InterviewSolution
Saved Bookmarks
InterviewSolution
This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 1751. |
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. (a) `Rate=(-d[N_(2)])/(dt)=(-1)/(3)(d[H_(2)])/(dt)=1/2(d[NH_(3)])/(dt)`B. (b) `Rate=(-d[N_(2)])/(dt)=(3d[H_(2)])/(dt)=(2d[NH_(3)])/(dt)`C. (c ) `Rate=(d[N_(2)])/(dt)=(1)/(3)(d[H_(2)])/(dt)=1/2(d[NH_(3)])/(dt)`D. (d) `Rate=(-d[N_(2)])/(dt)=(-d[H_(2)])/(dt)=(d[NH_(3)])/(dt)` |
| Answer» Correct Answer - a | |
| 1752. |
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)/(3) (d[H_(2)])/(dt) = (1)/(2) (d[NH_(3)])/(dt)`B. `Rate = -(d[N_(2)])/(dt)= -3 (d[H_(2)])/(dt)= 2 (d[NH_(3)])/(dt)`C. `Rate = (d[N_(2)])/(dt) = (1)/(3) (d[H_(2)])/(dt)= -(1)/(2) (d[NH_(3)])/(dt)`D. `Rate = -(d[N_(2)])/(dt)= -(d[H_(2)])/(dt) = (d[NH_(3)])/(dt)` |
|
Answer» Correct Answer - A The average rate and the instananeous rate are equal for only one instant in any time interval. As the time interval becomes smaller, the average rate becomes a better and better approximation of the instantaneous rate. The average rate will be the ssame as the instantaneous rate when the time interval is zero. Instanteous rate `=underset(Deltatrarr0)(lim)(-Delta[A])/(Deltat)` The instantaneous rate is the limit of the average rate as `Deltat` approaches zero. Using the notation of ther calculus, we write the expression for this limit as `-d[A]//dt` . The rate of most chemical reaction depends in some way on the concentration of one or more of the reaction proceeds, their concentrations decrease and the rate of the reaction decreases. The instantaneous rate changes continuously. At first, the instantaneous rate is higher than the average rate but at the end of the interval, the instantaneous rate is lower than the average rate. |
|
| 1753. |
Consider the chemical reaction `N_(2) (g) + 3H_(2)(g) to 2NH_(3) (g)`. The rate of this reaction can be expressed in terms of time derivated of concentration of `N_(2)(g) , H_(2) (g)` or `NH_(3) (g)` . Identify the correct relationship amongst the rate expressions .A. Rate = -d[`N_(2)`]/dt = `-1//3` d [ `H_(2)`] / dt = 1/2 [`NH_(3)`] /dtB. Rate = `-d[N_(2)`]/dt = `-3d[H_(2)]`/dt = 3d `[NH_(3)`]/dtC. Rate = d`[N_(2)`] /dt = 1/3d[`H_(2)`]/dt = 1/2 d `[NH_(3)]`/dtD. Rate = `-d[N_(2)]`/dt = `-d[H_(2)]`/dt = d`[NH_(3)]`/dt |
|
Answer» Correct Answer - c Individual rates become equal when eachof these are divided by their respective stoichiometric coefficient . Further the rate of the consumption of the reactant goes on decreasing and the rate of formation of the product goes on increasing with time hence -ve and +ve sign , respectively . |
|
| 1754. |
The reaction `X to` Product, follows first order kinetics . In 40 minutes the concentration of X changes from 0.1 to 0.025 M . The rate of reaction , when concentration of X is 0.01 M isA. `1.73 xx 10^(-4) M "min"^(-1)`B. `3.47 xx 10^(-5) M "min"^(-1)`C. `3.47 xx 10^(-4) M "min"^(-1)`D. `1.73 xx 10^(-5) M "min"^(-1)` |
|
