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.
| 151. |
Which of the following statements regarding rate constant is correct?A. Rate constant always depends on concentration of reactant.B. Rate constant is temperature dependent.C. For instantaneous reaction, rate constant will be very small.D. Rate constant will always depend on pressure or volume of the container. |
|
Answer» Correct Answer - B |
|
| 152. |
The rate law for hydrolysis of this acetamide, `CH_(3)CSNH_(2)`, is : `CH_(3)-underset(S)underset(||)C-NH_(2)+H_(2)Ooverset(H_(2)O)rarrCH_(3)-overset(O)overset(||)C-NH_(2)+H_(2)S` Rate= K [thioacetamide]`[H^(+)]` In which of the following solutions, will the rate of hydrolysis of thioacetamide (TA) is least at `25^(@)C` ?A. `0.1 "M TA" + 0.20"M HNO"_(3)`B. `0.1 "M TA" + 0.20"M CH"_(3)COOH_(3)`C. `0.1 "M TA" + 0.20"M HCOOH"`D. `0.15 "M TA" + 0.15"M HCl"` |
|
Answer» Correct Answer - B |
|
| 153. |
Assertion: The enthalapy of reaction remains constant in the presence of a catalyst. Reason: A catalyst participating in the reaction, forms different activated complex and lowers down the activation energy but the difference in energy of reactants lead to product formation.A. Both assertion and reason are correct and the reason is correct explanation of assertion.B. Both assertion and reason are correct but reason does not explain assertion.C. Assertion is correct but reason is incorrectD. Both assertion and reason are incorrect. |
|
Answer» Correct Answer - A |
|
| 154. |
Assertion (A) The enthalpy of reaction remains constant in the presence of a catalyst. Reason (R) A catalyst participating in the reaction froms different activated complex and lowers down the activation energy but the difference in energy of reactant and product remains the same.A. Both assertion and reason are correct and the reason in correct explanation of assertion.B. Both assertaion and reason are correct, but the reason does not explain essertionC. Assertion is correct but reason is incorrectD. Both assertion and reason are incrrect. |
|
Answer» Correct Answer - A Assertion ard reason both are correct and reason is the correct explanation of assertion. Enthalpy of reactior ie., difference of total enthalpy of reactants and product remains constant in the presence of a catalyst. As a catalyst participating in the reaction forms different activated complex and lowers down the activation erergy but the difference in energy of reactant and product remains same. |
|
| 155. |
In a reaction, the concentration of a reatant (A) changes from `0.200 "mol litre"^(-1) to 0.150 "mol litre"^(-1)` in 10 minutes. What is the average rate of reaction during this interval ? |
|
Answer» `Delta[A]=[A]_("initial")` `=[0.150-0.200]=0.050 "mole litre"^(-1)` `Deltat=10` minutes Average rate of reaction `=(-Delta[A])/(Deltat)=(-[-0.050])/(10)` `=(0.050)/(10)=0.050 "mol litre"^(-1)"min"^(-1)` |
|
| 156. |
Assertion (A) Rate constant determined form Arrhenius equations are fairly accurate for simple as well as complex molecules. Reason (R) Reatant molecules undergo chemical irrespective of their orientation during collison.A. Both assertion and reason are correct and the reason in correct explanation of assertion.B. Both assertaion and reason are correct, but the reason does not explain essertionC. Assertion is correct but reason is incorrectD. Both assertion and reason are incrrect. |
|
Answer» Correct Answer - C Assertion is correct, but reason is incorrcet. Rate constant determined form Arrhenius equation are farily accurate for simple and complex molecules because only those molecules which of have proper orientation during collision (i.e., effective collison) and sufficient kinetic lead the chemical change. |
|
| 157. |
Two reactiosn of same order have equal pre-exponential factors but their activation energies differ by 41.9 J/mol. Calculte the ratios between rate constant of these reactions at 600 K |
|
