Question

Problem 14.33 - Enhanced - with Feedback 1 Review | Constants i Peri Part B The iodide ion reacts with hypochlorite ion (the active ingredient in chlorine bleaches) in the following way: OCl +I- OI- +Cl- Calculate the rate constant with proper units. This rapid reaction gives the following rate data: Express the rate constant to two significant figures. Indicate the multiplication of units, as necess explicitly either with a multiplication dot or a dash (e.g. ft * Ib or ft-lb). ? [OC](M) 1.5 x 10-3 3.0 x 10-3 1.5 x 10-3 I](M) Initial Rate (M/s) 1.5 x 10-3 1.36 x 10-4 1.5 x 10-3 2.72 x 10-4 3.0 x 10-3 2.72 x 10-4 TE Å k = 60.4 Templates R o M 'S- You may want to reference (Pages 575 - 580) Section 14.3 while completing this problem. Submit Previous Answers Request Answer X Incorrect; Try Again; 9 attempts remaining Part Calculate the rate when [OCl"] = 1.5 x 10-3 and I-] = 5.6 x 10-4.

Need help with Part B and C

Answer 1

see experiment 1 and 2:

[OCl-] doubles

[I-] is constant

rate doubles

so, order of OCl- is 1

see experiment 1 and 3:

[OCl-] is constant

[I-] doubles

rate doubles

so, order of I- is 1

Rate law is:

rate = k*[OCl-]*[I-]

B)

rate = k*[OCl-]*[I-]

Put values from 1st row of table in rate law

rate = k*[OCl-]*[I-]

1.36*10^-4 = k*0.0015*0.0015

k = 60.44 M-1.s-1

Answer: 6.0*10^1 M-1.s-1

C)

rate = k*[OCl-]*[I-]

rate =60.44*0.0015*0.0056

rate = 5.08*10^-4 M/s

Answer: 5.1*10^-4 M/s