The first-order decomposition of Sp06(g) to form SO3(g), has a half-life of 1.262 x 1os at 45°C. How long will it take for 60% of a 0.34 M sample to decompose at 45°C?A. 1,68 x 10sO B. 406 sC. 728 SD. 9,34 x 104sE. 1.89 x 10 s

First, calculate the rate constant from half-life as follows: k = ln2t12= ln21.262×10-3s= 549.245 s-1Initial concentration, Ao = 0.34M Since 60% of the sample decomposes, the concentration remains after decomposition = 40% So, concentration remains after decomposition, A = [(40/100)×0.34] = 0.136M Using the integrated first-order rate expression: k = 1tlnAoA549.245 s-1 = 1tln0.34M0.136Mt = 1.668 ×10-3 s The answer is close to option-A i.e. 1.68 × 10-3 s. So, option-A is the correct answer.

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The first-order decomposition of S206(g) to form SO3(g), has a half-life of 1.262 x 103 s at 45°C. How long will it take for 60% of a 0.34 M sample to decompose at 45°C? O A. 1.68 x 103s O B. 406 s O C. 728 s O D.9.34 x 10-4s O E. 1.89 x 103s

The first-order decomposition of S206(g) to form SO3(g), has a half-life of 1.262 x 103 s at 45°C. How long will it take for 60% of a 0.34 M sample to decompose at 45°C? O A. 1.68 x 103s O B. 406 s O C. 728 s O D.9.34 x 10-4s O E. 1.89 x 103s

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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