ClO_2(g) + F_2(g) rightarrow ClO_2F(g) The following results were obtained for t
ID: 490415 • Letter: C
Question
ClO_2(g) + F_2(g) rightarrow ClO_2F(g) The following results were obtained for the unbalanced reaction represented above at 25 degree C. Write the rate law expression for the reaction. Calculate the value of the rate constant. In experiment 2, find the rate of decrease of [F_2]? Which of the following reaction mechanisms is consistent with the rate law developed in (a). Justify your choice. ClO_2 + F_2 rightarrow ClO_2F (fast) ClO_2F_2 rightarrow ClO_2F + F (slow) ClO_2 + F rightarrow ClO_2F (fast) F_2 rightarrow 2F (slow) 2(ClO_2 + F rightarrow ClO_2F) (fast)Explanation / Answer
Let Rate = k [ClO2]x[F2]y where x is order with respect to ClO2 and y = order with respect to F2
Comparing experiment 2 with experiment 1
rate2/rate1 = 9.6x10-3/2.4x 10-3 = 4 = (0.010/0.010)x (0.40/0.10)y
Thus 4 = 4y , thus y = 1
comparing rate 3 and rate 1
rate3/rate1 =9.6/2.4 = 2x .2
Thus x =1
Thus rate law is
rate= k [CLO2][F2]
b)Substituting values from experiment 1
2.4x10-3 = k (0.010) (0.10)
Therefore k = 2.4L/mol.s
c)From the stoichiometric expression
rate of decrese in [F2] = -1/2 rate of increase in [ClO2F]
= -1.2 x x10-3 M/s
d) As the rate law is
rate = k[ClO2][F2]
the mechanism consistent is first mechanism, in which the slow step has rate = k [ClO2F2] which inturn is [ClO2]{F2].