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  • How to solve this problem- - Chemical Kinetics and Surface Chemistry - BITSAT

For the reaction X+Y=Z the rate expression is Rate = k[x]^{2}[y]^\frac{1}{2}  If the concns of  x and y are both increased by a factor of 4, by what factor will the rate increase?

  • Option 1)

    4

  • Option 2)

    8

  • Option 3)

    16

  • Option 4)

    32

 

Answers (1)

best_answer

As we discussed inconcept

Rate of Law = Dependence of Rate on Concentration -

The representation of rate of a reaction in terms of concentration of the reactants is known as Rate Law

or

The Rate Law is the expression in which reaction rate is given in terms of molar concentration of reactants with each term raised to some power, which may/maynot be equal to stoichiometric of the reacting species in a balanced chemical equation 

- wherein

Formula: aA+bB\rightarrow cC+dD

              Rate=\frac{dR}{dT}

              =\alpha [A]^{x}.[B]^{y}

             =k[A]^{x}.[B]^{y}

 K= rate constant

 

 

X+Y\rightarrow Z

rate =k[x]^{2}[y]\frac{1}{2}

Now, the changes proposed in concentration are 

x\rightarrow 4x,\: \: y\rightarrow 4y

Then, new rate = k[4x]^{2}[4y]^{\frac{1}{2}}

=k16[x]^{2}2[y]^{\frac{1}{2}}

=32k[x]^{2}[y]^{\frac{1}{2}}

Therefore, rate will increase by a fraction of 32.

 


Option 1)

4

Incorrect option

Option 2)

8

Incorrect option

Option 3)

16

Incorrect option

Option 4)

32

Correct option

Posted by

divya.saini

View full answer

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