Answer» Correct Answer - `0.002` Use the relation `log_(10)k=log_(10)A-(E)/(2.303RT)" ".....(i)` `log_(10)k_(1)=log_(A)-(E_(1))/(2.303xx8.314xx600) " ".....(ii)` `log_(10)k_(2)=log_(A)-(E_(2))/(2.303xx8.314xx600) " ".....(iii)` `E_(1)-E_(2)=41.95//"mol"" ".....(iv)` Substract eq. (ii) from eq. (ii) to determine the ratio |
|
| 158. |
Rate constant of a reaction changes by `2%` by `0.1^(@)C` rise in temperataure at `25^(@)C`. The standard heat of reaction is `12.1kJ"mol"^(-1)` . Calculate `E_(a)` or reverse reaction. |
|
Answer» Correct Answer - `24.7kJ//"mol"` `log.((k_(2))/(k_(1)))=(E_(a))/(2.303RT)((1)/(T_(1))-(1)/(T_(2)))` `log.(102)/(100)=(E)/(2.303xx8.314)((1)/(298)-(1)/(298.1))` `E=1.463xx10^(5)J//"mol"=146.kJ//"mol"` `Delta=E_(f)-E_(b)` `131.6=146.3-E_(b)` `E_(b)=24.7 kJ//"mol"` |
|
| 159. |
The rate of decomposition for methyl nitrite and ethyl nitrite can be given in terms of rate constant `("in" sec^(-1)) K_(1)` and `K_(2)` respectively. The energy of activation for the two reaction are `152.30 kJ mol^(-1)` and `157.7 kJ mol^(-1)` as well as frequency factor are `10^(13)` and `10^(14)` respectively for the decomposition of methyl and ethyl nitrite. Calculate the temperature at which rate constant will be same for the two reactions. |
| Answer» Correct Answer - `282 K;` | |
| 160. |
The following data are for the decomposition of ammonium nitrite in aqueous solution.: `{:("Vol of" N_(2)(cm^(3)),6.25,9.0,11.42,13.65,35.2),("Time (min)",10,15,20,25,oo):}` The order of reaction is :A. 3B. 2C. 1D. zero |
|
Answer» Correct Answer - C `a=35.02,(a-x)=35.02-V_(t)` Apply, `k=(2.303)/(t)log.(35.02)/((35.02-V_(t))` |
|
| 161. |
The following data are for the decomposition of ammonium nitrite in aqueous solution.: `{:("Vol of" N_(2)(cm^(3)),6.25,9.0,11.42,13.65,35.2),("Time (min)",10,15,20,25,oo):}` The order of reaction is :A. ZeroB. OneC. TwoD. Three |
|
Answer» Correct Answer - B `K = (2.303)/(t)"log"(V_(oo))/(V_(oo) - V_(t))` gives constant value of `K`. With the given data. Hence it is 1st order. |
|
| 162. |
The rate constant, the activation energy and the Arrhenius parameter of a chemical reaction at `25^(@)C` are `3.0 xx 10^(-4) s^(-1), 104 "kJ mol"^(-1)` and `6.0 xx 10^(14)s^(-1)` respectively. The value of the rate constant as `T rarr oo` isA. `2.0 xx 10^(18)s^(-1)`B. `6.0 xx 10^(14) s^(-1)`C. InfinityD. `3.6 xx 10^(30)s^(-1)` |
|
Answer» Correct Answer - B The Arrhenius equation is `K = Ae^(-E_(a)//RT)` As `T rarr oo` the value `(Ea)/(RT) rarr 0` & `e^(-E_(a)//RT) rarr e^(-0) rArr 1` Hence `k rarr A` Hence `k = 6.0 xx 10^(14)s^(-1)` |
|
| 163. |
If the half-time for a particular reaction is found to be constant and independent of the initial concentration of the reactants, then the reaction is ofA. first orderB. zero orderC. second orderD. None of the above |
|
Answer» Correct Answer - A For a first order reaction the half-life period is independent of the initial concentration of the reactants. `t(1//2)=(o.693)/(K)` |
|
| 164. |
The rate constant of decomposition of methyl nitrite `(K_(1))` and ethyl nitrite `(K_(2))` follow the equations. `K_(1)(s^(-1))=10^(13)e^([(-152.3xx10^(3))/(RT)Jmol^(-1)])` and `K_(2)(s^(-1))=10^(14)e^([(-152.3xx10^(3))/(RT)Jmol^(-1)])` Calculate the temperature at which both have same rate of decomposition if `0.1M` of each is taken and both show `I` order kinetics. |
|
Answer» Rate= `K "[Reactant"]^(1)` Since [Reactant]=`0.1 M` and rates are same, thus at same temperature `K_(1)=K_(2)` `10^(13)e^((-152.3xx10^(3))/(RT))=10^(14) e^((-157.7xx10^(3))/(RT)` `e^(((+5.4xx10^(3))/(RT)))=10` `(5.4xx10^(3))/(RT)=2.303` `:. T=(5.4xx10^(3))/(8.314xx2.303)=282 K` |
|
| 165. |
The rate constant of a reaction is `1.5 xx 10^(7)s^(-1)` at `50^(@)C` and `4.5 xx 10^(7) s^(-1)` at `100^(@)C`. Evaluate the Arrhenius parameters `A` and `E_(a)`. |
|
Answer» We know, `2.303"log"_(10) K_(2)/K_(1)=E_(a)/R[(T_(2)-T_(1))/(T_(1)T_(2))]` `:. 2.303"log"_(10) (4.5xx10^(7))/(1.5xx10^(7))=E_(a)/8.314[(373-323)/(373xx323)]` `:. E_(a)=2.2xx10^(4) J mol^(-1)` Now `K=Ae^(-E_(a)//RT)` `:. 4.5xx10^(7)=Ae^(-(2.2xx10^(4))/(8.314xx373))` `:. A=5.42xx10^(10)` |
|
| 166. |
