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1701.

If the rate constant for a reaction is expressed as: rate = `k[A]^(2)[B]`, what is the order of the reacton?

Answer» Order of reaction =3
Discussion
1702.

The rate constant of a reaction is found to be `3xx10^(-3)"mol"L^(-1)"min"^(-1)`. The order of the reaction is :A. zeroB. 1C. 2D. `1.5`

Answer» Correct Answer - A
Unit of rate and rrte constant are same for zero order reaction.
Discussion
1703.

Consider the decomposition of hydrogen peroxide in the alkaline medium which is catalysed by iodide ions. This reaction takes place in two steps as given below step-1 `H_(2)O_(2) + I^(-1) to H_(2)O + IO^(-)(Slow)` Step II `H_(2)O_(2) + IO^(-) to H_(2)O + I^(-) +O_(2)` (fast) a) Write the rate law expression and determine the order of reactan w.r.t `H_(2)O_(2)` b) What is the molecularity of each individual step?

Answer» a) Rate law expression:
Rate = `k[H_(2)O_(2)][I^(-)]`
Order of reaction w.r.t `H_(2)O_(2)=1`
b) Molecularities of both step-I and step-II are two.
Discussion
1704.

A reaction is catalysed by `H^(+)` ions. The reaction has rate constant `3xx10^(-3) min^(-1)` in presence of acid `HA` and `2xx10^(-3) min^(-1)` in presence of acid `HB`. If both `HA` and `HB` are strong acid, which is//are correct?A. (a) `HA` is strong acid, than `HB`B. (b) Relative strength for `HA:HB=1.5`C. (c ) The reaction, `NaB+HA rarr NaA+HB` may possibleD. (d) `HB` is stromger acid than `HA`

Answer» Correct Answer - a, b, c
Discussion
1705.

Calculate the order of reaction for the reaction `2NH_(3)(g) to N_(2)(g) + 3H_(2)(g)` Given that half life period `(t_(1//2))` under a pressure of 50mm Hg is 3.52 and under a pressure of 100 mm Hg is 1.82

Answer» We know that
`n=1+(log(t_(1//2))1-log(t_(1//2))2)/(log p_(2)-log p_(1))`
`1+ (log 3.52 - log 1.82)/(log 100 - log 50)`
`1+(0.5465 - 0.2600)/(2-1.6989)`
`1+ (0.2865)/(0.3011) = 1+0.9515=2`
Discussion
1706.

The rate of first-order reaction is `1.5 xx 10^(-2) M "min"^(-1)` at `0.5 M` concentration of reactant. The half-life of reaction isA. `0.383 min`B. `23.1 min`C. `8.73 min`D. `7.53 min`

Answer» Correct Answer - B
Differential rate law for a first order reaction is Rate `=k[A]`
Thus,
`k=(Rate)/([A])`
`=(1.5xx10^(-2)molL^(-1)min^(-1))/(0.5M)`
`=3xx10^(-2)min^(-1)`
For a first order reaction
`t_(1//2)=(0/693)/(k)=(0.693)/(3xx10^(-2)min^(-1))`
`=23.1min`
Discussion
1707.

The rate of a first order reaction is `1.5 xx 10^(-2) mol L^(-1) "min"^(-1)` at 0.5 M concentration of the reactant . The half life of the reaction isA. 8.73 minB. 7.53 minC. 0.383 minD. 23.1 min

Answer» Correct Answer - d
Rate `((dx)/(dt))` = k.c, `1.5 xx 10^(-2) = k xx 0.5`
For first order k = `(1.5 xx 10^(-2))/(0.5) = 3 xx 10^(-2) "minute"^(-1)`
`t_(1//2) = (.693)/(k) = (.693)/(3 xx 10^(-2)) = 23.1` minute.
Discussion
1708.

The rate of first-order reaction is `1.5 xx 10^(-2) M "min"^(-1)` at `0.5 M` concentration of reactant. The half-life of reaction isA. `0.383min`B. `23.1min`C. `8.73 min`D. `7.53 min`

Answer» Correct Answer - B
for the first order reaction,
`rate((dx)/(dt))=K[A]`
[A] = concentration of reactant
k= rate constant
given that ,
`(dx)/(dt)=1.5xx10^(-2)molL^(-1)min""^(-1)`
`k=?and [A]=0.5M`
`therefore 1.5xx10^(-2)=kxx0.5`
`therefore k=(1.5xx10^(-2))/(0.5)`
`=3xx10^(-2) min^(-1)`
for first order reaction ,
Half -life period `t_(1//2)=(0.693)/(k)=(0.693)/(3xx10^(-2))`
`=23.1 min`
Discussion
1709.

