS EQUATION ,we know `"LOG"K_1/K_2=E_a/(2.303R)[1/T_1-1/T_2]` Here `K_2/K_1=2,T_1=298K,T_2=308K`.SUBSTITUTING the values,we get `"log"2=E_a/(2.303xx8.314JK^-1mol_^-1)[(T_2-T_1)/(T_1T_2)]` `E_a=`52897.7mol^-1=52.89kj` MOL.

Discussion

24531.	The rates of effusion of two gases A and B under the same conditions of temperature and pressure are in the ratio gamma_A : gamma_B= 2: 1. What would be the ratio of the rms speeds of molecules of A and B if T_A : T_B = 2 : 1?
Answer» SOLUTION :`2 SQRT2 : 1`

Discussion

24532.	The rate of most reactions become double when their temperature is raised from 298 K to 308 K. Calculate their activation energy. (Given, R = 8.314" J mol"^(-1) )
Answer» Solution :From Arrhenius EQUATION ,we KNOW `"log"K_1/K_2=E_a/(2.303R)[1/T_1-1/T_2]` Here `K_2/K_1=2,T_1=298K,T_2=308K`.Substituting the values,we get `"log"2=E_a/(2.303xx8.314JK^-1mol_^-1)[(T_2-T_1)/(T_1T_2)]` `E_a=`52897.7mol^-1=52.89kj` MOL.

Discussion

24533.	The rates of diffusion ofSO_2 , CO_2 , PCl_3 ,SO_3 are in the following order :
Answer» `PCl_3 GT SO_3 gt SO_2 gt CO_2` `CO_2 gt SO_3 gt PCl_3 gt SO_3` `SO_2 gt SO_3 gt PCl_3 gt CO_2` `CO_2 gt SO_2 gt SO_3 gt PCl_3` ANSWER :D

Discussion

24534.	The rates of a reaction starting with intial conentrations of 2xx10^(-3) M and 1xx10^(-3) M are equal to 2.40xx10^(-4) M s^(-1) and 0.60xx10^(-4) M s^(-1) respectively. Calculate the order of the reaction with respect to the reactant and also the rate constant.
Answer» Solution :If N is the order of reaction RATE `=k[A_(0)]^(n)=kC^(n)` For two different initialconcentrations, we have `(r_(1))/(r_(2))=((C_(1))/(C_(2)))^(n):.log""(r_(1))/(r_(2))=nlog""(C_(1))/(C_(2))` or `""n=(log(r_(1)//r_(2)))/(log(C_(1)//C_(2)))=(log(2.4xx10^(-4)//0.6xx10^(-4)))/(log(2xx10^(-3)//1xx10^(-3)))=(LOG4)/(log2)=2` Hence, orderof reaction = 2 Rate `=k[A_(0)]^(2)` `k=("Rate")/([A_(0)]^(2))=(2.4xx10^(-4)"mol L"^(-1)s^(-1))/((2xx10^(-3)"mol L"^(-1))^(2))=0.6xx10^(2)"L mil"^(-1)s^(-1)`

Discussion

24535.	The rates of diffusion of O_2 and H_2 at same P and T are the ratio :
Answer» 1:4 1:8 1:16 4:1 Answer :A

Discussion

24536.	The rate ofreaction at different times are as follows : [[time(sec),0,10,20,30],[rate(Msec^-),1.5xx10^-2,1.48xx10^-2,1.51xx10^-2,1.49xx10^-2]] The reaction is
Answer» ZERO prder 1st order 2nd order 3rd order Answer :A

Discussion

24537.	The rate of the reation is expressed in different way as followed : -1/m . (d[A])/(dt)=+(1)/(n)(d[B])/(dt)=- (1)/(p)(d[C])/(dt)=+ 1/q (d[D])/(dt) The reaction is
Answer» `na+mB rarr pC+qD` `pA + qB rarr mC +n D ` `pA +mB rarr nC +qD` `mA+pCrarr nB +qD ` Solution :A and C are reactants as their CONCENTRATION falls with time hence , (minus sign) . B and D are the product of the reaction because their concentration increases with time hence (PLUS sign) . 'm' and 'p' are the STOICHIOMETRIC coefficients of A and C while 'n' and 'q' are the coefficients of products B and D . So, the equation of the reaction BECOMES : `mA + pC to n B + qD`.