A reaction is said to be of…………. if its rate is entirely independent of the concentration of the reactants. |
| Answer» Correct Answer - zero order | |
| 167. |
The half`-` life periof of a radioactive element is 140 days. After 560 days, one gram of the element will reduce toA. `1/2(g)`B. `1/4 (g)`C. `1/8 (g)`D. `1/16 (g)` |
|
Answer» Correct Answer - d |
|
| 168. |
The half-life periof of a radioactive substance is `x` years. The fraction remaining after `2x` years is………….. . |
| Answer» Correct Answer - `1//4` | |
| 169. |
The time required for `10%` completion of a first order reaction at `298 K` is equal to that required for its `25%` completion at `308 K`. If the pre-exponential factor for the reaction is `3.56 xx 10^(9) s^(-1)`, calculate its rate constant at `318 K` and also the energy of activation. |
|
Answer» Correct Answer - `k = 9.3 xx 10^(-4) s^(-1)` `E_(a) = 76.6 kJ mol^(-1)` `t_(1) = (2.3)/(k_(1))log((100)/(100-25))` `t_(2) = (2.3)/(k_(2))log((100)/(100-25))` As `t_(1) = t_(2)`, ` (2.3)/(k_(1))log.((10)/(9)) = (2.3)/(k_(2))log.((4)/(3))` `:. (k_(2))/(k_(2)) = (log((4)/(3)))/(log((10)/(9))) = 2.73` Also, `log.(k_(2))/(k_(1)) = (E_(a))/(2.3 R)((T_(2)-T_(1))/(T_(1)T_(2)))` `log(2.73) = (E_(a))/(2.3 xx 8.314 J K^(-1) mol^(-1)) xx((10)/(298 xx 398))` `:. E_(a) = 76.6 xx 10^(3) J mol^(-1) = 76.6 kJ mol^(-1)` Further, `k = Ae^(-E_(a)//RT)`. or `ln k = ln A-(E_(a))/(RT)` `ln k = ln(3.56 xx 10^(9)s^(-1))` `- (76.6 K J mol^(-1))/(8.314 xx 10^(-3) K J mol^(-1) K^(-1) xx 318)` `:. k = 9.3 xx 10^(-4) s^(-1)`Correct Answer - `k = 9.3 xx 10^(-4) s^(-1)` `E_(a) = 76.6 kJ mol^(-1)` `t_(1) = (2.3)/(k_(1))log((100)/(100-25))` `t_(2) = (2.3)/(k_(2))log((100)/(100-25))` As `t_(1) = t_(2)`, ` (2.3)/(k_(1))log.((10)/(9)) = (2.3)/(k_(2))log.((4)/(3))` `:. (k_(2))/(k_(2)) = (log((4)/(3)))/(log((10)/(9))) = 2.73` Also, `log.(k_(2))/(k_(1)) = (E_(a))/(2.3 R)((T_(2)-T_(1))/(T_(1)T_(2)))` `log(2.73) = (E_(a))/(2.3 xx 8.314 J K^(-1) mol^(-1)) xx((10)/(298 xx 398))` `:. E_(a) = 76.6 xx 10^(3) J mol^(-1) = 76.6 kJ mol^(-1)` Further, `k = Ae^(-E_(a)//RT)`. or `ln k = ln A-(E_(a))/(RT)` `ln k = ln(3.56 xx 10^(9)s^(-1))` `- (76.6 K J mol^(-1))/(8.314 xx 10^(-3) K J mol^(-1) K^(-1) xx 318)` `:. k = 9.3 xx 10^(-4) s^(-1)` |
|
| 170. |
If `80%` of a radioactive element undergoing decay is left over after a certain periof of time `t` form the start, how many such periofs should elapse form the start for just over `50%` of the element to be left over? |
|
Answer» Correct Answer - 3 Let `a=100, (a-x)=80` (Amount left). `:.` Amount left `=a/((2)^(n)), [n= "number of half lives"]` `80=100/((2)^(n))implies(2)^(n)=10/8` `:. n log 2= log 10-3 log 2 =1-3xx0.3=0.1` `n=0.1/(log 2)=0.1/0.3=1/3` `n=1/3=(t(("Time for 80% amount"),("left or 20% decomposed")))/(t_(1//2)(("Time for 50% amount"),("left or decomposed")))implies t_(1//2)=3t` Alternatively `t_(30%("left"))/t_(1//2("left"))=(log(100/80))/0.3=1/3` `t_(1//2)=3t_(80%("left"))=3t`Correct Answer - 3 Let `a=100, (a-x)=80` (Amount left). `:.` Amount left `=a/((2)^(n)), [n= "number of half lives"]` `80=100/((2)^(n))implies(2)^(n)=10/8` `:. n log 2= log 10-3 log 2 =1-3xx0.3=0.1` `n=0.1/(log 2)=0.1/0.3=1/3` `n=1/3=(t(("Time for 80% amount"),("left or 20% decomposed")))/(t_(1//2)(("Time for 50% amount"),("left or decomposed")))implies t_(1//2)=3t` Alternatively `t_(30%("left"))/t_(1//2("left"))=(log(100/80))/0.3=1/3` `t_(1//2)=3t_(80%("left"))=3t` |
|
| 171. |
Assertion (A): Hydrolyiss of ethyl acetate in the presence of acid is a reaction of first order whereas in the presence of alkali, it is a reaction of second order. Reason (R ): Acid acts as catalyst only whereas alkali act as one of the reactant.A. If both (A) and (R) are correct but (R) is the correct explanation of (A) .B. If both (A) and (R) are correct but (R) is not the correct exaplanation of (A)C. If (A) is correct but (R) is incorrectD. If (A) is incorrect but (R) is correct |
| Answer» Correct Answer - A | |
| 172. |
For a reaction, `X(g) rarr Y(g) + Z(g)`. The half-life periof is `10 min`. In what periof of time would the concentration of `X` be reduced to `10%` of the original concentration ?A. `20` minB. `33` minC. `15` minD. `25` min |