The rate of first-order reaction is `1.5 xx 10^(-2) M "min"^(-1)` at `0.5 M` concentration of reactant. The half-life of reaction isA. 0.383 minB. 23.1 minC. 8.73 minD. 7.53 min

Answer» Correct Answer - B
For the first order reaction,
Rate =K[A]
Given that,
`(dx)/(dt)=1..5xx10^(-2)"mol L^(-1)min"^(-1)`
`K=? and [A]=0.5 M`
`therefore " "1.5xx10^(-2)=Kxx0.5`
`therefore" "K=(1.5xx10^(-2))/(0.5)=3xx10^(-2) "min"^(-1)`
"For first order reaction"
Half-life,period `t_(t//2)=(0.693)/(K)=(0.693)/(3xx10^(-2))=23.1"min`
Discussion
1710.

The rate constant fro a second order reaction is `8 xx 10^(-5) M^(-1) "min"^(-1)`. How long will in take a `1 M` solution to be reduced to `0.5 M`A. `8 xx 10^(-5)` minB. `8.665 xx 10^(3)` minC. `4 xx 10^(-5)` minD. `1.25 xx 10^(4)`min

Answer» Correct Answer - d
`8 xx 10^(-5) = (1)/(t) [(1)/(0.5) - (1)/(1)] , 8 xx 10^(-5) = (1)/(t)[2-1]`
`t = (1)/(8 xx 10^(-5)) = 0.125 xx 10^(5) = 1.25 xx 10^(4)` min.
Discussion
1711.

The half life of a first order reaction is 30 min and the initial concentration of the reactant is 0.1 M . If the initial concentration of reactant is doubled ,then the half life of the reaction will beA. 1800SB. 60 minC. 15 minD. 900s

Answer» Correct Answer - A
Discussion
1712.

The rate constant of a second order reactions `2A rarr` Products, is `10^(-4) "lit mol"^(-1) "min"^(-1)`. The initial concentration of the reactant is `10^(-2) "mol lit"^(-1)`. What is the half life (in min) ?A. 10B. 1000C. 100D. `10^(6)`

Answer» Correct Answer - d
Half-life period for a second order reaction
`t_(1//2) = (1)/("rate constant " xx "initial concentration")`
`= (1)/(10^(-4) xx 10^(-2)) = 10^(6)` min .
Discussion
1713.

A reaction, which is second order,has a rate constant of `0.002 L mol^(-1)s^(-1)`. If the initial cond. Of the reactant is `0.2M`. How long will it take for the concentration to become `0.0400M` ?A. `1000s `B. `400 s`C. `200 s`D. `10,000s`

Answer» Correct Answer - 4
`(1)/(C_(1))=(1)/(C_(0))+KL`
`(1)/(0.04)=(1)/(0.02)+0.002xxt.`
`rArr" "25=5+0.002xxt`
`rArr" "t=(20)/(2xx10^(-3))=10,000sec.`
Discussion
1714.

From the following data answer the questions Reaction `A+B rarr P` `{:([A]M,,[B]M,,Initially rate (M sec^(-1)),,,),(,,,,at 300K,at 400K,,),(2.5xx10^(-4),,3.0xx10^(-5),,5.0xx10^(-4),2.0xx10^(-3),,),(5.0xx10^(-4),,6.0xx10^(-5),,4.0xx10^(-3),,,),(1.0xx10^(-3),,6.0xx10^(-5),,1.6xx10^(-2),,,):}` The value of rate constant at 300 K is `(M^(-2)sec^(-1))`A. `2.667xx10^(8)`B. `2.667xx10^(5)`C. `2.667xx10^(4)`D. `2.667xx10^(9)`

Answer» Correct Answer - A
Discussion
1715.

From the following data answer the questions Reaction `A+B rarr P` `{:([A]M,,[B]M,,Initially rate (M sec^(-1)),,,),(,,,,at 300K,at 400K,,),(2.5xx10^(-4),,3.0xx10^(-5),,5.0xx10^(-4),2.0xx10^(-3),,),(5.0xx10^(-4),,6.0xx10^(-5),,4.0xx10^(-3),,,),(1.0xx10^(-3),,6.0xx10^(-5),,1.6xx10^(-2),,,):}` The energy of activation for reaction `(kcal//mol)` is `(log 2=0.3)`A. `1.68`B. `3.36`C. `6.72`D. `1.12`

Answer» Correct Answer - B
Discussion
1716.