Discussion

24538.	The rate of the reaction x+2yrarr product is 4xx10^(-3) " mol L"^(-1)s^(-1) , if [x] = [y] =0.2 M and rate constant at 400K is 2xx10^(-2)s^(-1) , what is the overall order of the reaction.
Answer» Solution :Rate `=K[x]^n[y]^m` `4xx10^(-3)"mol"L^(-1)s^(-1)=2xx10^(-2)s^(-1)(0.2 "mol L"^(-1))^n(0.2 "mol L"^(-1))^m` `(4xx10^(-3)"mol L"^(-1)s^(-1))/(2xx10^(-2)s^(-1))=(0.2)^(n+m)("mol L"^(-1))^(n+m)` `2.0("mol L"^(-1))=(0.2)^(n+m)("mol L"^(-1))^(n+m)` Comparing the powers on both sides , the overall order of the REACTION n + m = 1

Discussion

24539.	The rate of the reaction X to Y becomes 8 times when the concentration of the reactant 'X' is doubled . The rate law of the reaction is ………..
Answer» `- (d [X])/(dt) = k [X]^(2)` `- (d [x])/(dt) = k[X]^3` `- (d[x])/(dt) = k[X]^(4)` `- (d[x])/(dt) = k[X]^(8)` SOLUTION :`- (d[x])/(dt) = k[X]^(3)`

Discussion

24540.	The rate of the reaction X rarrY becomes 8 times when the concentration of the reactant 'X' is doubled . The rate law of the reaction is ............
Answer» `-(d[X])/(dt)=K[X]^2` `-(d[x])/(dt)=k[X]^3` `-(d[x])/(dt)=k[X]^4` `-(d[x])/(dt)=k[X]^8` ANSWER :B

Discussion

24541.	The rate of the reaction A to products , at the intitial concentrations of 3.24 xx 10^(-2)M is nine times its rate at constant initial concentration of 1.2 xx 10^(-3)M . The order of the reaction is
Answer» `(1)/(2)` `(3)/(4)` `(3)/(2)` `(2)/(3)` Solution :9r = `k (3.24 xx 10^(-2))^(ALPHA ) "" …. (i)` `r = k(1.2 xx 10^(-3))^(alpha) "" …. (ii)` DIVIDE EQ. (i) by (ii) `(9r)/(r) = (k(3.24 xx 10^(-2))^(alpha))/(k (1.2 xx 10^(-3))^(alpha)) implies 3^(2) = 3^(3alpha)` `THEREFORE alpha = (2)/(3)`.

Discussion

24542.	The rate of the reaction A toProducts, at the initial concentration of 3.24xx10^(-2) M is nine times its rate at another initial concentration of 1.2xx10^(-3) M. The order of the reaction is
Answer» `(1)/(2)` `(3)/(4)` `(3)/(2)` `(2)/(3)` SOLUTION :`9 r = K(324 xx 10^(-2))^(n)""…(i)` `r = k(1 2 xx 10^(-3))^(n)""…(ii)` DIVIDING (i) by (ii), ` 9 = ((3.24 xx 10^(-2))/(1*2 xx 10^(-3)))^(n) or 9 = (27)^(n)` or `(3)^(2) = (3^(3))^(n) = (3)^(3N).` Hence, `3 n = 2 or n = 2//3`.

Discussion

24543.	The rate of the reactionA + B to 3Cgets increased by 72 times when the concentration of A is tripled and that of B is doubled . The order of the reaction with respect to A and B are -- and --- respectively
Answer» 1,2 2,3 3,2 2,2 ANSWER :B

Discussion

24544.	The rate of the reaction A rarr Products at the initial concentration of 3.24 xx10^(-2) M is nine times its rate at another initial concentration of 1.2xx10^(-3) M. The order of the reaction is
Answer» `(1)/(2)` `(3)/(4)` `(2)/(3)` `(3)/(2)` Solution :( C) If rate at conc .` 1.2 xx10^(-2) ` M is r then `r = k(1.2xx10^(-3))^(a)"" cdots(i) ` `9r= k(3.24xx10^(-2))^(a) "" cdots(II)` DIVIDING eq (ii) by eq. (i) `9=((3.24xx10^(2))^(a))/((1.2xx10^(-3))^(a))=27^(a)` or` 3^(2)=3^(3A)` `:. 3a=2 "or" a =2//3`

Discussion

24545.	The rate of the reaction,A+B+Crarr products, is given byr=(-d[A])/(dt)=k[A]^(1//2)[B]^(1//3)[C]^(1//4). The order of the reaction is
Answer» `1/3` `1/4` `1/2` `13/12` SOLUTION : ORDER of REACTION `=1/2+1/3+1/4=(6+4+3)/(12)=(13)/(12)`

Discussion

24546.	The rate of the reaction 2N_(2)O_(5) to 4NO_(2)+O_(2) can be written in three ways (-d[N_(2)O_(5)])/(dt)=k[N_(2)O_(5)],(d[N_(2)O_(5)])/(dt)=k'[N_(2)O_(5)](d[O_(2)])/(dt)=k''[N_(2)O_(5)] The relationship between k and k' and between k and k'' are
Answer» `K'=2K,k'=k` `k'=2k,k"=k//2` `k'=2k,k"=2k` `k'=k,k"=k` ANSWER :B