|
Answer» Correct Answer - B `x_((g)) rarr y_((g)) + z_((g))` The reaction is a first-order reaction hence, `K = (0.693)/(t_(1//2)) = (2.303)/(t) "log"(a)/(a-x)` `= (0.693)/(10 min)` `= (2.303)/(t) "log"(a)/(a//10)` `= (0.693)/(10) = (2.303)/(t)log 10` `t = (2.303 xx 10)/(.693) = 33 min` |
|
| 173. |
If one starts with `1` Curie `(Ci)` of radioactive substance `(t_(1//2)=15 hr)` the activity left after a periof of two weeks will be about `0.02x muCi`. Find the value of `x`. |
|
Answer» Correct Answer - 9 `k=0.693/(15 hr)=0.0462 hr^(-1)` `k=2.3/(14xx24 hr) log c_(0)/c_(t)` `0.0462 hr^(-1)=2.3/(14xx24 hr) log.(1 Ci)/c_(t)` Solve for `c_(t):` `:. c_(t)=1.82xx10^(-7) Ci~~0.18 muCi=0.02x muCi` `:. x=9`Correct Answer - 9 `k=0.693/(15 hr)=0.0462 hr^(-1)` `k=2.3/(14xx24 hr) log c_(0)/c_(t)` `0.0462 hr^(-1)=2.3/(14xx24 hr) log.(1 Ci)/c_(t)` Solve for `c_(t):` `:. c_(t)=1.82xx10^(-7) Ci~~0.18 muCi=0.02x muCi` `:. x=9` |
|
| 174. |
The ionization constant of `overset(o+)(NH_(4))` ion in water is `5.6 xx 10^(-10)` at `25^(@)C`. The rate constant the reaction of `overset(o+)NH_(4)` and `overset(ɵ)(OH)` ion to form `NH_(3)` and `H_(2)O` at `25^(@)C` is `3.4 xx 10^(10)L mol^(-1) s^(-1)`. Calculate the rate constant for proton transfer form water to `NH_(3)`. |
|
Answer» Correct Answer - Rate constant `= 6.07 xx 10^(5) s^(-1)` `NH_(3) + H_(2)O underset(k_(b))overset(k_(f))hArr NH_(4)^(o+)+overset(Θ )(OH) (k_(b) = 3.4 xx 10^(10))` ...(i) `NH_(4)^(o+) + H_(2)O hArr NH_(4)OH +H^(o+) (k_(a) = 5.6 xx 10^(-10))` `k_(base)(NH_(3))=(k_(f))/(k_(b))=(k_(w))/(k_(acid)(NH_(4)^(o+)))` `( :. k_(acid)xxk_(base)=k_w)` `(k_(f))/(3.4xx10^(10))=(10^(-14))/(5.6xx10^(-10))impliesk_(f)= 6.07xx10^(5)` (Alternative methof) We are given that `NH_(4)^(o+) +H_(2)O hArr NH_(3) + H_(3)O^(o+) (k_(a) = 5.6 xx 10^(-10))` `NH_(4)^(o+) + overset(Θ)(OH) underset(k_(b))overset(k_(f))hArr NH_(3) + H_(2)O (k_(f) = 3.4 xx 10^(10) L mol^(-1) s^(-1))` The second equation can be generalized as followes: `{:(NH_(4)^(o+) + H_(2)O hArr NH_(3) + H_(3)O^(o+) (k_(a))),(" "2H_(2)O hArr H_(3)O^(o+) + overset(Θ )(OH) (k_(w))),(ul(bar("Subtract" : NH_(4)^(o+) + overset(Θ )(OH) hArr NH_(3)+H_(2)O))):}` ...(ii) `k_(eq)=(k_(a))/(k_(w))` and also `k_(eq)=(k_(f))/(k_(b))` `:. (k_(f))/(k_(b))=(k_(a))/(k_(w))` `k_(b)=k_(f)[["Note here that equation(ii)is reversed",],["of Eq.(i) So" k_(b)"is calculated",]]` `=(3.4xx10^(10)Lmol^(-1)s^(-1))((1.0xx10^(-14)mol^(2)L^(-2))/(5.6xx10^(-10)molL^(-1)))` `= 6.07xx10^(5)s^(-1)`Correct Answer - Rate constant `= 6.07 xx 10^(5) s^(-1)` `NH_(3) + H_(2)O underset(k_(b))overset(k_(f))hArr NH_(4)^(o+)+overset(Θ )(OH) (k_(b) = 3.4 xx 10^(10))` ...(i) `NH_(4)^(o+) + H_(2)O hArr NH_(4)OH +H^(o+) (k_(a) = 5.6 xx 10^(-10))` `k_(base)(NH_(3))=(k_(f))/(k_(b))=(k_(w))/(k_(acid)(NH_(4)^(o+)))` `( :. k_(acid)xxk_(base)=k_w)` `(k_(f))/(3.4xx10^(10))=(10^(-14))/(5.6xx10^(-10))impliesk_(f)= 6.07xx10^(5)` (Alternative methof) We are given that `NH_(4)^(o+) +H_(2)O hArr NH_(3) + H_(3)O^(o+) (k_(a) = 5.6 xx 10^(-10))` `NH_(4)^(o+) + overset(Θ)(OH) underset(k_(b))overset(k_(f))hArr NH_(3) + H_(2)O (k_(f) = 3.4 xx 10^(10) L mol^(-1) s^(-1))` The second equation can be generalized as followes: `{:(NH_(4)^(o+) + H_(2)O hArr NH_(3) + H_(3)O^(o+) (k_(a))),(" "2H_(2)O hArr H_(3)O^(o+) + overset(Θ )(OH) (k_(w))),(ul(bar("Subtract" : NH_(4)^(o+) + overset(Θ )(OH) hArr NH_(3)+H_(2)O))):}` ...(ii) `k_(eq)=(k_(a))/(k_(w))` and also `k_(eq)=(k_(f))/(k_(b))` `:. (k_(f))/(k_(b))=(k_(a))/(k_(w))` `k_(b)=k_(f)[["Note here that equation(ii)is reversed",],["of Eq.(i) So" k_(b)"is calculated",]]` `=(3.4xx10^(10)Lmol^(-1)s^(-1))((1.0xx10^(-14)mol^(2)L^(-2))/(5.6xx10^(-10)molL^(-1)))` `= 6.07xx10^(5)s^(-1)` |
|
| 175. |
How will the rate of reaction `2SO_(2)(g) + O_(2)(g) rarr 2SO_(3)(g)` change if the volume of the reaction vessel is halved?A. (a) It will br `1//16th` of its initial value.B. (b) it will be `1//4th` of its initial value.C. It will be `8` times of its initial value.D. It will be `4` times of its initial value. |