Most reactions occure more rapidly at high temperatures than at low temperature. This is consistent with an increase in which property at higher temperatures? (P) Activation energy (Q) Collision energy (R) Rate constantA. P onlyB. Q onlyC. P and Q onlyD. Q and R only

Answer» Correct Answer - C
Discussion
1717.

A human body excretes certain material through sweating by law similar to radioactivity if technitium is injected in some form in human body the body exretes half the amount in 24 hours A patient is given an injection containing `.^(98)Tc` The isotope is radioactive with half life of 8 hours The activity just after the injection is `32muCi` What will be activity after 48 hrs of the overall excreted material till that time?A. `0.125muCi`B. `0.375 muCi`C. `0.5 muCi`D. `1 muCi`

Answer» Correct Answer - B
Discussion
1718.

The activation energy for a hypothetical reaction `A rarr X` is `12.49 kcal mol^(-1)`. If temperature is raised to `305` form `295 K`, the reaction rate increased by `0.002 kcal L^(-1) mol^(-1)` is almost equal toA. 0.6B. 1C. 0.5D. 0.2

Answer» Correct Answer - B
Discussion
1719.

The activation energy for a hypothetical reaction `A rarr X` is `12.49 kcal mol^(-1)`. If temperature is raised to `305` form `295 K`, the reaction rate increased by `0.002 kcal L^(-1) mol^(-1)` is almost equal toA. `60%`B. `100%`C. `50%`D. `20%`

Answer» Correct Answer - B
Discussion
1720.

The rate of reaction increases by the increase of temperature becauseA. colliison frequency is increased.B. Energy Products decreases.C. Fraction of molecules possesisng energy `ge E_(T)` (threshold energy) increases.D. Meachanism of a reaction is changed.

Answer» Correct Answer - C
Discussion
1721.

In a reaction carried out at `500 K, 0.001%` of the total number of colliisons are effective. The energy of activation of the reaction is approximatelyA. `15.8 kcal mol^(-1)`B. `11.5 kcal mol^(-1)`C. `12.8 kcal mol^(-1)`D. zero

Answer» Correct Answer - B
The fraction of molecules having energy equal to or greater than `E_(a)` is :
`x = (n)/(N) = e^(-E_(a)//RT) (x = (n)/(N) = 0.001% = (0.001)/(100) = 10^(-5))`
`:. log x = (-E_(a))/(2.3 xx RT) = (R = 2 cal mol^(-1) K^(-1), T = 500 K)`
`log 10^(-5) = (-E_(a))/(2.3 xx 2 xx 500)`
`E_(a) = 11.5 xx 103 cal mol^(-1) = 11.5 kcal mol^(-1)`
Discussion
1722.

For profucing effective colliisons, the colliding molecules must haveA. A certain minimum amount of energyB. Energy equal to or greater than thresholdC. Proper orientationD. threshold energy and proper orientation both

Answer» Correct Answer - D
Discussion
1723.

For producing effective collisions, the colliding molecules must haveA. certain minimm amount of energyB. energy equal to or greater than the threshold energyC. Proper orientationD. threshold energy and proper orientation both

Answer» Correct Answer - D
Discussion
1724.

For producing effective collisions, the colliding molecules must haveA. minimum potential energyB. sufficient kinetic energyC. sufficient potential energyD. maximum energy of activation

Answer» Correct Answer - B
Discussion
1725.

For the elementary reaction `MrarrN`, the rate of disappearance of `M` increases by a factor of `8` upon doubling the concentration of `M`. The order of the reaction will respect to `M` isA. 4B. 3C. 2D. 1

Answer» Correct Answer - b
Discussion
1726.

For the reaction `X+3Y rarrZ`, which form of differential rate law is incorrect?A. `dX//dt = dY//3dt`B. `3dZ//dt = -dY//dt`C. `dZ//dt = -dX//dt`D. `dX//dt = dZ//dt`

Answer» Correct Answer - D
`X+3Y rarr Z`
Rate `= (-d[X])/(dt) = (-d[Y])/(3 dt) = (d[Z])/(dt)`
Discussion
1727.