Answer» Correct Answer - c Since 0.1 M of X changes to `0.025` M in 40 minutes , `t_(1//2)` of reaction = 40/2 = 20 minutes Rate of reaction of X = k[X] = `(0.693)/(t_(1//2)) xx [X] = (0.693)/(20) xx 0.01 = 3.47 xx 10^(-4) M "min"^(-1)`. |
|
| 1755. |
The decomposition of a substance follows first order kinetics. If its concentration is reduced to 1/8 th of its initial value in 12 minutes, the rate constant of the decomposition system isA. `((2.303)/(12) log (1)/(8)) "min"^(-1)`B. `((2.303)/(12) log 8) "min"^(-1)`C. `((0.693)/(12)) "min"^(-1)`D. `((1)/(12) log 8) "min"^(-1)` |
|
Answer» Correct Answer - B `k = (2.303)/(12) "log" (a)/(a - x)` (for first order) `k = (2.303)/(12) "log" (1)/(1//8) = ((2.303)/(12) "log" 8) "min"^(-1)` |
|
| 1756. |
A first order reaction takes 40 min for `30%` decomposition. Calculate `t_(1//2)`. (Given `log 7 =0.845)`A. 77.7 minB. 52.5 minC. 46.2minD. 22.7 min |
|
Answer» Correct Answer - A 30% decomposition means X = 30% of `R_(0)` of `R = R_(0) - 0.3 R_(0) = 0.7 R_(0)` For first order, `k = (2.303)/(t) "log" ([R_(0)])/([R])` `= (2.303)/(40) "log" (10)/(7) "min"^(-1) = 8.918 xx 10^(-3) "min"^(-1)` `t_(1//2) = (0.693)/(k) = (0.693)/(8.918 xx 10^(-3) "min"^(-1)) = 77.7 "min"` |
|
| 1757. |
A first order reaction takes 40 minutes for `30%` decomposition of rectants. Calculate its half life period. |
|
Answer» Step-I: Calculation of rate constant: a=`100%, (a-x)=70%, t=40` min For first order reaction, `k=(2.303)/t log (a)/(a-x) =(2.303)/(40"min") log 100/70=(2.303 xx 0.1549)/(40"min") =0.0089 min^(-1)` Step-II Calculation of half life period `(t_(1//2))` `t_(1//2) = 0.693/k =(0.693)/(0.0089 min^(-1)) = 78 min` |
|
| 1758. |
A first order reaction takes 40 minutes for `30%` decomposition. Calculate its half life period. |
|
Answer» For the first order reaction, `k=2.303/k xx a/(a-x)` `a=100%, x=30%, (a-x) = (100-30) =70%,` t=40 min `a=100%, x=30%, (a-x)=(100-30)=70%`, t=40min `k =(2.303)/(40min)log 100/70 = 2.303/(40 min) xx 0.1549 = 0.00892 min^(-1)` `t_(1//2) = 0.693/k= 0.693/(0.00892 min^(-1)) = 77.7 min` |
|
| 1759. |
A first-order reaction which is `30%` complete in `30` minutes has a half-life period ofA. `24.2` minB. `58.2` minC. `102.2` minD. `120.2` min |
|
Answer» Correct Answer - B `k = (2.303)/(r )"log"(a)/(a-x)` `(0.693)/(T) = (2.303)/(t) "log"(100)/(100 - 30)` `T = 58.2 min` |
|
| 1760. |
`A(g)overset(Delta)rarrP(g)+Q(g)+R(g),` follows first order kinetics with a half-life of 69.3 s at `500^(@)C.` Starting from the gas A enclosed in a container at `500^(@)C` and at a pressure of 0.4 atm, the total pressure of the system after 230 s will beA. 1.15 atmB. 1.32 atmC. 1.22 atmD. 1.12 atm |
|
Answer» Correct Answer - D `A(g)rarrP(g)+Q(g)+R(g)` `{:(At " t"=0,0."4atm",0,0,0),(At" time "t,(0.4-x),x,x,x):}` `p_(t)=(0.4-x)+x+x+x=0.4+2x` `orx=(p_(t)-0.4)/(2)` `thereforep_(A)=p_(0)-x=p_(0)-(p_(t)-0.4)/(2)` `=(2xx0.4-p_(t)+0.4)/(2)=(1.2-p_(t))/(2)` `k=(2.303)/(t)"log"((p_(0))/(p_(A)))` `(0.693)/(t_(1//2))=(2.303)/(230)"log"((0.4xx2)/(1.2-p_(t)))` `(0.693)/(69.3)=(2.303)/(230)"log"((0.8)/(1.2-p_(t)))` `"log"((0.8)/(1.2-p_(t)))=(0.693xx230)/(69.3xx2.303)=0.9987` `(0.8)/(1.2-p_(t))=Antilog(0.9987)=10` `therefore" " 12-10p_(t)=0.8orp_(t)=1.12" atm"` |
