The iodine ion reacts with hypochlorite ion (the activeingredient in chlorine bl

Question

The iodine ion reacts with hypochlorite ion (the activeingredient in chlorine bleaches) in the following way: OCl- + I- ----> OI- +Cl-. This rapid reaction gives the following ratedata: [OCl-],M                   I-,M                       Rate, M/s 1.5 x10-3                     1.5x10-3                   1.36 x 10-4 3.0 x10-3                1.5x10-3                   2.72 x 10-4 1.5 x10-3                     3.0x10-3                    2.72 x 10-4 A) write the rate law for this reaction B) Calculate the rate constant C) Calculate the rate when [OCl-] = 2.0 x10-3 and [I-] = 5.0 x 10-4M. The iodine ion reacts with hypochlorite ion (the activeingredient in chlorine bleaches) in the following way: OCl- + I- ----> OI- +Cl-. This rapid reaction gives the following ratedata: [OCl-],M                   I-,M                       Rate, M/s 1.5 x10-3                     1.5x10-3                   1.36 x 10-4 3.0 x10-3                1.5x10-3                   2.72 x 10-4 1.5 x10-3                     3.0x10-3                    2.72 x 10-4 A) write the rate law for this reaction B) Calculate the rate constant C) Calculate the rate when [OCl-] = 2.0 x10-3 and [I-] = 5.0 x 10-4M. A) write the rate law for this reaction B) Calculate the rate constant C) Calculate the rate when [OCl-] = 2.0 x10-3 and [I-] = 5.0 x 10-4M.

Explanation / Answer

From the chart you can see that as both reactant are doubledthe rate also doubles. This means that it is a second orderrate reation and the equation is like this. A.   Rate=k*[OCl-]*[I-] B. Use the data from any of the row to calculate the kat this temperature. 1.36*10^-4=k*1.5*10^-3*1.5*10^-3 k=60.44 C. Use the rate constant from B to calculate the newrate. rate=60.44*2.0*10^-3*5.0*10^-4 rate=6.0*10^-5

Related Questions

Navigate

• Browse (All)
• Browse T
• Subjects

---

---

Integrity-first tutoring: explanations and feedback only — we do not complete graded work. Learn more.