|
Answer» Correct Answer - C `V = (1)/(2)`, then concentration `= 2` times, Assuming the rate law as: `r_(1) = k[SO_(2)]^(2)[O_(2)]` `r_(2) = k[2SO_(2)]^(2)[2O_(2)]` `r_(2) = 8r_(1)`Correct Answer - C `V = (1)/(2)`, then concentration `= 2` times, Assuming the rate law as: `r_(1) = k[SO_(2)]^(2)[O_(2)]` `r_(2) = k[2SO_(2)]^(2)[2O_(2)]` `r_(2) = 8r_(1)` |
|
| 176. |
The hydrolyiss of an ester was carried out with `0.1 M H_(2)SO_(4)` and `0.1 M HCl` separately. Which of the following expresison between the rate constants is expected ? The rate expresison being rate `= k[H^(o+)]["ester"]`A. `k_(HCl) = k_(H_(2)SO_(4))`B. `k_(HCl) gt k_(H_(2)SO_(4))`C. `k_(HCl) lt k_(H_(2)SO_(4))`D. `k_(H_(2)SO_(4)) = 2k_(HCl)` |
|
Answer» Correct Answer - B `[H_(2)SO_(4)] = 0.1 M = 0.1 xx 2 = 0.2 N` `[HCl] = 0.1 N` In case of `H_(2)SO_(4)` `r_(1) = k[H^(o+)]["Ester"]` `k_(H_(2)SO_(4)] = (r_(1))/(2 N xx ["Ester"])` In case of `HCl`, `r_(2) = k[H^(o+)]["Ester"]` `k_(HCl) = (r_(2))/(1 N xx ["Ester"])` Hence `K_(HCl) gt K_(H_(2)SO_(4)`Correct Answer - B `[H_(2)SO_(4)] = 0.1 M = 0.1 xx 2 = 0.2 N` `[HCl] = 0.1 N` In case of `H_(2)SO_(4)` `r_(1) = k[H^(o+)]["Ester"]` `k_(H_(2)SO_(4)] = (r_(1))/(2 N xx ["Ester"])` In case of `HCl`, `r_(2) = k[H^(o+)]["Ester"]` `k_(HCl) = (r_(2))/(1 N xx ["Ester"])` Hence `K_(HCl) gt K_(H_(2)SO_(4)` |
|
| 177. |
At `380^(@)C` , the half-life periof for the first order decompoistion of `H_(2)O_(2)` is `360 min`. The energy of activation of the reaction is `200 kJ mol^(-1)`. Calculate the time required for `75%` decompoistion at `450^(@)C`. |
|
Answer» Correct Answer - `20.358 min` The first order reaction Half life `=(0.693)/(k)` or `360=(0.693)/(k)` `k_(380^(@)C)=(0.693)/(360)` `k=Ae^(-E_(a)//RT)` or `log k = log A-(E_(a))/(2.303RT)` `:. log.(k_(2))/(k_(1))= (E_(a))/(2.303TR)[(1)/(T_(1))-(1)/(T_(2))]` or `log.(k_(450^(@)C))/((0.693)/(360))=(200xx1000)/(2.303xx8.314)[(1)/(653)-(1)/(723)]` or `k_(450^(@)C)=6.18xx10^(2-)min^(-1)` For `75%` decompoistion at `723 K` `k_(450^(@)C)= (2.303)/(t) log.(a)/(a-x)` or `6.81xx10^(-2)=(2.303)/(t)log.(100)/(25)` or `t= 20.358min`Correct Answer - `20.358 min` The first order reaction Half life `=(0.693)/(k)` or `360=(0.693)/(k)` `k_(380^(@)C)=(0.693)/(360)` `k=Ae^(-E_(a)//RT)` or `log k = log A-(E_(a))/(2.303RT)` `:. log.(k_(2))/(k_(1))= (E_(a))/(2.303TR)[(1)/(T_(1))-(1)/(T_(2))]` or `log.(k_(450^(@)C))/((0.693)/(360))=(200xx1000)/(2.303xx8.314)[(1)/(653)-(1)/(723)]` or `k_(450^(@)C)=6.18xx10^(2-)min^(-1)` For `75%` decompoistion at `723 K` `k_(450^(@)C)= (2.303)/(t) log.(a)/(a-x)` or `6.81xx10^(-2)=(2.303)/(t)log.(100)/(25)` or `t= 20.358min` |
|
| 178. |
The half-life periof for catalystic decompoistion of `AB_(3)` at `50 mm` is found to be `4` hr and at `100 mm` it is `2.0 hr`. The order of reaction isA. `3`B. `1`C. `2`D. `0` |
|
Answer» Correct Answer - C When initial pressure of `AB_(3) = 100 mm` `t_(1//2) = 2 hr`. It means when concentration is doubled, half life is halved. So, it is a second order reaction. For second order reaction: `t_(1//2) prop (a)^(-1)`Correct Answer - C When initial pressure of `AB_(3) = 100 mm` `t_(1//2) = 2 hr`. It means when concentration is doubled, half life is halved. So, it is a second order reaction. For second order reaction: `t_(1//2) prop (a)^(-1)` |
|
| 179. |
The half time of first order decomposition of nitramide is `2.1` hour at `15^(@)C`. `NH_(2)NO_(2(aq.))rarr N_(2)O_((g))+H_(2)O_((l))` If `6.2 g` of `NH_(2)NO_(2)` is allowed to decompose, calculate: (i) Time taken for `NH_(2)NO_(2)` is decompose `99%`. (ii) Volume of dry `N_(2)O` produced at this point measured at STP. |
|
Answer» `(i) k=(0693)/(t_(1//2))=(0.693)/(2.1)=0.33hr^(-1)` Applying kinetic eqution of rist order reaction, `k=(2.303)/(t)log_(10).(a)/((a-x))` `or t=(2.303)/(0.33)log_(10).(100)/((10099))=13.96hrs` (ii) No. of moles of `NH_(2)NO_(2)` decomposed `=0.99xx(6.2)/(62)=0.099` No. of moels of `N_(2)O` formed =0.099 Volume f `N_(2)O` at STP `=0.099xx225400 mL=2217.6mL` |
|
| 180. |