For a chemical reaction `2X+YrarrZ`, the rate of appearance of Z is 0.05 mol `L^(-1)`. The rate of disappearance of X will beA. 0.05 mol `L^(-1) min^(-1)`B. 0.05 mol `L^(-1) h^(-1)`C. 0.1 mol `L^(-1) min^(-1)`D. 0.25 mol `L^(-1) min^(-1)`

Answer» Correct Answer - C
`2X + Y rarr Z`
Hence, rate expression for the reaction will be
`-(1)/(2)(d[X])/(dt)=(d[Y])/(dt)=+(d[Z])/(dt)`
`therefore" "-(d[X])/(dt)=2xx0.05 " mol "L^(-1)" min"^(-1)`
`=0.1 " mol "L^(-1)" min"^(-1)`
Discussion
1728.

For producing effective collisions, the colliding molecules must haveA. Have energy aqual to or greater than the threshold energyB. Have proper orientationC. Acquire the energy of activationD. All of these

Answer» Correct Answer - d
These are the characteristics of effective collisions.
Discussion
1729.

In acidic medium, the rate of reaction between `BrO_(3)^(ɵ)` and `Br^(ɵ)` is given by the expression `(-d[BrO_(3)^(ɵ)])/(dt) =k[BrO_(3)^(ɵ)][Br^(ɵ)][H^(o+)]^(2)`A. The rate constant of overall reaction is `4s^(-1)`.B. The rate of reaction is independent of the concentration of acid.C. The change in `pH` of the solution will no affect the rate.D. Doubling the concentration of `H^(o+)` ions will increase the reaction rate by `4` times.

Answer» Correct Answer - D
Order w.r.t. `[H^(o+)] = 2`.
Hence, doubling the concentration of `[H^(o+)]` ions will increase the reaction rate by `4` times.Correct Answer - D
Order w.r.t. `[H^(o+)] = 2`.
Hence, doubling the concentration of `[H^(o+)]` ions will increase the reaction rate by `4` times.
Discussion
1730.

For the reaction `A + B rarr C`, it is found that doubling the concentration of `A` increases the rate by `4` times, and doubling the concentration of `B` doubles the reaction rate. What is the overall order of the reaction ?A. `4`B. `3//2`C. `3`D. `1`

Answer» Correct Answer - C
`A + B rarr C`
On doubling the concentration of `A` rate of reaction increases by four times, Rate `prop [A]^(2)`
However on doubling the concentration of `B`, rate of reaction increases two times, Rate `prop [B]`
Thus, overall order of reaction `= 2 + 1 = 3`.
Discussion
1731.

For a chemical reaction `2X + Y rarrZ`, the rate of appearance of `Z` is `0.05 mol L^(-1) min^(-1)`. The rate of diappearance of `X` will beA. `0.05 mol L^(-1) hr^(-1)`B. `0.05 mol L^(-1) min^(-1)`C. `0.1 mol L^(-1) min^(-1)`D. `0.25 mol L^(-1) min^(-1)`

Answer» Correct Answer - C
`2X+Y rarr Z`
Rate `= (-d[X])/(2 dt) =(-d[Y])/(dt) = (+d[Z])/(dt)`
`:. (-d[X])/(dt) = 2 xx (d[Z])/(dt) = 2 xx 0.05 = 0.1 mol L^(-1) min^(-1)`Correct Answer - C
`2X+Y rarr Z`
Rate `= (-d[X])/(2 dt) =(-d[Y])/(dt) = (+d[Z])/(dt)`
`:. (-d[X])/(dt) = 2 xx (d[Z])/(dt) = 2 xx 0.05 = 0.1 mol L^(-1) min^(-1)`
Discussion
1732.

For a chemical reaction `A rarr B`, it is found that the rate of reaction doubles when the concentration of `A` is increased `4` times. What is the order of reaction ? Suggest the rate law also.A. `4`B. `0`C. `1//2`D. `1`

Answer» Correct Answer - C
`r_(1) = k[A]^(n)` …(i)
`r_(2)=2r_(1) = k[4A]^(n)` …(ii)
Divide Eq. (ii) by, Eq. (i),
`(2)^(1) = (4)^(n) = (2)^(2n)`
`:. 2n = 1`
`n = (1)/(2)`Correct Answer - C
`r_(1) = k[A]^(n)` …(i)
`r_(2)=2r_(1) = k[4A]^(n)` …(ii)
Divide Eq. (ii) by, Eq. (i),
`(2)^(1) = (4)^(n) = (2)^(2n)`
`:. 2n = 1`
`n = (1)/(2)`
Discussion
1733.

