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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.
| 951. |
In a reaction , `A + B rarr` Product, rate is doubled when the concentration of `B` is doubled, and rate increases by a factor of `8` when the concentration of both the reactants (A and B) are doubled, rate law for the reaction can be written asA. `Rate = k[A][B]^(2)`B. `Rate = k[A]^(2)[B]^(2)`C. `Rate = k[A][B]`D. `Rate = k[A]^(2)[B]` |
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Answer» Correct Answer - D Let the rate law be `Rate=k[A]^(x)[B]^(y)` Let the initial concentrations `A` and `B` be `a` and `b` respectively. According to first observation `R_(1)k[a]^(x)[b]^(y)` `R_(2)=2R_(1)=k[a]^(x)[2b]^(y)` Dividing the `2^(nd)` equation by `1^(st)` , we have `2=2^(y)` or `2^(1)=2^(2)` Hence, `y=1` According to the second observation ltbr. `R_*(1)=k[a]^(x)[b]^(y)` `R_(2)8R_(1)=k[2a]^(x)[2b]^(y)` Dividing the `2^(nd)` equation by `1^(st)` , we get `8=2^(x).2^(y)` `2^(3)=2^(x+y)` or `x+y=3` Substituting the value of `y=1` , we have `x+1=3` or `x=3-1=2` Thus, the correct rate law is `Rate=K[A]^(2)[B]` |
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| 952. |
The rate of the first-order reaction `X rarr` products is `7.5 xx 10^(-4) mol L^(-1) "min"^(-1)`. What will be value of rate constant when the concentration of `X` is `0.5 mol L^(-1)` ?A. `3.75xx10^(-4) min ^(-1)`B. `2.5xx10^(-5) min ^(-1)`C. `0.1xx10^(-4)`D. `0.3xx10^(-4)` |
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Answer» Correct Answer - C Rate `=kxx["concentration "]` ` therefore k= ("Rate")/("[concentraion ]")=(7.5xx10^(-4))/(0.5)=1.5xx10^(-3)` |
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| 953. |
The rate of the first-order reaction `X rarr` products is `7.5 xx 10^(-4) mol L^(-1) "min"^(-1)`. What will be value of rate constant when the concentration of `X` is `0.5 mol L^(-1)` ?A. `3.75 xx 10^(-4)s^(-1)`B. `2.5 xx 10^(-5) s^(-1)`C. `1.5 xx 10^(-3) s^(-1)`D. `8.0 xx 10^(-4) s^(-1)` |
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Answer» Correct Answer - B Rate of reaction `= 7.5 xx 10^(-4) "mol L"^(-1) "min"^(-1)` `= (7.5 xx 10^(-4))/(60) "mol L"^(-1) s^(-1)` `= 1.25 xx 10^(-5) "mol L"^(-1) sec^(-1)` Rate of reaction `= k["concentration of react ant X"]` Or `k = (1.25 xx 10^(-5))/(0.5) = 2.5 xx 10^(-5) s^(-1)` |
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| 954. |
The rate of the first-order reaction `X rarr` products is `7.5 xx 10^(-4) mol L^(-1) "min"^(-1)`. What will be value of rate constant when the concentration of `X` is `0.5 mol L^(-1)` ?A. `8xx10^(-4)`B. `1.5xx10^(-3)`C. `2.5xx10^(-5)`D. `3.75xx10^(-4)` |
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Answer» Correct Answer - B 1st order reaction , `7.5xx10^(-4) = K [0.5]` ` therefore k= (7.5xx10^(-4))/(t_(1//2))=1.5xx10^(-3)` |
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| 955. |
Rate of a reactionA. increases with increase in temperatureB. decreases with increase in temperatureC. does not depend on temperatureD. does not depend on concentration |
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Answer» Correct Answer - A Factors affecting rate of reaction . |
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| 956. |
Number of reactant molecules taking part in a reaction isA. Order of reactionB. Molecularity of the reactionC. Rate of reactionD. Conplex reaction |
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Answer» Correct Answer - B Definition of molecularity of reaction . |
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| 957. |