From the following data for the reaction between A and B Calculate the following (i) The order of the reaction with respect to A and with respect to B (ii) The rate constant at 300K (iii) The energy of activation and (iv) The pre-exponential factor. |
|
Answer» (i) Let the rat law be : Rate= `k[A]^(x)[B]^(y)` `"From expt".(1),5.0xx10^(-4)=k[2.5xx10^(-4)]^(x)[3.0xx10^(-5)]^(y)" "....(i)` `"From expt".(2),4.0xx10^(-4)=k[5.0xx10^(-4)]^(x)[6.0xx10^(-5)]^(y)" "....(ii)` Dividing by eq. (i), `(4.0xx10^(-3))/(5.0xx10^(-4))=2^(x)*2^(y)=8` `"From expt".(3),1.6xx10^(-2)=k[1.0xx10^(-3)]^(x)[6.0xx10^(-5)]^(y)" "....(iii)` `"From expt".(1),5xx10^(-4)=k[2.5xx10^(-4)]^(2)[3.0xx10^(-5)]` or `k=(5xx10^(-4))/([2.5xx10^(-4)]^(2)[3.0xx10^(-5)])=2.67xx10^(8)L^(2)"mol"^(-2)s^(-1)` (iii) From Arrhenius eqauation, `log_(10).(2.0xx10^(-3))/(5.0xx10^(4))=(E_(a))/(2.3030xx8.314)xx(20)/(300xx320)` `E_(a)=(2.303xx8.314xx300xx320)/(20)xxlog_(10)4` (iv) Applying `log_(10) k=log_(10) A-(E_(a))/(2.3030RT)` `log_(10).(A)/(k)=(55333)/(2.3030xx8.314xx300)=9.633` or `(A)/(k)=4.29xx10^(9)` or `A=4.29xx10^(9)xx2.67xx10^(8)` `=1.145xx10^(18)` |
|
| 181. |
At a certain temperature the half change periof for the catalystic decompoistion of ammonia was found as follows: `|{:("Pressure (Pa)",6667,13333,26666),("Half life periof in hours",3.52,1.92,1.0):}|` Calculate the order of reaction. |
|
Answer» `((t_(1//2))_(1))/((t_(1//2))_(2))=((a_(2))/(a_(1)))^(n-1)` where, n is order of reaction From the given data, `(3.25)/(1.92)=((13333)/(6667))^(n-1)" " (a prop"initial pressure")` `log .(3.52)/(1.92)=(n-1)log2=0.3010xx(n-1)` `0.2632=0.30120xx(n-1)` `n=1.87~~2` Similar calculations are made between first and third observation. n comes equal to `1.908(~~2)` Thus, the reaction is of second. order. |
|
| 182. |
Cane sugar is gradualaly converted into dextrose and laevuloe by dilute acid. The rate of inversion is observed by measuring the polarisation angle, at various times, when the following results are obtained: `{:("Time (min)",0,10,20,30,40,100,oo),("Angle",32.4,28.8,25.5,22.4,19.6,-6.1,-14.1):}` Show that the reaction is of first order. Calculate the value of t, when the solution is optically inactive. |
|
Answer» in case, the inversion of cane sugar is a first order change, then `k=(2.303)/(t)log_(10).([A]_(0))/([A])=(2.303)/(t)log_(10).(r_(0)-r_(oo))/(r_(t)-r_(oo))` When `t=10, r_(0)=32.4, r_(t)=28.8, r_(oo)=-14.1` `k=(2.3030)/(10)log_(10).(32.4-(-14.1))/(28.8-(-14.1))` `k=(2.3030)/(10)log_(10).(46.5)/(42.9)=0.008 "min"^(-1)` When `t=20mr_(0)=32.4,r_(t)=25.5,r_(oo)=-14.1` `k=(2.3030)/(20)log_(10).(32.4-(-14.1))/(22.5-(-14.1))=(2.3030)/(20)log.(46.5)/(39.6) 0.008"min"^(-1)` When `t=30,r_(0)=32.4,r_(t)=22.4,r_((oo)=-14.1` `k=(2.3030)/(30)log_(10).(32.4-(-14.1))/(22.5-(-14.1))=(2.3030)/(30)log.(46.5)/(36.5) =0.008 "min"^(-1)` Thus, the reaction is of first order as the value of k is constant. Solution will be optically inactive when half of the came sugar in inverted. |
|
| 183. |
It has been proposed that the conversion of ozone into `O_(2)` proceeds in two steps: `O_(3(g)) hArrO_(2(g))+O_((g))` `O_(3(g))+O_((g))rarr 2O_(2(g))` (a) Write the equation for overall reaction. (b) Identify the intermediate, if any. (c ) Derive molecularity for each step of mechanism. |
|
Answer» (a) Addition of the two steps gives the overall reaction: `2O_(3(g))rarr 3O_(2(g))` (b) Intermediate is `O_((g))`. (c ) The first step is unimolecular. The second step is bimolecular. |
|
| 184. |
Consider the proposed mechanism for the destruction of ozone in the stratosphere: `O_(3) +Cl rarr ClO. +O_(2)` `ClO +O_(3) rarr Cl. +2O_(2)` Which of the statements about this mechanism is correct?A. Cl. Is a catalystB. `O_(2)` is in intermediate.C. Equal amounts of Cl. and CIO. Are present.D. The number of moles fo `O_(2)` produced equals the number of moles of `O_(3)` consumed. |
|
Answer» Correct Answer - A |
|
| 185. |