|
| 1761. |
Assertion: The photochemical reaction `H_(2) + Cl_(2) rarr 2HCl` and `H_(2) + Br_(2) rarr 2HBr` have equal quantum efficiencies. Reason: Both the reaction proceed by different mechanism.A. If both assertion and reason are true and reason is the correct explanation of the assertion.B. If both assertion and reason are true and reason Iis not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
|
Answer» Correct Answer - D `H_(2) + Cl_(2) rarr 2HCl` has much higher quantum efficiency than `H_(2) + Br_(2) rarr 2HBr`. The first step of secondary process of `H_(2) + Cl_(2) rarr 2HBr` is exothermic while the same for `H_(2) + Br_(2) rarr 2HBr` is endothermic. |
|
| 1762. |
Assertion (A) : Both `H_(2)(g)+Cl_(2)(g) rarr 2HCl(g)` and `H_(2)(g)+Br_(2)(g) rarr 2HBr(g)` have the same order of reaction. Reason (R ): Both reaction proceed by the same mechanism.A. If both `(A)` and `(R )` are correct, and `(R )` is the correct explnation of `(A)`.B. If both `(A)` and `(R )` are correct, but `(R )` is noth the correct explanation of `(A)`.C. If `(A)` is correct, but `(R )` is incorrect.D. If both (A) and (R ) are incorrect. |
|
Answer» Correct Answer - D Correct (A) `H_(2)+Cl_(2) rarr 2HCl` is zero order while `H_(2)+Br(2) rarr` is of order `1.5`. Correct (R ) The two reactions proceed by different mechanism.Correct Answer - D Correct (A) `H_(2)+Cl_(2) rarr 2HCl` is zero order while `H_(2)+Br(2) rarr` is of order `1.5`. Correct (R ) The two reactions proceed by different mechanism. |
|
| 1763. |
For the reaction `H_(2) + Cl_(2) overset("Sunlight")(rarr) 2HCl` taking place on water, the order of reaction isA. `1`B. `2`C. `3`D. `0` |
|
Answer» Correct Answer - D The rate of this photochemical reaction is independent of the concentration, therefore, it is zeroorder reaction. |
|
| 1764. |
The half -life period for a first order reaction is 69.3 S . Its rate constant isA. `10^(-2) S^(-1)`B. ` 10^(-4) S^(-1)`C. `10S^(-1)`D. `10^(2)S^(-1)` |
|
Answer» Correct Answer - A Rate constant `=(0.693)/(t_(1//2)) =(0.693)/(69.3)=10^(-2) S^(-1)` |
|
| 1765. |
The rate constant of a reaction is `0.69 xx 10^(-1)` and the initial concentration is `0.2 "mol l"^(-1)`. The half-life period isA. `400 sec`B. `600 sec`C. `800 sec`D. `1200 sec` |
|
Answer» Correct Answer - B The unit of rate constant shows that reaction is of first-order. For first-orde reaction, half life is independent of initial conc. of the reactant. Thus, `t_(1//2) = (0.693)/(0.69 xx 10^(-1)) = (0.693 xx 60)/(0.69 xx 10^(-1)) = 600 sec` |
|
| 1766. |
The rate constant of a first-order reaction is `3 xx 10^(-6)` per second. If the initial concentration is `0.10 m`, the initial rate of reaction isA. `3 xx 10^(-5) ms^(-1)`B. `3 xx 10^(-6) ms^(-1)`C. `3 xx 10`D. `3 xx 10^(-7) ms^(-1)` |
|
Answer» Correct Answer - D Given: Rate constant of the first-order reaction `(K) = 3 xx 10^(-6)` per sec and initial concentration `[A] = 0.10 M`. We know that initial rate constant `K[A] =3 xx 10^(-6) xx 0.10 = 3 xx 10^(-7) ms^(-1)` |