for first order reactionA. `-k=(2.303)/(t) log ""((a-x))/(a)`B. `lamda = (2.303)/(t) log ""(N_(t))/(N_(0))`C. `-K=(2.303)/(t)log ""(a)/(a-x)`D. `lamda = 2.303 t log ""(a) /(a-x)` |
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Answer» Correct Answer - A Rate law equation |
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| 958. |
Which of the following rate expression does not represent the rate of the reaction ? `2N_(2)O_(5(g)) to 4NO_(2(g)) + O_(2(g))`A. `-(1)/(2) (d[N_(2)O_(5)])/(dt)`B. `(1)/(2) (d[N_(2)O_(5)])/(dt)`C. `([O_2])/(dt)`D. `(1)/(4) (d[NO_(2)])/(dt)` |
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Answer» Correct Answer - B Rate of reaction `=-(1)/(2) (d[N_(2)O_(5)])/(dt) ` `=+(1)/(4) (d[NO_(2)])/(dt ) =+ (d[O_(2)])/(dt)` |
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| 959. |
The first order integrated rate equation isA. `k= t _(l_(n))(a)/(a-x)`B. `k=(1)/(l) l_(n) (a-x)/(a)`C. `k=(1)/(t) l_(n) (a) /(a-x)`D. `k=tl_(n) (a-x)/(a)` |
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Answer» Correct Answer - C Derivation of rate law. |
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| 960. |
The effect of catalyst in a chemical reaction is to lower theA. Heat of reactionB. equilibrium concentrationC. activation energyD. concentration of reactants |
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Answer» Correct Answer - D Action of catalyst |
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| 961. |
Select the correct statement out of P, Q and R for zero order reaction. (P) Quantity of the product formed is directly proportional to time (Q) Larger the initial concentration of the reactant, greater the half-life period (R) If `50%` reaction takes place in 100 minutes, `75%` reaction takes place in 150 minutes.A. P onlyB. P and Q onlyC. Q and R onlyD. P, Q and R |
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Answer» Correct Answer - D |
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| 962. |
The rate of chemical reaction is directly proportional toA. Active Masses of reactandB. Equilibrium constantC. Active masses of productD. Pressure |
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Answer» Correct Answer - A Factors affecting rate of chemical reaction . |
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| 963. |
For which order reaction, the unit of rate constant is `"time"^(-1)` ?A. Zero orderB. first orderC. Second orderD. thrid order |
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Answer» Correct Answer - B `k =(2.303)/(t)log""(A_(0))/(A_(t)) therefore (1)/(sec)xx(conc.)/(conc.)=(1)/(sec )=sec^(-1)` |
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| 964. |
unit of rate isA. `Mol dm^(-3)`B. ` mol dm ^(3)`C. `(Mol dm ^(-3))/( sec)`D. `(Mol dm^(3))/(sec)` |
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Answer» Correct Answer - C `"Rate "=("molar concentration ")/("time") therefore (mol dm^(3))/(sec)` |
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| 965. |
For an exothermic reaction an activation energy of 70 KJ `"mole " ^(-1 ) ` and the enthalpy change of reaction is 30 KJ ` "mole " ^(-1) ` . The order of the reaction isA. `70KJ "mole"^(-1)`B. `30 KJ "mole "^(-1)`C. `40 KJ "mole" ^(-1)`D. `100 KJ "mole "^(-1)` |
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Answer» Correct Answer - D `Delta H=E_(a)-E_(a)^(r ) therefore -30 =70 -E_(a )^(r ). therefore E_(a)^(r ) = 100 ` |
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| 966. |
Dependance of rate on concentration is expressed byA. Rate lawB. order of reactionC. MolecularityD. Law of mass action |
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Answer» Correct Answer - A Rate law . |
