In a multi-step reaction, the rate is determined by conisdering the………….step.

In a multistep reaction, the fastest step is the rate determining step.

(A) In a multiple reaction, the rate determining step is the fast step. (R ) In multistep reaction order for each step can be defined.

True or False : The rate of a multistep reaction is determined by the fastest step in the sequence.

The rate law expresison is given for a typical reaction, n_(1)A + n_(2) B rarrP as r = k[A]^(n)[B]^(n2) . The reaction completes only in one step and A and B are present in the solution. If the reaction occurs in more than one step, then the rate law is expressed by consdering the slowest step, i.e., for S_(N)l reaction r = k[RX] . If the eraction occurs in more than one step and the rates of the steps involved are comparable, then steady state approximation is conisdered, i.e., the rate of formation of intermediate is always equal to the rate of decompoistion of the intermediate. Conisder the reaction: [[I_(2) underset(k_(1))overset(k_(2))hArr2I("rapid equilibrium")],[H_(2)+2I overset(k_(3))rarr 2HI("slow")]] If we increase the concentration of I_(2) two times, then the rate of formation of HI will

The rate law expresison is given for a typical reaction, n_(1)A + n_(2) B rarrP as r = k[A]^(n)[B]^(n2) . The reaction completes only in one step and A and B are present in the solution. If the reaction occurs in more than one step, then the rate law is expressed by consdering the slowest step, i.e., for S_(N)l reaction r = k[RX] . If the eraction occurs in more than one step and the rates of the steps involved are comparable, then steady state approximation is conisdered, i.e., the rate of formation of intermediate is always equal to the rate of decompoistion of the intermediate. Conisder the reaction: [[I_(2) underset(k_(1))overset(k_(2))hArr2I("rapid equilibrium")],[H_(2)+2I overset(k_(3))rarr 2HI("slow")]] If we increase the concentration of I_(2) two times, then the rate of formation of HI will

The rate law expresison is given for a typical reaction, n_(1)A + n_(2) B rarrP as r = k[A]^(n)[B]^(n2) . The reaction completes only in one step and A and B are present in the solution. If the reaction occurs in more than one step, then the rate law is expressed by consdering the slowest step, i.e., for S_(N)l reaction r = k[RX] . If the eraction occurs in more than one step and the rates of the steps involved are comparable, then steady state approximation is conisdered, i.e., the rate of formation of intermediate is always equal to the rate of decompoistion of the intermediate. Conisder the reaction: [[I_(2) underset(k_(1))overset(k_(2))hArr2I("rapid equilibrium")],[H_(2)+2I overset(k_(3))rarr 2HI("slow")]] Which of the following expresison is correct?

The rate law expresison is given for a typical reaction, n_(1)A + n_(2) B rarrP as r = k[A]^(n)[B]^(n2) . The reaction completes only in one step and A and B are present in the solution. If the reaction occurs in more than one step, then the rate law is expressed by consdering the slowest step, i.e., for S_(N)l reaction r = k[RX] . If the eraction occurs in more than one step and the rates of the steps involved are comparable, then steady state approximation is conisdered, i.e., the rate of formation of intermediate is always equal to the rate of decompoistion of the intermediate. Conisder the reaction: [[I_(2) underset(k_(1))overset(k_(2))hArr2I("rapid equilibrium")],[H_(2)+2I overset(k_(3))rarr 2HI("slow")]] Which of the following expresison is correct?

In Cannizzaro's reaction, what is the rate determining step ?