|
| 1767. |
The rate constant of a first order reaction is `3 xx 10^(-6)` per second. If the initial concentration is `0.10 M`, the initial rate of reaction isA. `3 xx 10^(-5) Ms^(-1)`B. `3 xx 10^(-6) Ms^(-1)`C. `3 xx 10^(-8) Ms^(-1)`D. `3 xx 10^(-7) Ms^(-1)` |
|
Answer» Correct Answer - D Given: Rate constant of the first order reaction `(K) = 3 xx 10^(-6)` per sec and initial concentration `[A] = 0.10 M`. We know that initial rate constant `K[A] = 3 xx 10^(-6) xx 0.10 = 3 xx 10^(-7) ms^(-1)`. |
|
| 1768. |
Aqueous `AB_(2)` decomposes according to the first order reaction: `AB_(2)(aq) rarr A(g) + 2B(l)` After `20 min` the volume of `A(g) ` colledcted during such a reaction is `20 mL`, and that collected after a very long time is `40 mL`. The rate constant is :A. `A. 3.45 xx 10^(-3) min^(-1)`B. `B. 3.45 xx 10^(-2) min^(-1)`C. `C. 1.435 xx 10^(-2) min^(-1)`D. `D. 6.93 min^(-1)` |
|
Answer» Correct Answer - B `{:(,AB_(2)(aq) ,underset("first order")rarr,A(g),2B(l)),(t=0,a,,0,0),(t=20 min,(a-x),,x,2x),(t=oo min,0,,a,2a):}` Note: At constant `T` and `P`, mole `prop` pressure. `:. x prop 20` `alpha prop 40` `:. k = (2.3)/(t)log.((a)/(a-x))` `= (2.3)/(20 min) log. (40)/((40-20))` `= (2.3)/(20) xx log2 = (2.3 xx 0.3)/(20) = 3.45 xx 10^(-2) min^(-1)` |
|
| 1769. |
A substance X decomposes in a second-order reaction. A solution that is initially `1.00` M in X requires `0.50` h for its concentration to decrease to `0.50 M`. How much time will it take for a solution of X to decrease in concentration from `1.00` M to `0.25` M?A. `0.50` hB. `1.0` hC. `1.5` hD. `2.0` h |
|
Answer» Correct Answer - C |
|
| 1770. |
The rate constant of a first order reaction is `3 xx 10^(-6)` per second . If the initial concentration is `0.10` M , the initial rate of reactionA. `3 xx 10^(-5) M s^(-1)`B. `3 xx 10^(-6) M s^(-1)`C. `3 xx 10^(-8) Ms^(-1)`D. `3 xx 10^(-7) M s^(-1)` |
|
Answer» Correct Answer - d Given : Rate constant of the first order reaction `(k) = 3 xx 10^(-6)` per sec and initial concentration `[A] = 0.10` M . We know that initial rate constant `k [A] = 3 xx 10^(-6) xx 0.10 = 3 xx 10^(-7) Ms^(-1)`. |
|
| 1771. |
A substance A decomposes by a first order reaction starting initially with [A] = 2.00 m and after 200 min [A] = 0.15 m . For this reaction what is the value of kA. `1.29 xx 10^(-2) "min"^(-1)`B. `2.29 xx 10^(-2) "min"^(-1)`C. `3.29 xx 10^(-2) "min"^(-1)`D. `4.40 xx 10^(-2) "min"^(-1)` |
|
Answer» Correct Answer - a Given A(a) = `2.00` m, t = 200 min and a(a-x) = 0.15 m we know `k = (2.303)/(t) "log" (a)/(a-x) = (2.303)/(200) "log" (2.00)/(0.15)` `= (2.303)/(200) xx (0.301 + 0.824) = 1.29 xx 10^(-2) "min"^(-1).` |
|
| 1772. |
A First order reaction is half completed in 45 minutes . How long does it need 99.9 % of the reaction to be completedA. 5 hoursB. 7.5 hoursC. 10 hoursD. 20 hours |
|
Answer» Correct Answer - b `k = (0.693)/(45) "min"^(-1) = (2.303)/(t_(99.9%)) "log" (a)/(a-0.999 a)` or `t_(99.9%) = (2.303 xx 45)/(0.693) "log" 10^(3) = 448` min = 7.5 hrs. |
|
| 1773. |