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| 967. |
for a raction ,` I^(-) +OCI^(-) to IO^(-) +Cl^(-) ` in an aqueous medium , the rate of reaction is given by `(d[IO]^(-))/(dt)=k([I^(-)][OCI^(-)])/([OH^(-)])` the overall order of reaction isA. `-1`B. `0`C. `1`D. `2` |
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Answer» Correct Answer - C `(d[IO]^(-))/(dt)=K([I^(-)]^(1)[OCI^(-)]^(1))/([OH^(-)]^(1))` order of reaction `=1+1-1=1` |
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| 968. |
The rate for the decomposition of `NH_(3)` on platinum surface is zero order. What are the rate of production of `N_(2)` and `H_(2)` if `K = 2.5 xx 10^(-4) mol litre^(-1)s^(-1)`? |
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Answer» `2NH_(3) rarr N_(2)+3H_(2)` `-1/2(d[NH_(3)])/(dt)=(d[N_(2)])/(dt)=1/3(d[H_(2)])/(dt)` Rate `=K [NH_(3)]^(0)` or `(d[NH_(3)])/(dt)=2.5xx10^(-4) mol litre^(-1) s^(-1)` `(d[N_(2)])/(dt)=1/2xx2.5xx10^(-4)` `=1.25xx10^(-4) mol litre^(-1) s^(-1)` `(d[H_(2)])/(dt)=3/2xx2.5xx10^(-4)` `=3.75xx10^(-4) mol litre^(-1) s^(-1)` |
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| 969. |
At `25^(@)C`, the rate constant for the reaction `I^(-) + ClO^(-) rarr IO^(-) + Cl^(-)` is `0.0606 "litre mol"^(-1) sec^(-1)`. If a solution is intially `1.0 M` in `I^(-)` and `5.0 xx 10^(-4) M` in `ClO^(-)`. Can you calculate the `[CO^(-)]` after `300 sec`? If yes, then how much ? If no, then why ? |
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Answer» No doubt the unit of rate constant shows that it is II order. But it is necessary to know the rate law expression which from the statement of the problem, colud be any of the following (given below): Rate `=K[I][ClO^(-)]` or Rate `=K[I^(-)]^(2)` or Rate `=K[ClO^(-)]^(2)` |
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| 970. |
`2A+B to 3C ` for the reaction instant rate of reaction is `A. `+(1)/(2)(d[A])/(dt)=+(d[B])/(dt)=+(1)/(3)(d[C])/(dt)`B. `-(1)/(2)(d[A])/(dt)=-(d[B])/(dt)=+(1)/(3)(d[C])/(dt)`C. `+2(d[A])/(dt)=+(d[B])/(dt)=+(1)/(3)(d[C])/(dt)`D. `-2(d[A])/(dt)=-(d[B])/(dt)=+3(d[C])/(dt)` |
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Answer» Correct Answer - B Rate of reaction `=-(1)/(2) (d[A])/(dt) =-(d[B])/(dt)` `=+(1)/(3)(d[C ])/(dt)` |
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| 971. |
For the reaction of `4A+B to 2C +D.`. Which of the following statements is correct ?A. The rate of formation of C and D are equalB. The rate of formation of D is one half the rate of consumption of AC. The rate of appearance of C is one half the rate of disappearance of BD. The rate of disappearance of B is one fourth of the rate of disappearance of A |
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Answer» Correct Answer - D `-(1)/(4) (d[A])/(dt)=-(d[B])/(dt) =+(1)/(2) (d[C])/(dt)=+(d[D])/(dt)` |
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| 972. |
For the following chemical reaction` 2x +yhArr Z ` the expression of equilibrium constant will beA. `K_(c ) =([x]^(2)[Y])/(Z)`B. `k_(c)=([x][y]^(2))/([Z])`C. `k_(c) =([Z])/([x]^(2)[Y])`D. `K_(c ) =([Z])/([X][Y]^(2))` |
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Answer» Correct Answer - C `k=(["product "])/(["Reactant"])` |
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| 973. |
The reaction of `O_(3)` with chlorine atom is given as : `O_(3)(g)+Cl(g)rarrO_(2)(g)+ClO(g),k_(1)=5.2xx10^(9)L"mol"^(-1)sec^(-1)` `ClO(g)+O(g)rarrCl(g)+O_(2)(g),k_(2)=2.6xx10^(10)L"mol"^(-1)sec^(-1)` Which of theses values is closest to the rate constant of the overall reaction ? `O_(3)(g)O(g)rarr2O_(2)(g)`A. `5.2xx10^(9)`B. `2.6xx10^(10)`C. `3.1xx10^(10)`D. `1.4xx10^(20)` |