Discussion

24547.	The rate of the reaction 2N_(2)O_(5) to 4 NO_(2) + O_(2) can be written in three ways (-d [N_(2)O_(5)])/(dt) = k [N_(2)O_(5)] (d [NO_(2)])/(dt) = k' [N_(2)O_(5)] (d[O_(2)])/(dt ) = k'' [N_(2)O_(5)] The relationship between k and k' and between k and k'' are
Answer» k' = 2K , k''= 2k k' = k , k''= k k' = 2k , k''= k k' = 2k , k''= k/2 Solution :`-(1)/(2) (d(N_(2)O_(5)))/(DT) = (1)/(4) (d(NO_(2)))/(dt) = (d(O_(2)))/(dt)` `(1)/(2) k (N_(2)O_(5)) = (1)/(2) k' (N_(2)O_(5)) = k'' (N_(2)O_(5))` `(k)/(2) = (k')/(4) = k'' implies k' = 2k , k'' = (k)/(2)`.

Discussion

24548.	The rate of the reaction 2SO_(2) +O_(2) rarr 2SO_(3) may be expressed as : -(d[O_(2)])/(dt)= 2.5 xx10^(-4) "mol L"^(-1) "sec"^(-1) The rate of reaction when expressed in terms of SO_(3) will be :
Answer» `-5.0xx10^(-4) "mol L"^(-1) "SEC"^(-1)` `-1.25xx10^(-4) "mol L"^(-1) "sec"^(-1)` `3.75xx10^(-4) "mol L"^(-1) "sec"^(-1) ` `5.0xx10^(-4) "mol L"^(4) "sec"^(-1)` Solution :(D) Rate `= -(1)/(2) (d[SO_(2)])/(DT) =-(d[O_(2)])/(dt)` `(d[SO_(2)])/(dt)=2 (d[O_(2)])/(dt)` `=2xx2.5 xx10^(-4) ` `= 5.0 xx10^(-4)mol L^(-1) sec^(-I)`

Discussion

24549.	The rate of the reaction 2 NO+Cl_(2)to2NOCl is given by the rate equation : rate =k[NO]^(2)[Cl_(2)]. Thevalue of the rate constant can be increased by
Answer» increasing the temperature increasing the concentration of NO increasing the concentration of `Cl_(2)` doing all of these Solution :(a) : The rate CONSTANT of a reaction depends only on temperature and does not depend uponconcentrations of the REACTANTS.

Discussion

24550.	The rate of the reaction 2 NO + Cl_(2) to 2 NOCl 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 Solution :Concentration do not AFFECT rate CONSTANT

Discussion

Explore topic-wise InterviewSolutions in .

The ratio of 'd' values in NaCl crystal is

The ratio of Cu and Sn metals in bronze

The ratio of CN and OS of x in [x(NH_(3))_(5)SO_(4)]Cl is

The ratio of cis and trans-isomers of the complex [Ma_(2)bcde]^(n+-) is:

The ratio of closed packed atoms to tetrahedral holes in cubic close packing is:

The ratio of close packed atoms to tetahedral hole in cubic packing is ………….

The ratio of cations to anion in a octahedral close packing is :

The ratio of cations to anion in a closed pack tetrahedral is-

The ratio of cations to anion in a closed pack tetrahedral is :

The ratio of cationic radius to anionic radius in an ionic crystal is greater than 0.732. its coordination number is

The ratio of C^(14)//C^(12) in dead tissue is less than that in fresh tissue.

The ratio of carbon, hydrogen and oxygen in 2- methyl benzoic acid is:

The ratio of Boyle's temperature and critical temperature for a gas is :

The ratio of average speed of an oxygen molecular to the r.m.s. speed of a N_(2) molecular at the same temperature is :

The ratio of average speed and most probable speed is :

The ratio of area covered by second orbit to the first orbit in hydrogen atom is :-

The ratio of any colligative property for KCl solution to that of sugar solution of same molality is

The ratio of activities of two radionuclides X and Y in a mixture at time t=0 was found to be 8:1 after two hour's the ratio of activities become 1:1. If the t_(1//2) of radionuclide X is 20 min find the t_(1//2)[in minutes] of radionuclide Y.