A first order reaction is half completed in `45` minutes. How long does it need `99.9%` of the reaction to be completedA. `5` hoursB. `7.5` hoursC. `10` hoursD. `20` hours |
|
Answer» Correct Answer - B `k = (0.693)/(45)"min"^(-1) = (2.303)/(t_(99.9%))"log"(a)/(a-0.999a)` or `t_(99.9%) = (2.303xx45)/(0.693)log10^(3) = 448 min ~~ 7.5 hrs` |
|
| 1774. |
Set -I (without catalyst) `{:(,"Reaction",,"Temperature",,"E(activation)",,"k"),(,ArarrB,,T_(1)K,,Ea_(1),,k_(1)),(,A rarr B,,T_(2)K,,Ea_(2),,k_(2)):}` Set-II (with catalyst) (consider `+ve` catalyst only) `{:(,"Reaction",,"Temperature",,"E(activation)",,"k"),(,A rarr B,,T_(1)K,,Ea_(3),,k_(3)),(,A rarr B,,T_(2)K,,Ea_(4),,k_(4)):}` For the Set-I:A. If `T_(1) gt T_(2),k_(1)gt k_(2)` alwaysB. If `T_(1) gt T_(2),k_(1) gt k_(2)` (for exothermic reaction)C. If `T_(1) gt T_(2), k_(1) lt k_(2)` (for endothermic reaction)D. `Ea_(1) ne Ea_(2)` |
|
Answer» Correct Answer - A |
|
| 1775. |
Set -I (without catalyst) `{:(,"Reaction",,"Temperature",,"E(activation)",,"k"),(,ArarrB,,T_(1)K,,Ea_(1),,k_(1)),(,A rarr B,,T_(2)K,,Ea_(2),,k_(2)):}` Set-II (with catalyst) (consider `+ve` catalyst only) `{:(,"Reaction",,"Temperature",,"E(activation)",,"k"),(,A rarr B,,T_(1)K,,Ea_(3),,k_(3)),(,A rarr B,,T_(2)K,,Ea_(4),,k_(4)):}` For the Set-I:A. `Ea_(1)gt Ea_(2)if T_(1) gt T_(2)`B. `Ea_(1) lt Ea_(2) if T_(1) gt T_(2)`C. `Ea_(1) = Ea_(2)`D. `Ea_(1)=0.5 Ea_(2)` |
|
Answer» Correct Answer - C |
|
| 1776. |
For a first order reaction the rate constant is `6.909 "min"^(-1)`. The time taken for 75 % conversion in minutes isA. `(2)/(3) log2`B. `(2)/(3) log4`C. `(3)/(2) log2`D. `(3)/(2) log4` |
| Answer» Correct Answer - A | |
| 1777. |
In the formation of sulphur trioxide by the contact process, `2SO_(2)+O_(2) hArr 2SO_(3)`, the rate of reaction was measured as `(d[O_(2)])/(d t) = 3.0 xx 10^(-4) "mol" L^(-1)s^(-1)`. The rate of reaction expressed in terms of `SO_(3)` will beA. `3.0xx10^(-4) "mol" L^(-1)s^(-1)`B. `6.0xx10^(-4) "mol" L^(-1)s^(-1)`C. `1.5xx10^(-4) "mol" L^(-1)s^(-1)`D. `4.5xx10^(-4) "mol" L^(-1)s^(-1)` |
| Answer» Correct Answer - B | |
| 1778. |
In the formation of sulphur trioxide by the contact process, `2SO_(2)+O_(2) hArr 2SO_(3)`, the rate of reaction was measured as `(d[O_(2)])/(d t) = 3.0 xx 10^(-4) "mol" L^(-1)s^(-1)`. The rate of reaction expressed in terms of `SO_(3)` will beA. `3.0 xx 10^(-4) mol L^(-1) s^(-1)`B. `6.0 xx 10^(-4) mol L^(-1) s^(-1)`C. `1.5 xx 10^(-4) mol L^(-1) s^(-1)`D. `4.5 xx 10^(-4) mol L^(-1) s^(-1)` |
|
Answer» Correct Answer - B `-(d[O_(2)])/(d t) = +(1)/(2)(d[SO_(3)])/(d t)` `3 xx 10^(-4) = (1)/(2) (d[SO_(3)])/(d t)` `(d[SO_(3)])/(d t) = 6 xx 10^(-4) M//sec`. |
|
| 1779. |
Assertion: The acid hydrolysis of ester takes place more rapidly in `D_(2)O` than `H_(2)O`. Reason: `D_(3)O^(+)` is stronger acid tham `H_(3)O^(+)` or `H_(2)D^(+)O`.A. If both assertion and reason are true and the reason is the correct explanation of the assertion.B. If both assertion and reason are true but the reason is not the correct explanation of the assertion.C. If assertion is true but reason is false.D. If assertion is false but reason is true. |