| Answer» Lowest value of k shows that the step is rate determining | |
| 974. |
The unit for the rate constant for the second order reaction [Concentration : Mol `"litre"^(-1)`, time : s] areA. `Mol^(-1)"litre"^(-1)`B. `Mol "litre"^(-2)S^(-1)`C. `s^(-1)`D. `mol"litre"^(-1)S^(-1)` |
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Answer» Correct Answer - A Rate `=(dx)/(dt)=k(conc.)^(n)` `k=(dx)/(dt)=k(conc.)^(n)` `k=(Mol Lit^(-1))^(1-n)S^(-1)`,) ( For second order reaction n=2) `k=Mol Lit^(-1)S^(-1) ` `K=Mol ^(-1) Lit S^(-1)` |
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| 975. |
The rate constant of a reaction is `1.2xx10^(-5)mol ^(-2)"litre"^(2) S^(-1)` the order of the reaction isA. zeroB. 1C. 2D. 3 |
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Answer» Correct Answer - D `(dx)/(dt[C])=k` Order or reaction `=(1)/(C^(n-1))=C^(1-n)` |
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| 976. |
The rate constant for a first order reaction whose half life is 480 sec, is :A. `1.44xx10^(-3)S^(-1)`B. `1.44S^(-1)`C. `0.72xx10^(-3)S^(-1)`D. `2.88xx10^(-3)S^(-1)` |
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Answer» Correct Answer - A `k=(0.693)/(t_(1//2)) = (0.693)/(480) =1.44 xx10^(-3)S^(-1)` |
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| 977. |
Which of the following statements are in accordance with the Arrhenius equation?A. Rate of a reaction increases with increases in temperaturesB. Rate of a reaction increases with decrease in activation energyC. Rate constant decreases exponentially with increase in temperatureD. Rate of reaction decreases with decreases in activation energy. |
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Answer» Correct Answer - A::B |
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| 978. |
The rate constant of first order reaction whose half-life is 480 s, isA. `1.44 xx 10^(-3)s^(-1)`B. `1.44 s^(-1)`C. `0.72 xx 10^(-3)s^(-1)`D. `2.88 xx 10^(-3)s^(-1)` |
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Answer» Correct Answer - A For the first order reaction `implies " "k=(0.693)/(t_(1//2))=(0.693)/(480 sec)=1.44 xx 10^(-3)s^(-1)` |
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| 979. |
Show by uisng rate laws how much the rate of reaction `2NO(g) + O_(2)(g) rarr 2NO(g)` will change if the volume of the reaction vessel is diminished to `1//3` of its initial volume.A. 1/3 timesB. 2/3 timesC. 3 timesD. 6 times |
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Answer» Correct Answer - c For the following reaction `, 2NO_((g)) + O_(2(g)) to 2NO_(2)(g)` When the volume of vessel change into `(1)/(3)` then concentration of reactant become three times. The rate of reaction for first order reaction `prop` concentration . So rate of reaction will increases three times . |
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| 980. |
In the reaction `A rarr `products, time required to complete `50%` reaction was found to increase 9 times when the initial concentration of the reactant was decreased to one third. The rate law equation is `:`A. `-(d(A))/(dt)=K(A)^(1//2)`B. `-(d(A))/(dt)=K(A)`C. `-(d(A))/(dt)=K(A)^(2)`D. `-(d(A))/(dt)=K(A)^(3)` |
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Answer» Correct Answer - D `t_(1//2)prop(1)^(n-1)` `(9)/(1)=((3)/(1))^(n-1)` `n-1=2` `n=3` |
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| 981. |
