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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

Monjur Alahi

The rate of reaction depends on the concentration of the reacting species, i.e., reactants in rate equations or rate law, consider the gaseous reactions:
`aA bB to "Products"`
The rate laws is `r=k[AP][Y]^(y)`
Here (x+y) is the order with respect to that reactant will be taken as zero.
In the reaction `A+B to "Products"`, if the concentration of B is kept fixed and the concentration of A is increased 3 times, the reaction rate increases 27 times. If the concentration of both A and B are doubled, then the increases 8 times. The order w.r.t. A is:
A. 3
B. 2
C. 1
D. zero


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1 Answers

Salim Ahmed Kawsar

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

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For a reaction between gaseous compounds, `2A + B rarrC + D` , the reaction rate law is rate k[A][B]. If the volume of the container is made `1//4^(th

Correct Answer - C Concentration will increase by 4 times of each reactant `therefore ` Rate will become ` (4)^(2)` I .e 16 times

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a) write dimensions of first order reaction. b) Give reason for no change in concentration of `H_(2)` and `Cl_(2)` with respect to time in the reactio

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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...

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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

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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...

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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

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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)`...

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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

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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

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