A reaction `A_(2)+B_(2)rarr 2AB` occurs by the following mechanism: `Ararr A+A`… (slow) `A+B_(2)rarr AB+B`… (fast) `A+Brarr AB` … (fast) Its order would beA. `3//2`B. `1`C. ZeroD. `2`

Answer» Correct Answer - B
Slowest step is
`A_(2) rarr A + A (slow)`
Hence, `r = k[A_(2)]`
Therefore, `OR = 1`Correct Answer - B
Slowest step is
`A_(2) rarr A + A (slow)`
Hence, `r = k[A_(2)]`
Therefore, `OR = 1`
Discussion
1734.

A reaction `A_(2)+B_(2)rarr 2AB` occurs by the following mechanism: `A_(2)rarr A+A`… (slow) `A+B_(2)rarr AB+B`… (fast) `A+Brarr AB` … (fast) Its order would beA. `3//2`B. `1`C. ZeroD. `2`

Answer» Correct Answer - b
The rate expression is derived for slowest step of mechanism.
Discussion
1735.

If doubling the concentration of a reactant `X` in a reaction `X+Y rarr` Products, increases the rate times and tripling its concentration increases the rate nine times, this indicates that the rate of reaction is proportional to the ………….of the concentration of.............and thus rate is given by........... .

Answer» Correct Answer - square, `X, k[X]^(2)`
Discussion
1736.

For the reaction A + B `to` products , it is observed that : (i) On doubling the initial concentration of A only , the rate of reaction is also doubled and (ii) On doubling the intial concentration of both A and B , there is a change by a factor of 8 in the rate of the reaction The rate of this reaction is given byA. Rate = `k [A]^(2) [B]`B. Rate = ` k [A] [B]^(2)`C. Rate = k `[A]^(2) [B]^(2)`D. Rate = `k [A] [B]`

Answer» Correct Answer - b
When concentration A is doubled , rate is doubled . Hence order with respect to A is one.
When concentration of both A and B are doubled , rate increased by 8 times hence total order is 3 .
`therefore` Rate = k `[A]^(1) [B]^(2)`
Order = 1 + 2 = 3 .
Discussion
1737.

A following mechanism has been proposed for a reaction: `2A+Brarr D+E` `A+B rarr C+D` (slow) `A+C rarr E` (fast) The rate law expresison for the reaction isA. `r = k[A]^(2)[B]`B. `r = k[A][B]`C. `r = k[A]^(2)`D. `r = k[A][C]`

Answer» Correct Answer - B
Slow step is
`A+B rarr C+D (slow)`
Hence, `r = k[A][B]` Correct Answer - B
Slow step is
`A+B rarr C+D (slow)`
Hence, `r = k[A][B]`
Discussion
1738.

For the reaction `A + B rarr` products, doubling the concentration of `A` the rate of the reaction is doubled, but on doubling the concentration of `B` rate reamins unaltered. The overall order of the reaction isA. `1`B. `0`C. `2`D. `3`

Answer» Correct Answer - A
For first order reactions rate is depend on the concetration of one reactant.
Discussion
1739.

For the decompoistion of `HI` at `1000 K(2HI rarr H_(2)+I_(2))`, following data were obtained: `|{:([HI] (M),"Rate of decomposition of HI" (mol L^(-1) s^(-1))),(0.1,2.75 xx 10^(-8)),(0.2,11 xx 10^(-8)),(0.3,24.75 xx 10^(-8)):}|` The order of reaction isA. 1B. 2C. 0D. `1.5`

Answer» Rate `k=[HI]^(n)`
`11xx10^(-8)=k[0.2]^(n)" "....(i)`
`2.75xx10^(8)=k[0.1]^(n)" "....(ii)`
Divinding eq. (i) by eq. (ii),
`4=2^(n),n=2`
Discussion
1740.

The rate constant `(K)` for the reaction, `2A+Brarr` Product was found to be `2.5xx10^(-5) litre mol ^(-1) sec^(-1)` after `15 sec, 2.60xx10^(-5) litre mol^(-1) sec^(-1)` after `30` sec and `2.55xx10^(-5) litre mol^(-1) sec^(-1)` after `50 sec`. The order of reaction is:A. `2`B. `3`C. ZeroD. `1`

Answer» Correct Answer - a
`K` does not change with time, Also unit of `K` suggest it to be II order.
Discussion
1741.