|
Answer» Correct Answer - A Both assertion and reason are true and the reason is the correct explanation of the assertion. |
|
| 1780. |
Which of the foloowingf statements are correct for temperature dependence of an exothermic reversible reaction ?A. On increasing temperature, `K_(f)` increases while `K_(b)` decreases.B. On increasing temperature, both `K_(f)` and `K_(b)` increases.C. On increasing temperature, increases in `K_(b)` must be more than increase in `K_(f)`D. On increasing temperature percentage increase in `K_(b)` will be more than percentage increase in `K_(f)` |
|
Answer» Correct Answer - B::D |
|
| 1781. |
In Arrhenius equation, what does the factor `e^(Ea)//RT` correspond to? |
| Answer» This factors corresponds to the molecules which have kinetic energy greater than `E_(a)`. | |
| 1782. |
Consider the Arrhenius equation given below and mark the correct option. `k=A e^(-(Ea)/(RT)`A. Rate constant increases exponentially with the increasing activation energy and decreasing temperature.B. Rate constant decreases exponentially with increasing activation energy and decreasing temperature.C. Rate constant increases exponentially with decreasing activation energy and decreasing temperature.D. Rate constant increases exponentially with decreasing activation energy and increasing temperature |
|
Answer» Correct Answer - D |
|
| 1783. |
The half life period of a first order reaction is 10 minutes . The time required for the concentration of the reactant to change from `0.08 M to 0.02`is :A. 10 minB. 20 minC. 30 minD. 40 min |
| Answer» Correct Answer - B | |
| 1784. |
Find out the percentage of the reactant molecules crosisng over the energy barrier at `325 K`. Given: `Delta H_(325 K) = 0.12 kcal`, `E_(a(b)) = 0.02 kcal`A. `80.65%`B. `70.65%`C. `60.65%`D. `50.65%` |
|
Answer» Correct Answer - A `Delta H = E_(a(f)) - E_(a(b))` `E_(a(f)) = 0.12 xx 10^(3) + 0.02 xx 10^(3) cal` `= 0.14 xx 10^(3) cal` [The value of `R` in calories `= 1.98 cal ~~ 2.0 cal`] Fraction of molecules crosisng over the barrier `= (n)/(N) = x = e^(-E_(a(f)//RT))` `:. log x = (-E_(a(f)))/(2.303 RT) = (0.14 xx 10^(3) cal)/(2 xx 325)` `:. x = 0.8065` `%` of molecules crosisng over the barrier `= 0.8065 xx 100 = 80.65%` |
|
| 1785. |
Find out the percentage of the reactant molecules crosisng over the energy barrier at `325 K`. Given: `Delta H_(325 K) = 0.12 kcal`, `E_(a(b)) = 0.02 kcal` |
| Answer» Correct Answer - `80.65%;` | |
| 1786. |
When a catalyst is used an equilibrium process,A. it increases the rate of forward reactionB. it decreases the rate of backward reactionC. it decreases activation energy of forward process and decreases activation energy of backward processD. it fastens the attainment of equilibrium by lowering activation energy. |
|
Answer» Correct Answer - D The equilibrium is attained faster in case of reaction in which catalyst is used as it lowers the activation energy. |
|
| 1787. |