Half life id independent of the concentration of `A`. After `10 mi n` volume of `N_(2)` gas is `10 L` and after complete reaction is `50 L`. Hence, the rate constant isA. `(2.303//10) log 5 "min"^(-1)`B. `(2.303//10) log 1.25 "min"^(-1)`C. `(2.303//10) log 2 "min"^(-1)`D. `(2.303//10) log 4 "min"^(-1)` |
|
Answer» Correct Answer - B `K = (2.303)/(10)"log"((50)/(50 - 10)) = (2.303)/(10)log 1.25 "min"^(-1)` |
|
| 186. |
Calculate the half life of a first order reaction from their rate constants given below `:` `a. 200s^(-1)`,`b. 2 mi n^(-1)`,`c.4years^(-1)` |
|
Answer» (i) Half life , `t_((1)/(2)) = (0.693)/(k)` `= (0.693)/(200 s^(-1))` =`3.47 xx 10^(-3)s` (approximately) (ii) Half life , `t_((1)/(2)) = (0.693)/(k)` = `(0.693)/(4 "years"^(-1))` = `0.173` years (approximately) |
|
| 187. |
Graph between `log k` and `1//T` [`k` rate constant `(s^(-1))` and `T` and the temperature `(K)`] is a straight line with `OX =5, theta = tan^(-1) (1//2.303)`. Hence `-E_(a)` will be A. `2.303 xx 2 cal`B. `2//2.303 cal`C. `2 cal`D. None |
|
Answer» Correct Answer - C `logk=logA-(E_(a))/(2.303 RT)(y=c+mx)` Slope `=(-E_(a))/(2.303R)=(1)/(2.303)(given)(tan theta=(1)/(2.303))` `-E_(a)= 2.303Rxxslope` `=2.303xx(R )/(2.303)=R=2 cal` |
|
| 188. |
For a zero order reaction :A. `t_(1//2)propa`B. `t_(1//2)prop (1)/(a)`C. `t_(1//2)propa^(2)`D. `t_(1//2)prop (1)/(a^(2))` |
| Answer» Correct Answer - A | |
| 189. |
`Ararr` Product, `[A]_(0) = 2M`. After `10 min` reaction is `10%` completed. If `(d[A])/(dt) = k[A]`, then `t_(1//2)` is approximatelyA. `0.693 min`B. `69.3 min`C. `66.0 min`D. `0.0693 min` |
|
Answer» Correct Answer - C `k(2.3030=)/(10) log.(100)/(90)=0.001054 min^(-1)` `t_(1//2)=(2.3030)/(k)=(2.303)/(0.001054)~~66min` |
|
| 190. |
For a chemical reaction `A to B` , the rate of the reaction is `2 xx 10^(-3) mol dm^(-3) s^(-1)` . When the initial concentration is `0.05` mol `dm^(-3)` . The rate of the same reaction is `1.6 xx 10^(-2)` mol `dm^(-3) s^(-1)` when the initial concentration is `0.1` mol `dm^(-3)` . The order of the reaction is |
|
Answer» Correct Answer - b Let the rate equation for the reaction be , rate = `k [A]^(n)` where , k = rate constant , n = order of reaction [A] = concentration of reactant given , `"rate"_(1) = 2 xx 10^(-3) "mol" dm^(-3) s^(-1)` `[A_(0)] = 0.05 ` mol `dm^(-3)` `"rate"_(2) = 1 . 6 xx 10^(-2)` mol `dm^(-3) s^(-1)` `(A_(0)) = 0.1` mol `dm^(-3)` `therefore 2 xx 10^(-3) = k [0.05]^(n) " ".... (i)` `1.6 xx 10^(-2) = k[0.1]^(n) " ".... (ii)` Divide (i) by (ii) `implies (2 xx 10^-3)/(1.6 xx 10^(-2)) = ([0.05]^(n))/([0.1]^(n)) = ([0.05]^(n))/(2^(n)[0.05]^(n))` `implies (1)/(2^(n)) = (1)/(8) implies n = 3, " " therefore` Order of reaction = 3 . |
|
| 191. |
The rate of the first order `Ararr` Products, is `7.5xx0^(-4) "mol" L^(-1)s^(-1)` , when the concentration of A is 0.2 mol `L^(-1)` . The rate constant of the reaction is :A. `2.5xx10^(-5)s^(-1)`B. `8.0xx10^(-4)s^(-1)`C. `6.0xx10^(-4)s^(-1)`D. `3.75xx10^(-3)s^(-1)` |
| Answer» Correct Answer - D | |
| 192. |
The graph between concentration `(X)` of the Product and time of the reaction `Ararr B` is of the type `1`. Hence, graph between `-(d[A])/(dt)` and time will be of the type: A. B. C. D. |
| Answer» Correct Answer - C | |
| 193. |
For a chemical reaction `Ararr B`,the rate of the reaction is `2.0xx10^(-3) sec^(-1)`, when the initial concentration is `0.05 mol dm^(-3)`. The rate of the same reaction is `1.6xx10^(-2) mol dm^(-3) sec^(-1)`. When the initial concentration is `0.1` mol `dm^(3)`, find the order of reaction. |
|
Answer» Correct Answer - 3 Let the rate equation for the reaction be rate `=K[A]^(n)` `r_(1)=2.0xx10^(-3) mol dm^(3) mol dm^(-3) sec^(-1)`, `[A_(0)]=0.05 mol dm^(-3)` `:. 2.0xx10^(-3)=K[0.05]^(n) …(1)` and `r_(2)=1.6xx10^(-2) mol dm^(-3) sec^(-1)`, `[A_(0)]=0.1 mol dm^(-3)` `:. 1.6xx10^(-2)=K[0.1]^(n)` eq. (1) divide by. `(2)`, `(2.0xx10^(-3))/(1.6xx10^(-2))=([0.05]^(n))/([0.1]^(n))implies 1/8=(1/2)^(n)` `implies n=3` `:.` Order of reaction `=3` |
|
| 194. |
a) What will be the initial rate of reaction if rate constant is `10^(-3)s^(-1)` at concentration of 0.2 mol `L^(-1)`? How much of the reactant will be converted into product in 200 minutes? Assume reaction to be of first order. b) The half life of 1st order reaction `A to B` is 600s. What percentage of A remains after 30 minutes? |