For the decompoistion of `HI` at `1000 K(2HI rarr H_(2)+I_(2))`, following data were obtained: `|{:([HI] (M),"Rate of decomposition of HI" (mol L^(-1) s^(-1))),(0.1,2.75 xx 10^(-8)),(0.2,11 xx 10^(-8)),(0.3,24.75 xx 10^(-8)):}|` The order of reaction isA. `1`B. `2`C. `0`D. `1.5`

Answer» Correct Answer - B
(b) Rate `=k[HI]^(n)`
`2.75 xx 10^(-8) = k[0.1]^(n)` …(i)
`11 xx 10^(-8) = k[0.2]^(n)` …(ii)
Dividing Eq. (ii) by Eq. (i),
`4 = 2^(n) rArr n = 2`
Discussion
1742.

The first order rate constant for the decompoistion of `C_(2)H_(5)I` by the reaction. `C_(2)H_(5)I(g)rarrC_(2)H_(4)(g)+HI(g)` at `600 K is 1.60xx10^(-5)s^(-1)`. Its energy of activation is `209 kJ mol^(-1)`. Calculate the rate constant at `700 K`

Answer» `log k_(2)/k_(1) = E_(a)/(2.303)[(T_(2)-T_(1))/(T_(1)T_(2))]`
`E_(a) = 209 kJmol^(-1) = 209 xx 10^(3)Jmol^(-1), T_(1)=600K, T_(2)=700K`,
`log k_(2)/k_(1)=(209 xx 10^(3)Jmol^(-1) xx 100K)/(2.303 xx (8.314 JK^(-1)mol^(-1)(600K) xx (700K)) = 2.599`
`k_(2)/k_(1) = "Antilog" 2.599 =397.2`
`k_(2)=397.2 xx k_(1) =397.2 xx (1.6 xx 10^(-5)s^(-1)) =6.36 xx 10^(-3)s^(-1)`
Discussion
1743.

Diazonium salt decomposes as `C_(6)H_(5)N_(2)^(+)Cl^(-) rarr C_(6)H_(5)Cl + N_(2)`. At `0^(@)C`, the evolution of `N_(2)` becomes two times faster when the initial concentration of the salt is doubled. Therefore, it isA. A first order reactionB. A second order reactionC. Independent of the initial concentration of the saltD. A zero order reaction

Answer» Correct Answer - a
As doubling the initial conc., doubles the rate of reaction , order = 1 .
Discussion
1744.

Diazonium salt decomposes as `C_(6)H_(5)N_(2)^(o+)Cl^(ɵ) rarrC_(6)H_(5)Cl+N_(2)` at `10^(@)C`, the evolution of `N_(2)` becomes two first faster when the initial concentration of the salt is doubled. Thus, it isA. A first order reactionB. A second order reactionC. Independent of the initial concentration of reactantD. A zero order reaction.

Answer» Correct Answer - A
`r_(1) = k["salt"]^(n)`
`r_(2) = 2r_(1) = k[2 "salt"]^(n)`.
`(2r_(1))/(r_(1)) = (2)^(n)`
`(2)^(1) = (2)^(n)`
`n = 1` Correct Answer - A
`r_(1) = k["salt"]^(n)`
`r_(2) = 2r_(1) = k[2 "salt"]^(n)`.
`(2r_(1))/(r_(1)) = (2)^(n)`
`(2)^(1) = (2)^(n)`
`n = 1`
Discussion
1745.

If `I` is the intenisty of an absorbed light and `c` is the concentration of `AB` for the photochemical process. `AB+hv rarr AB^(**)`, the rate of formation of `AB^(**)` is directly proportional toA. `c`B. `I`C. `I^(2)`D. `c xx I`

Answer» Correct Answer - B
For photochemical reactions, the rate of reaction is directly proportional to the intenisty of light.Correct Answer - B
For photochemical reactions, the rate of reaction is directly proportional to the intenisty of light.
Discussion
1746.

Diazonium salt decomposes as `C_(6)H_(5)N_(2)^(+)Cl^(-) rarr C_(6)H_(5)Cl + N_(2)`. At `0^(@)C`, the evolution of `N_(2)` becomes two times faster when the initial concentration of the salt is doubled. Therefore, it isA. a first order reactionB. a second order reactionC. independent of the initial concentration of the salt.D. a zero order reaction

Answer» Correct Answer - A
Discussion
1747.

