The rate law for a reaction between the substances A and B  is given by rate = k\left [ A \right ]^{n}\left [ B \right ]^{m}.

On doubling the concentration of A and halving the concentration of B , the ratio of the new rate to the earlier rate of the reaction will be as

  • Option 1)

    \frac{1}{2^{m+n}}\; \; \;

  • Option 2)

    (m+n)\; \;

  • Option 3)

    \; (n-m)\; \;

  • Option 4)

    \; 2^{(n-m)}

 

Answers (1)

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

 

 R_{1}=K\left [ A \right ]^{n}\: \left [ B \right ]^{n} - - - - - - \left ( i \right )

R_{2}=K\left [ 2A \right ]^{n}\:\left [ \frac{B}{2} \right ]^{m} - - -- - -- - \left ( ii \right )

\frac{R2}{R1}=\left ( 2 \right )^{n} \times\left ( \frac{1}{2} \right )^{m}

=2^{\left ( n-m \right )}


Option 1)

\frac{1}{2^{m+n}}\; \; \;

Incorrect option

Option 2)

(m+n)\; \;

Incorrect option

Option 3)

\; (n-m)\; \;

Incorrect option

Option 4)

\; 2^{(n-m)}

Correct option

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