Writte the rate law and order for the following reaction:
`AB_(2)+C_(2)to AB_(2)C+C` (slow)
`AB_(2)+C to AB_(2)C` (fast)

Write the rate law and order for the following reaction: AB_2 +C_2 to AB_2 C +C (slow) AB_2 + C to AB_2 C ( fast)

The reaction , 2AB(g) + 2C(g) to A_(2(g)) + 2BC_((g)) proceeds according to the mechanism . I. 2AB hArr A_(2)B_(2) (fast) II. A_(2)B_(2)+C to A_(2)B + BC (slow ) III. A_(2)B+C to A_(2)+BC (fast) what will be the initial rate taking [AB] = 0.2 M and [C] = 0.5 M ? The K_(c) for the step I is 10^(2) M^(-1) and rate constant for the step II is 3.0 xx 10^(-3) mol^(-1) min^(-1)

A chemical reaction A + 2B rArr AB_(2) follows in two steps A + B rArr AB(slow) AB + B rArr AB_(2) (fast) Then the order of the reaction is

In hypothetical reaction A_2+B_2 = 2AB follows the mechanism as given below: A_2 iff A+A (fast reaction) A+B_2 to AB+B (slow reaction) A+B to AB (fast reaction) give the rate law and order of reaction.