If `I` is the intenisty of an absorbed light and `c` is the concentration of `AB` for the photochemical process. `AB+hv rarr AB^(**)`, the rate of formation of `AB^(**)` is directly proportional toA. (a) `C`B. (b) `I`C. (c ) `I^(2)`D. (d) `C. I`

Answer» Correct Answer - b
In photo initiated primary process, rate of reaction is directly proportional to intensity of light uses.
Discussion
1748.

For the decomposition of azoisopropane to hexane and nitrogen at 54 K, the following data are obtained. Calculate the rare constant.

Answer» The decomposition of azoisopropane to hexane and nitrogen at
543 K is represented by the following equation.
`(CH_(3))_2 CHN=NCH(CH_(3))_(2(g))toN_(2(g))+C_(6)H_(14(g))`
`{:("At "t=0," "P_(0),0,0),("At "t=t,P_(0)-P,P,P):}`
After time, t, total pressure,
`P_(t)=(P_(0)-p)+p+p`
`rarr P_(t)=P_(0)+p`
`rArr p=P_(t)-P_(0)`
Therefore, `P_(0)-p=P_(0)-(P_(0)-P_(0))`
`=2P_(0)-P_(t)`
For a first order reaction,
`k =2.303/t"log"(P_(0))/(P_(0)-P)`
` =2.303/t"log"(P_(0))/(2P_(0)-P_(t))`
When `t =360 s_(1),`
` k=(2.303)/(720" s")"log" (35.0)/(2xx35.0-63.0)`
When t `720 s_(1)`
=`2.235xx10^(-3)s^(-1)`
Hence, the average value of rate constant is
`k=((2.175xx10^(-3))+(2.35xx10^(-3)))/2s^(-1)`
` 2.21xx10^(-3)s^(-1)`
Note: There is a slight variation in this answer and the one given in the NCERT textbook.
Discussion
1749.

Azoisopropane decomposes acording to the reaction: `(CH_(3))_(2)CHN = NCH(CH_(3))_(2)(g) overset(250-290^(@)C)rarr N_(2)(g) + C_(6)H_(14)(g)` It is found to be a first order reaction. If the initial pressure is `P_(0)` and pressure of the mixture at time `t` is `(P_(t))`, then the rate constant `(k)` would beA. `k = (2.303)/(t) log.(P_(0))/(2P_(0) -P_(t))`B. `k = (2.303)/(t) log.(P_(0) - P_(t))/(P_(0))`C. `k = (2.303)/(t) log.(P_(0))/(P_(0) - P-(t))`D. `k = (2.303)/(t) log.(2P_(0))/(2P_(0) - P_(t))`

Answer» Correct Answer - A
`{:(,(CH_(3))_(2)CH=NCH(CH_(3))_(2), rarr, N_(2) +, C_(6)H_(14)), (t=0,P_(0),,0,0),(t=t,P_(0)-x,,x,x):}`
`P_(t) = P_(0) - x + x + x = P_(0) + x`
`x = (P_(t)-P_(0))`
`(a-x) = (P_(0)-x) = P_(0) - (P_(t) - P_(0))`
`= 2P_(0) - P_(t)`
`k = (2.303)/(t) log.(a)/(a-x)`
`= (2.303)/(t) log.((P_(0)))/((2P_(0) - P_(t)))`
Discussion
1750.

Conisder the chemical reaction `N_(2)(g) + 3H_(2)(g) rarr 2NH_(3)(g)` The rate of this reaction can be expressed in terms of time derivatives of the concentration of `N_(2)(g), H_(2)(g)`, or `NH_(3)(g)`. Identify the correct relationship among the rate expresisons.A. `"rate "=- d[N_(2)]//dt =-1//3d [H_(2)]//dt =1//2d [NH_(3)]//dt`B. `"rate "=-d[N_(2)]//dt =-3d[H_(2)]//dt =2d[NH_(3)]//dt`C. `"Rate "=d[N_(2)]//dt =-3//3d [H_(2)]//dt = 1//2d[NH_(3)]//dt`D. `"rate "=d[N_(2)]//dt =-d[H_(2)]//dt 1//2s[NH_(3)]//dt`

Answer» Correct Answer - A
From the given equation , we have
`-(d)/(dt)[N_(2)]=-(1)/(3) (d)/(dt) [H_(2)]=(1)/(2) (d)/(dt)[NH_(3)]`
Discussion