The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentration of the reactants raised to some powers fo

Monjur Alahi

The rate law expresses the relationship of the rate of a reaction to the rate constant and the concentration of the reactants raised to some powers for the general reaction.
`aA+bBtocC+dD`
Rate law takes the form
`r=k[A]^(x)[B]^(y)`
where x and y are number that must be determined experimentely k is the rate constant and [A] and [B] are concentration of A & B respectively.
Gaseous reaction `AtoB+C` follows first order kinetics concentration of A changes from A changes from 1M to 0.25M in 138.6 min.Find the rate of reaction when conc. of A is 0.1 M.
A. `10^(-2) M "min"^(-1)`
B. `10^(-3) M "min"^(-1)`
C. `10^(-5) M "min"^(-1)`
D. `10^(-2) M "min"^(-1)`


0 like 0 dislike
1 views
Share with your friends

1 Answers

Salim Ahmed Kawsar

Call

Correct Answer - B
`r=k[A]=0.693/69.3xx0.1=10^(-3) M, t_(1//2)=69.3` min

0 like 0 dislike
1 views
Talk Doctor Online in Bissoy App
For the reaction `A + B rarr C + D`, doubling the concentration of both the reactants increases the reaction rate by `8` times and doubling the initia

Correct Answer - C c) The data suggests that reaction rate ( r) `propto [A]^(2)` and is `propto [B]`. `therefore r=k[A]6(2)[B]`

2 Answers 1 views
In a second order reaction, when the concentration of both the reactants are equal, the reaction is completed `20%` in 500 s. How long would it take f

Correct Answer - A a) For the second order reaction, `k=1/(t xx a). (X/(a-x))` According to available data, in the first case t=500s , a=100 a-x `=80%` of 100 = 80...

2 Answers 1 views
The rate law for a reaction between the substances A and B is given by Rate = `k[A]^(n)[B]^(m)` On doubling the concentration of A and halving the con

Correct Answer - C c) `Rate_(1) = k[A]^(n)[B]^(m)` `Rate_(2) = k[2A]^(n)[1//2B]^(m)` `(Rate_(2))/(Rate_(1))= (k[2A]^(n)[1//2B]^(m))/(k[A]^n)[B]^(m)` `=[2]^(n([1//2])^(m))= 2^(n).2^(-m)` `=2(n-m)`

2 Answers 1 views
The half life period of a reaction is halved as the initial concentration of the reactants is doubled. The order of reaction is:

Correct Answer - B c) `t_(1//2) propto 1/([A_(0)]^(n-1)),(t_(1//2))_(1)/(t_(1//2))_(2) = [A_(0)]_(2)^(n-1)/([A_(0)]_(1)^(n-1))={([A_(0)]_(2))/([A_(0)]_(1))}^(n-1)` `t/(t_(1//2))=(2a)/(a)^(n-1)` or `2=(2)^(n-1)` or n-1 or n=2

2 Answers 1 views
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 o

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...

2 Answers 1 views
The rate of reaction depends on the concentration of the reacting species, i.e., reactants in rate equations or rate law, consider the gaseous reactio

Correct Answer - A a) in the first experiment, Rate `propto [A]^(x)` `[3]^(3)=27 propto [3]^(x)` x=3

2 Answers 1 views
Assertion: Rate of reaction doubles when the concentration of the reactant is doubled if it is a first order reaction. Reason: Rate constant also beco

Correct Answer - C c) Correct reason. Rate constant remains constant for the reaction and is independent of the initial concentration of the reason.

2 Answers 1 views
For each of the following reactions, express the given rate of change of concentration of the reactants or Products in that reaction: (a) `H_(2)O_(2)

We have, (a) `H_(2)O_(2)+2H^(+)+3I^(-)rarrI_(3)^(-)+2H_(2)O` The equantity in this case is : `-(d[H_(2)O_(2)])/(dt)=-(1)/(2)(d[H^(+)])/(dt)=-(1)/(3)(d-[I^(-)])/(dt)=(d[I_(3)^(-)])/(dt)=(1)/(2)(d[H_(2)O])/(dt)` So, `-(d[I^(-)])/(dt)=-3(d[H_(2)O_(2)])/(dt)=-(2)/(3)(d[H^(+)])/(dt)=3(d[I_(2)^(-)])/(dt)=(3)/(2)(d[H_(2)O])/(dt)` and `-(d[H^(+)])/(dt)=-2(d[H_(2)O_(2)])/(dt)=(2)/(3)(d[I^(-)])/(dt)=2(d[I_(3)^(-)])/(dt)=(d[H_(2)O])/(dt)` (b) `16H^(+)=2MnO_(4)^(-) +10I^(-)rarr2Mn^(2+)+8H_(2)O+5I_(2)` The equality in this case : `(1)/(16)(d[H^(+)])/(dt)=(1)/(2)(d[MnO_(4)^(-)])/(dt)=-(1)/(4)(d[I_(2)])/(dt)` `(1)/(16)(d[H^(+)])/(dt)=(1)/(2)(d[MnO_(4)^(-)])/(dt)=-(1)/(10)(d[I^(-)])/(dt)` `=(1)/(2)(d[Mn^(2+)])/(dt)=(1)/(8)(d[H_(2)O])/(dt)=(1)/(5)(d[I_(2)])/(dt)` So, `-(d[MnO_(4)^(-)])/(dt)=-(1)/(8)(d[H^(+)])/(dt)=-(1)/(5)(d[I^(-)])/(dt)`...

2 Answers 1 views
Consider the following statement : 1 The rate of reaction is always proportinal to the concentration of reactants. 2 The order of an elementary chemic

Correct Answer - C Statement (1) cannot be correct because the rate of zero order reactions does not depend on concentrartion of reactant

2 Answers 1 views
In a reversible chemical reaction having two reactants in equilibrium, if the concentration of the reactants are doubled then the equilibrium constant

Correct Answer - D Consider a hypothetical change. `A+BhArrC+D` For the reaction. `K_(eq)=([C][D])/([A][B])` For the above reaction if concentration of reactions are doubled then the rate of forward reaction increases for...

2 Answers 1 views
X