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Consider the reaction, 2A+B\rightarrow products. When concentration of B alone was doubled, the half­-life did not change. When the concentration of A alone was doubled, the rate increased by two times. The unit of rate constant for this reaction is

  • Option 1)

    s^{-1}\;

  • Option 2)

    \; L\; mol^{-1}s^{-1}\;

  • Option 3)

    \; no\; unit\; \;

  • Option 4)

    \; mol\, L^{-1}s^{-1}

 

Answers (1)

best_answer

As we learnt in 

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

 

2A+ B\rightarrow Product

-\frac{d\left [ R \right ]}{dt}=K\left [ A^{1} \right ]\left [ B^{1} \right ]

Reaction is first ordered and unit of rate constant is sec^{-1}

Reaction is 2nd ordered and unit of rate constant is Lmol^{-1}}sec^{-1}


Option 1)

s^{-1}\;

Inorrect Option

Option 2)

\; L\; mol^{-1}s^{-1}\;

Correct Option

Option 3)

\; no\; unit\; \;

Inorrect Option

Option 4)

\; mol\, L^{-1}s^{-1}

Inorrect Option

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Aadil

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