PORTAGE LEARNING CHEM 104 EXAMINATION TEST 2026 QUESTIONS WITH SOLUTIONS GRADED A+

PORTAGE LEARNING CHEM 104
EXAMINATION TEST 2026 QUESTIONS WITH
SOLUTIONS GRADED A+


◉ Show your work and be sure to give answer with the correct units.
The following rate data was obtained for the reaction:
2 ClO2 + 2 OH- → ClO3- + ClO2- + H2O


Experiment # [ClO2] [OH-] rate
1 0.060 0.030 2.48 x 10-2
2 0.020 0.030 2.76 x 10-3
3 0.020 0.090 8.28 x 10-3
Determine the reaction order with respect to (1) ClO2 and (2) OH-, (3)
write the rate law and then (4) determine the value of the rate constant,
k. Answer: Rate = k [ClO2]x [OH-]y
If data from experiments 1 and 2 was substituted into the equation we
would obtain
2.48 x 10-2 = k [0.060]x [0.030]y
2.76 x 10-3 k [0.020]x [0.030]y
and if we cancel all common terms we would obtain

, 2.48 x 10-2 = k [0.060]x [0.030]y
2.76 x 10-3 k [0.020]x [0.030]y
2.48 x 10-2 = [0.060]x 9 = (3)x
2.76 x 10-3 [0.020]x


(1) which yields x = 2 as the order of reaction with respect to ClO2
If data from experiments 2 and 3 was substituted into the equation we
would obtain
2.76 x 10-3 = k [0.020]x [0.030]y
8.28 x 10-3 k [0.020]x [0.090]y
and if we cancel all common terms we would obtain
2.76 x 10-3 = k [0.020]x [0.030]y
8.28 x 10-3 k [0.020]x [0.090]y


2.76 x 10-3 = [0.030]y 0.333 = (0.333)y
8.28 x 10-3 [0.090]y


(2) which yields y = 1 as the order of reaction with respect to OH-.
(3) the overall rate law can now be written as follows:
Rate = k [ClO2]2 [OH-]1
(4) and using the data from experiment 1 we can determine the rate
constant as follows:
2.48 x 10-2 = k [0.060]2 [0.030]