Ratio of the amounts of H_(2)S needed to precipitate all the metal ions from 100 ml of 1M AgNO_(3) and 100 ml of 1M CuSO_(4) will be :

The ratio of amounts of H_(2)S needed to precipitate all the metal ions from 100 ml of 1 M AgNO_(3) and 100 ml of 1 M CuSO_(4) will be

The ratio fo fraction of molecules present in the velocity range 450 m/sec to (450+0.01) m/sec for O_(2) at 100K and SO_(2) present at 200 K is :

The ratio gamma((C_(P))/(C_(V))) for iner gases is

The ratio C_p//C_vfor noble gases is

The ratio between the root mean square velocity of H_(2) at 50K and that of O_(2) at 800K is

The ratio between the root mean square velocity of H_(2) at 50 K and that of O_(2) at 800 K is

The ratio between the root mean square speed of H_2 at 50k and that of O_2 at 800k is :

The ratio between the maximum tolerated dose of a drug and a minimum curative dose is called

The van der Walls' equation explains the behaviour of :

The rates of reaction invrease with increase of temperature because

The rates of most the reactions double when the temperature is raised from 298 K to 308 K Calculate their activation energy.

The rates of effusion of two gases A and B under the same conditions of temperature and pressure are in the ratio gamma_A : gamma_B= 2: 1. What would be the ratio of the rms speeds of molecules of A and B if T_A : T_B = 2 : 1?

The rate of most reactions become double when their temperature is raised from 298 K to 308 K. Calculate their activation energy. (Given, R = 8.314" J mol"^(-1) )

The rates of diffusion ofSO_2 , CO_2 , PCl_3 ,SO_3 are in the following order :

The rates of a reaction starting with intial conentrations of 2xx10^(-3) M and 1xx10^(-3) M are equal to 2.40xx10^(-4) M s^(-1) and 0.60xx10^(-4) M s^(-1) respectively. Calculate the order of the reaction with respect to the reactant and also the rate constant.

The rates of diffusion of O_2 and H_2 at same P and T are the ratio :

The rate ofreaction at different times are as follows : [[time(sec),0,10,20,30],[rate(Msec^-),1.5xx10^-2,1.48xx10^-2,1.51xx10^-2,1.49xx10^-2]] The reaction is

The rate of the reation is expressed in different way as followed : -1/m . (d[A])/(dt)=+(1)/(n)(d[B])/(dt)=- (1)/(p)(d[C])/(dt)=+ 1/q (d[D])/(dt) The reaction is

The rate of the reaction x+2yrarr product is 4xx10^(-3) " mol L"^(-1)s^(-1) , if [x] = [y] =0.2 M and rate constant at 400K is 2xx10^(-2)s^(-1) , what is the overall order of the reaction.

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

The rate of the reaction X rarrY becomes 8 times when the concentration of the reactant 'X' is doubled . The rate law of the reaction is ............

The rate of the reaction A to products , at the intitial concentrations of 3.24 xx 10^(-2)M is nine times its rate at constant initial concentration of 1.2 xx 10^(-3)M . The order of the reaction is

The rate of the reaction A toProducts, at the initial concentration of 3.24xx10^(-2) M is nine times its rate at another initial concentration of 1.2xx10^(-3) M. The order of the reaction is

The rate of the reactionA + B to 3Cgets increased by 72 times when the concentration of A is tripled and that of B is doubled . The order of the reaction with respect to A and B are -- and --- respectively

The rate of the reaction A rarr Products at the initial concentration of 3.24 xx10^(-2) M is nine times its rate at another initial concentration of 1.2xx10^(-3) M. The order of the reaction is

The rate of the reaction,A+B+Crarr products, is given byr=(-d[A])/(dt)=k[A]^(1//2)[B]^(1//3)[C]^(1//4). The order of the reaction is

The rate of the reaction 2N_(2)O_(5) to 4NO_(2)+O_(2) can be written in three ways (-d[N_(2)O_(5)])/(dt)=k[N_(2)O_(5)],(d[N_(2)O_(5)])/(dt)=k'[N_(2)O_(5)](d[O_(2)])/(dt)=k''[N_(2)O_(5)] The relationship between k and k' and between k and k'' are

The rate of the reaction 2N_(2)O_(5) to 4 NO_(2) + O_(2) can be written in three ways (-d [N_(2)O_(5)])/(dt) = k [N_(2)O_(5)] (d [NO_(2)])/(dt) = k' [N_(2)O_(5)] (d[O_(2)])/(dt ) = k'' [N_(2)O_(5)] The relationship between k and k' and between k and k'' are

The rate of the reaction 2SO_(2) +O_(2) rarr 2SO_(3) may be expressed as : -(d[O_(2)])/(dt)= 2.5 xx10^(-4) "mol L"^(-1) "sec"^(-1) The rate of reaction when expressed in terms of SO_(3) will be :

The rate of the reaction 2 NO+Cl_(2)to2NOCl is given by the rate equation : rate =k[NO]^(2)[Cl_(2)]. Thevalue of the rate constant can be increased by

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