The rate constant for a first order reaction at `300^(@)C` for which `E_(a)` is 35 kcal `mol^(-1)` and frequency constant is `1.45 xx 10^(11) s^(-1)` isA. `10 xx 10^(2) s^(-1)`B. `5.37 xx 10^(10) s^(-1)`C. `5 xx 10^(-4) s^(-1)`D. `7.94 xx 10^(-3) s^(-1)` |
|
Answer» Correct Answer - D log `k = log A - (E_(a))/(2.303RT)` log k = log `(1.45 xx 10^(11)) - (35 xx 10^(3))/(2.303 xx 2 573)` log k = 11.16 - 13.26 = - 2.1 Taking antilog, `k = 7.94 xx 10^(-3) s^(-1)` |
|
| 1788. |
The graph of the effect of catalyst on activation energy is given below. Fill up the blanks X and Y with appropriate statements A. X = energy of activation without catalyst, Y = energy of activation with catalystB. X = path of reaction with catalyst, Y = path of reaction without catalystC. X = energy of activation with catalyst, Y = energy of activation without catalystD. X = energy of endothermic reaction, Y = energy of exothermic reaction |
|
Answer» Correct Answer - C Energy of activation is lowered when a catalyst is used |
|
| 1789. |
Which of the following is correct plot for effect of catalyst on activation energy.A. B. C. D. |
| Answer» Correct Answer - B | |
| 1790. |
A first order reaction has a rate constant of 0.0051 `min^(-1)`. If we begin with 0.10 M concentration of the reactant, What concentration of reactant will remain in solution after 3 hours? |
|
Answer» For a first order reaction, `k=2.303/tloga/(a-x)` k = 0.0051 `min^(-1)`, a-0.10M, t=3hr =`3 xx 60=180min`. `loga/(a-x) = (k xx t)/(2.03) =((0.0051 min^(-1)) xx (180 min))/(2.303) = 0.3986` `a/(a-x) = "antilog"(0.3986) = 2.053` `(a-x) = a/(2.503) = (0.10M)/(2.503)= 0.039 M` |
|
| 1791. |
For a two step reaction, `A hArr R+B " "R+C overset(k_(2)) rarr P` (Where , R is a reactive intermediate whose concentration is maintained at some low steady state throughout the reaction).If the concentration of C is very high then the order of reaction for formation of "P" is :A. 2B. 0C. 1D. `(1)/(2)` |
|
Answer» Correct Answer - C |
|
| 1792. |
Statement-2: In the reaction, `NO_(2)(g)+CO(g)rarrCO_(2)(g)+NO(g)` Rate =`k[NO_(2)]^(2)` The rate of reaction does not depend on the concentration of CO. Because Statement-2m: Carbon monoxide is involved in fast step.A. Statement-1 is trre, Statement-2 is true, Statement-2 is a correct explantion for Statement-2B. Statement-1 is true, Statement -2 is true, Statement-2 is not a correcte explanation for Statement-2C. Statement-1 is true, Statement-2 is false.D. |
| Answer» Correct Answer - A | |
| 1793. |
For a reaction `A + 2B rarr C`, rate is given by `R = K [A] [B]^(2)`. The order of reaction is:A. (a) `3`B. (b) `6`C. (c ) `5`D. (d) `7` |
|
Answer» Correct Answer - a Sum of power raised on concentration terms to express rate equation. |
|
| 1794. |
Statement-2: The reciprocal of time in which `66%` of the reactant is conveted to product is equal to the rate constant of first order reaction. Because Stetment-2 : The rate constant for first order reaction depends on initial concentration of reactants.A. Statement-1 is trre, Statement-2 is true, Statement-2 is a correct explantion for Statement-2B. Statement-1 is true, Statement -2 is true, Statement-2 is not a correcte explanation for Statement-2C. Statement-1 is true, Statement-2 is false.D. Both statement-1 and statement-2 are false |