Which of the following options regarding characteristics of zero order reaction is correct? (P) Time for `50%` dissociation will increase as initial concentration increase. (Q) In same time interval same `%` if reactant gets consumed. (R) The graph of log of concentration of reactant vs. time will be linear. (S) Average rate between two time interval and instantaneous rate at the two intervals will be same as long as reaction is occurring at the two instants.A. Only P is correctB. Q and R are the only incorrect statementsC. Only R is incorrectD. Only S is correct |
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Answer» Correct Answer - B |
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| 982. |
For a first order reaction, the concentration decreases to `30%` of its initial value in `5.0` min. What is the rate constant?A. `0.46^(-1)`B. `0.24"min"^(-1)`C. `0.14"min"^(-1)`D. `0.060"min"^(-1)` |
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Answer» Correct Answer - B |
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| 983. |
The rate constant for the reaction `2N_(2)O_(5)to4NO_(2)+O_(2), "is"" " 3.0xx10^(-5)S^(-1)`. If the rate is `2.40xx10^(-5)"mol"L^(-1)s^(-1)` then the concentration of `N_(2)O_(5)("in mol" L^(-1))` isA. 1.4B. 1.2C. 0.04D. 0.8 |
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Answer» Correct Answer - D For a first order reaction `Rate =k[N_(2)O_(5)]` `2.4xx10^(-5)=3.0xx10^(-5)[N_(2)O_(5)]` `[N_(2)O_(5)]=(2.4xx10^(-5))/(3.0xx10^(-5))=0.8 "mol" L^(-1)` |
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| 984. |
In a reaction, `2Ato "products",` the concentration of A decreases from 0.5 mol `L^(-1) to 0.4 mol L^(-1)` in 10 min. The rate during this interval isA. `0.05 "mol"L^(-1)"min"^(-1)`B. `0.42 """mol"" "L^(-1)"min"^(-1)`C. `0.005 """mol"" "L^(-1)"min"^(-1)`D. `0.5 """mol"" "L^(-1)"min"^(-1)` |
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Answer» Correct Answer - C Rate of reaction =`-(1)/(2)(d[A])/(dt)` `=-(d[A])/(dt)=-([A]_(2)-[A]_(1))/(t_(2)-t_(1))=-(0.4-0.5)/(10-0)=(0.1)/(10)` `0.01 "mol"L^(-1) "min"^(-1)` `therefore Rate of reaction =(1)/(2)xx0.01` `=0.005 "mol" L^(-1) "min"^(-1)` |
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| 985. |
The reaction,`2NO(g)+O_(2)(g)hArr2NO_(2)(g),` is of first order. If the volume of reaction vessel is reduced to `(1)/(3)`, the rate of reaction would beA. `(1)/(3) "times"`B. `(2)/(3) "times"`C. 3 timesD. 6 times |
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Answer» Correct Answer - C `2NO(g)+O_(2)(g)to2NO_(2)(g)` When the volume of reaction vessel is reduced to `(1)/(3) ` then concentration of reactant become three. The rate of reaction for first order reaction is directly proportional to concentration. Therefore, rate of reaction will increase threee times. |
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| 986. |
On increasing the pressure three fold, the rate of reaction of `2H_(2)S+O_(2)to"products would incerease"`A. 3 timesB. 39timesC. 12 timesD. 27 times |
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Answer» Correct Answer - D `2H_(2)S+O_(2) to products` Rate `=kxxP_(H_(2)S)^(2)xxP_(O_2)=x` On increasing the pressure threee fold Rate `=k[3P_(H_2S)^(2)]^(2)xx[3P_(O_2)]` `=kxx9P_(H_2S)^(2)xx3P_(O_2)` `=kxx27xxP_(H_2S)^(2)xxP_(O_2)` Hence , rate will increase 27 times. |
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| 987. |
Which of the following statements is correct ?A. The rate of a reaction decrease with passage of time as the concentration of reactants decreaseB. The rate of a reaction is same at any time during the reactionsC. The rate of a reaction is indepent of temperature changeD. The rate of a reaction decreases with increase in concentration ofreactant(s). |
