Question

Need help with Part B

KAssignment 16 Chap 14: Integrated Rate Law and Half Life Problem 14.42 - Enhanced - with Feedback 5 of 7 Review I Constants I Periodic Table Molecular iodine, I2 (g), dissociates into iodine atoms at 625 K with a first-order rate constant of What is the half-life for this reaction? -1 0.271 S Express the half-life in seconds to three significant figures. You may want to reference (Pages 582-587) Section 14.4 while completing this problem. t1/2 2.56 s Previous Answers Submit Submit Correc The integrated rate law for this first-order reaction (In(I2]t/[12]0) = --kt) can be substituted with variables corresponding to the half-life, where the ratio in the natural log becomes 0.5 (for half the initial concentration) and t1/2 replaces t. The equation is then rearranged to solve for the half-life as follows: In(2 n(0.5) 0.693 t1/2 0.271 s- Part B If you start with 0.050 M of I, at this temperature, how much will remain after 2.56s assuming that the iodine atoms do not recombine to form I, ? Express the molarity to two significant figures.

Answer 1

2.56 s is half life of the reaction after this, the amount of I₂ reduced to half so moles of I₂ reduced to half i.e 0.050/2 = 0.025 moles

So molarity of of solution is 0.025 M