Learning Goal: To understand reaction order and rate constants. For the general equation aA+bBcC+dD, the rate law is expressed as follows: rate=k[A]m[B]n where m and n indicate the order of the reaction with respect to each reactant and must be determined experimentally and k is the rate constant, which is specific to each reaction. For a particular reaction, aA+bB+cCdD, the rate law was experimentally determined to be rate=k[A]0(B]t[C]2=k[B][C]2 A. This equation is zero order with respect to A. Therefore, changing the concentration of A has no eflect on the rate because A with always equal 1. B. This equation is first order with respect to B. This means that it the concentration of B is doubled, the rale will double. If [B] is reduced by hat, the rate will be halved. If [B] is tripled, the rate will triple, and 00 on. C. This equation is second order with respect to C. This means that if the concentration of C is doubled, the rate will quadruple. If [C] is tripled, the rate will increase by a factor of 9 , and so on. Overall reaction order and rate-constant units The sum of the individual orders gives the overall reaction order. The example equation above is third order overal because 0+1+2=3. For the units of rate to come out to be M/s, the units of the rate constant for third-order reactons must be M2s1 since M/s=(M2s1)(M3) For a second-order reaction, the rate constant has units of M181 because M/s=(M151)(M2). In a first-order ceaction, the rate constant has the units s1 because M/s=(s1)(M1). Analyzing a specific reaction Consider the following reaction: O2+2NO2NO2rate=k[O2][NO]2 What would happen to the rate if [O2] were doublod? View Avallable Hint(s) stay the same. The tate would double. triple. quadruple. Part D What would happen to the rate if [NO] were doubled? View Avaliable Hint(s) stay the same. The rate would double. triple. quadruple