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Answer» Correct Answer - A Rate of reactions is defined as rate of decrease of concentration of any one of reactant with passage of time . `"Rate of reaction"=("Rate of disappearance of reactant")/("Time taken")` `r =(-dx)/(dt)` Thus, as the concentration of reactant decreases with passage of time, rate of rectant decreases. |
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| 988. |
Rate of a reaction can be defined asA. change in concentration of a reactant in unit timeB. change in concentration of a product in unit timeC. Both (a) and (b)D. None of the above |
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Answer» Correct Answer - C Rate of a reaction can be defined as a change in concentration of a reactant or product in unit time. |
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| 989. |
Identify the incorrect statements .A. rusting of iron in the presence of the air and moisture, is a slow reactionB. inversio of a cane sugar occurs at a moderate rateC. hydrolysis of starch is a fast reactionD. ionic reactions are the examples of fast reactions |
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Answer» Correct Answer - C Hydrolysis of starch proceeds with a moderate rate. |
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| 990. |
For a general reaction `RtoP,"then" r_("inst")` will be?A. `(d[R])/(dt)=-(d[P])/(dt)`B. `(d[R]-d[P])/(dt)`C. `(d[R])/(dt)=+(d[P])/(dt)`D. `(d[R]+d[P])/(dt)` |
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Answer» Correct Answer - C `r_("inst") =(d[R])/(dt)=(d[P])/(dt)` |
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| 991. |
The rate of a certain reaction is given by , rate `=K [H^(+) ]^(n)` . The rate increases 100 times when the pH changes from 3 to 1 . The order (n) of the reaction is ______.A. 3B. 0C. 1D. 1.5 |
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Answer» Correct Answer - C Given,rate `k[H^(+)]^n `initial pH= `3so[H^(+)]=1xx10^(-3)`,initial rate= `r_(1)`Final pH= `1so[H^](+)=1xx10^(-1)`final rate `=r_(2)=100r_(1)` On substituting values, we get `r_(1)=k[1xx10^(-3)]^(n) and" " r_(2)=100r_(1)=k[10^(-1)]^(n)" "...(ii)`On dividing Eq. (i),we get `(r_(1))/(100r_(1))=[(1xx10^(-3))/(1 xx10^(-1))]^(n)` `1/100=[(10^(-2))/1]^(n)implies[(1)/(100)]^(1)=[(1)/(100)]^(n)` `therefore" "n=1`Thus, reaction is of first order. |
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| 992. |
For a gaseous reaction at constant temperatureA. concentration`prop(1)/("partial pressure of species")`B. concentration`prop"partial pressure of species"`C. concentration=partial pressure of speciesD. None of the above |
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Answer» Correct Answer - B For a gaseous rection at constant temperature , concentration `prop` partial pressure of species. |
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| 993. |
Unit of rate of a reaction isA. concentration `tmie^(-1)`B. `"concentration^(-1)"` tmieC. concentration tmieD. `"concentration^(-1) tmie^(-1)"` |
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Answer» Correct Answer - A Units of rate of reaction is concentration `time ^(-1)`. |
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| 994. |
In the reaction alternative. `2Hl(g)toH_(2)(g)+l_(2)(g)` Choose the correct alternative.A. Rate of reaction `=(1)/(2)(Delta[HI])/(Deltat)=(Delta[H_(2)])/(Deltat)=(Delta[I_(2)])/(Deltat)`B. Stoichiometric coefficients of HI (reactant ) and `H_(2) and l_(2)` (products) are not same.C. Rate of consumption of HI=2 (rate of formation of `H_(2) Or l_(2))`D. All of the above |