|
Answer» a) Calculation of initial rate of reaction for the first order reaction, Reaction rate=`k[A]=(10^(-3)min^(-1)) xx (0.2 mol L^(-1))` `=2.0 xx 10^(-4)mol L^(-1)min^(-1)` Calculation of amount of reactant changing into products. For the first order reaction, `t=(2.303)/k log ([A]_(0))/([A]), (200 min) = (2.303)/(10^(-3)min^(-1)) log ([A]_(0))/([A])` `log([A])_(0)/[A] = (200 min xx 10^(-3) min^(-1))/(2.303) = 0.0868` `[A]_(0)/[A] = "Antilog" 0.0868=1.221` or `[A]=([A]_(0))/(1.221) = (0.2 mol L^(-1))/(1.221) = 0.164 mol L^(-1)` `therefore` Amount of rectant changing into product =`0.2-0.164=0.036 mol L^(-1)` b) Calculation of the percentage of A left after 30 minutes. For the first order reaction, `t=(2.303)/k log ([A]_(0))/([A]), (30 xx 60s) = (2.303)/(0.001155 s^(-1)) log ([A]_(0))/([A])` `log ([A]_(0))/[A] = (30 xx 60s xx 0.001155 s^(-1))/(2.303) = 0.9027` `([A]_(0))/([A]) = "Antilog" 0.9027 = 7.9928, [A] = ([A]_(0))/(7.9928) = 100/(7.9928) = 12.51%` |
|
| 195. |
For a first - order reaction, the rate of reaction is `1.0xx10^(-2) mol L^(-1) s^(-1)` and the initial concentration of the reactant is 1 M. The half-life period for the reaction isA. `6.93xx10^(-3)s^(-1)`B. `0.693 s^(-1)`C. `69.3 s`D. `0.693 s^(-1)` |
|
Answer» Correct Answer - C Consider a simple general first order reaction `Ararr`products Differential rate law is `Rate=k[A]` or `k=("Rate")/([A])=(1.0xx10^(-2)molL^(-1)s^(-1))/(1.0molL^(-1))` `k10^(-2)s^(-1)` For a first order reaction, half is given as `t_(1//2)=(0.693)/(10^(-2))s^(-1))` `=0.693xx10^(2)s` `=69.3s` |
|
| 196. |
The rate constant for a first order reaction is `7.0xx10^(-4) s^(-1) `. If initial concentration of reactant is 0.080 M, what is the half life of reaction ?A. 990SB. `79.2S`C. `12375 S`D. `10.10xx10^(-4)S` |
|
Answer» Correct Answer - A `k=(0.693)/(t_(1//2))` ` therefore t_(1//2) =(0.693)/(7.0xx10^(-4))=990 sec.` |
|
| 197. |
In a reaction the initial concentration of the reactants increase four fold and the rate becomes eight times its initial value . The order of the reaction isA. `2.0`B. `3.5`C. `2.5`D. `1.5` |
|
Answer» Correct Answer - d Rate = `KC^(n)` Or r = `KC^(n) " "… (i)` `8r = K(4C)^(n) " " … (ii)` Dividing Eq. (ii) by (i) , we get `2^(3) = 2^(2n)` or 2 n =3 n = 1.5. |
|
| 198. |
In a certain reaction, `10%` of the reactant decomposes in one hour, `20%` in two hours, `30%` in theee hours, and so on. The dimenison of the velocity constant (rate constant) areA. `hr^(-1)`B. `Mol L^(-1) hr^(-1)`C. `L mol^(-1) s^(-1)`D. `Mol s^(-1)` |
|
Answer» Correct Answer - B form the question, it is clear `t_(1//2) prop a`. Hence, zero order reaction. So, dimenison of `k = mol L^(-1) hr^(-1)`. For zero order: `t = (x)/(k) or k = (x)/(t)` If `t = t_(10%) = 10 min, x = 10`, Then, `k = (10)/(10) = 1 mol L^(-1) t^(-1)`. If `t = t_(20%) = 20 min, x = 20`, Then, `k = (20)/(20) = 1 mol L^(-1) t^(-1)`. ismilarly, for `30%` and so on. Thus, reaction is of zero order. |
|
| 199. |
In a certain reaction, `10%` of the reactant decomposes in one hour, `20%` in two hours, `30%` in theee hours, and so on. The dimenison of the velocity constant (rate constant) areA. `"hour"^(-1)`B. `"mole litre"^(-1)sec^(-1)`C. `"litre mole"^(-1)sec^(-1)`D. `"mole sec"^(-1)` |
|
Answer» Correct Answer - B |
|
| 200. |
Assertion: In rate law, unlike in expression for equilibrium constants, the exponents for the concentration donot necessarily match the stoichometric coefficients. Reason: It is the mechanism and not the balanced chemical equation for the overall change that governs the reaction rate.A. If both assertion and reason are correct explanation for assertion.B. If both assertion and reason are correct but reason is not correct explanation for assertion.C. If assertion is correct but reason is incorrect.D. If both assertion and reason are incorrect. |
|
Answer» Correct Answer - A a) Reason is the correct explanations for assertion. |
|