| Answer» Correct Answer - C | |
| 1795. |
For the reaction `H_(2)(g)+I_(2)(g) hArr 2HI(g)`, the rate of reaction is expressed asA. (a) `(-d[H_(2)])/(dt)=(-d[I_(2)])/(dt)=(-d[HI])/(dt)`B. (b) `(d[H_(2)])/(dt)=(d[I_(2)])/(dt)=(d[HI])/(dt)`C. (c ) `1/2 (d[H_(2)])/(dt)=1/2(d[I_(2)])/(dt)=(-d[HI])/(dt)`D. (d) `-2(d[H_(2)])/(dt)=-2(d[I_(2)])/(dt)=(d[HI])/(dt)` |
|
Answer» Correct Answer - d `(d[H_(2)])/(dt)=(-d[I_(2)])/(dt)=1/2(d[HI])/(dt)` |
|
| 1796. |
Which statement is true about a reactant that appears in the balanced equations for a reaction but does not appear in the rate equation ?A. It is an inhibitorB. It is not part of the reactionC. Its concentration is too law to be important.D. It takes part in the reaction after the rate-determining step. |
|
Answer» Correct Answer - D |
|
| 1797. |
For the reaction `H_(2)(g)+I_(2)(g) hArr 2HI(g)`, the rate of reaction is expressed asA. `-(Delta[I_(2)])/(Delta t) = - (Delta[H_(2)])/(Delta t) = (1)/(2)(Delta[HI])/(Delta t)`B. `(Delta[I_(2)])/(Delta t) = - (Delta[H_(2)])/(Delta t) = (Delta[HI])/(2Delta t)`C. `(Delta[H_(2)])/(Delta t) = (1)/(2) (Delta[I_(2)])/(Delta t) = -(Delta[HI])/(Delta t)`D. None of these |
|
Answer» Correct Answer - A For given reaction: `-(Delta[I_(2)])/(Delta t) = -(Delta[H_(2)])/(Delta t) = (1)/(2) (Delta[HI])/(Delta t)` |
|
| 1798. |
The reaction `C_(3)H_(7)I+Cl^(-) rarr C_(3)H_(7)O +I^(-)` is thought to occur in the polar solvent `CH_(3)OH` by the mechanism: Step 1 :`C_(3)H_(7)I rarr C_(3)H_(7)^(+) + I^(-)` (slow) Step 2 : `C_(3)H_(7)^(+) + Cl^(-) rarr C_(3)H_(7)Cl` (fast) Which species is an intermediate in this reaction?A. `CH_(3)OH`B. `C_(3)H_(7)^(+)`C. `I^(-)`D. `Cl^(-)` |
|
Answer» Correct Answer - B |
|
| 1799. |
The oxidation of `SO_(2)` to `H_(2)SO_(4)` in acid rain is thought to occur by the following mechanism. `SO_(2)(aq)+2H_(2)O(l) rarr HSO_(3)^(-)(aq) +H_(3)^(+)(aq)` `2HSO_(3)^(-)(aq) +O_(2)(aq)rarr S_(2)O_(7)^(2-)(aq)+H_(2)O(l)` `S_(2)O_(7)^(2-)(aq) +3H_(2)O(l) rarr 2SO_(4)^(2-)(aq)+2H_(3)O^(+)(aq)` Which species in this mechanism can be given the following designation?A. `{:("Reactant","catayst","Intermediate"),(SO_(2)(aq),H_(2)O(l),HSO_(3)^(-)(aq)H_(3)O^(+)(aq)):}`B. `{:("Reactant","catayst","Intermediate"),(SO_(2)(aq),HSO_(3)^(-)(aq),S_(2)O_(7)^(2-)(aq)):}`C. `{:("Reactant","catayst","Intermediate"),(SO_(2)(aq)H_(2)O(l),S_(2)O_(7)^(2-)(aq),HSO_(3)^(-1)(aq)):}`D. `{:("Reactant","catayst","Intermediate"),(SO_(2)(aq)H_(2)O(l),"none",HSO_(3)^(-)(aq)S_(2)O_(7)^(2-)(aq)):}` |
|
Answer» Correct Answer - D |
|
| 1800. |
The rate constant of the reaction `2H_(2)O_(2)(aq) rarr 2H_(2)O(l) + O_(2)(g)` is `3 xx 10^(-3) min^(-1)` At what concentration of `H_(2)O_(2)`, the rate of the reaction will be `2 xx 10^(-4) Ms^(-1)` ?A. `6.67 xx 10^(-3) (M)`B. `2 (M)`C. `4 (M)`D. `0.08 (M)` |
|
Answer» Correct Answer - C `-(d[H_(2)O_(2)])/(d t) = k[H_(2)O_(2)]^(1)` `:. 2 xx 10^(-4) = (3 xx 10^(-3))/(60) xx [H_(2)O_(2)]` or `[H_(2)O_(2)] = 4 (M)` |
|