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Answer» Correct Answer - D For the given reaction , `2HI(g)to H_(2)(g)+I_(2)(g)` `(dx)/(dt)=(1)/(2)(d[HI])/(dt)=+(d[H_(2)])/(dt)=+(d[I_(2)])/(dt)` Stoichiometric coefficient (HI) `ne` Stoichiometric coefficient `H_(2) or I_(2)` Rate of consumption of HI=2 (rate of formation of ` H_(2) or I_(2)` Hence, all the statements are true. |
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| 995. |
For the reaction `A +2B to C`, the reaction rate is doubled if the concentration of A is doubled. The rate becomes four times when the concentration of both A and B are made four times. The order of reaction is:A. 3B. 0C. 1D. 2 |
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Answer» Correct Answer - C `A + 2B to C` Since the reaction rate becomes double when the concentration of A is made twice order w.r.t A=1 Since the reaction rate becomes four times when concentration of both A and B are made four times. The reaction rate is independent of the concentration of B. Order w.r.t. B=0 Overall order of reaction =1 |
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| 996. |
Rate of a reaction can be expressed by following rate expression, Rate =`K[A]^(2)[B]`, if conentration of A is incereased by 3 times and concentration of B is incereased by 2 times, how many times rate of reaction increses?A. 9 timesB. 27 timesC. 18 timesD. 8 times |
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Answer» Correct Answer - C Given, `R_(1)=K[A]^(2)[B]` According to question , `R_(2) R_(2)=k[3A]^(2)[2B]` =`kxx9[A]^(2) 2[B]` = `18xxk[A]^(2)[B]=18R_(1)` |
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| 997. |
A + 2B `rarr C, the rate equation for this reaction is given as Rate = k[A] [B]. If the concentration of A is kept the same but that of B is doubled what will happen to the rate itelf?A. HalvedB. SameC. DeubledD. Quadrupled |
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Answer» Correct Answer - C Rate `k[A] [B]` Rate is first order with respect to B. So, it doubles on doubling the concentration of B, while keeping the concentration of A is same. |
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| 998. |
The rate constant and half - life of a first order reaction are related to each other as _____.A. `t_(1//2) =(0.693)/(K)`B. `t_(1//2)=0.693K`C. `K=0.693t_(1//2)`D. `kt_(1//2)=(1)/(0.693)` |
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Answer» Correct Answer - A `T_(1//2) =(0.693)/(k) ` |
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| 999. |
The reaction takes place in two steps as `(i) NO_(2) Cl _((g)) overset(K_(1)) to NO_(2(g)) +Cl_((g))` `(ii ) NO_(2) Cl_((g)) +Cl_((g)) overset(K_(2)) to NO_(2(g)) +Cl_(2(g))` Identify the reaction intermediate .A. `NO_(2)Cl_((g))`B. `NO_(2(g))`C. `Cl_(2(g))`D. `Cl_((g))` |
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Answer» Correct Answer - D Common in continuous reaction is chlorine |
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| 1000. |
`2A rarr B + C`. It would be a zero-order reaction whenA. the rate of reaction is proportional to square of concentration of AB. the rate of reaction remains same at any concentration of AC. the rate remains unchanged at any concentration of B and CD. the rate of reaction doubles if concentration of B is incereased to double |
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Answer» Correct Answer - B For the reaction `3ArarrB+c` the differential rate law is Rate `=k[A]^(x)` where `x` the order of reaction. Zero order reaction means that the rate of the reaction is proportional to zero power of the concentration of reactant i.e., `x=0` . Thus, Rate `=k` Hence, the rate of reaction remains the same at any concentration of `A` . |
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