Question

TO DO TODAY STUDY for ches Problem #3 Consider the following reaction: MONDAY A (4) 3 (4) AG 4,930 A reaction vessel was charged with compound A (4.11 M) and compound (6.71 M) and allowed to react ku. What was the concentration of compound A at that time? 251 k. Alter a period of 11 hours, was determined 2.11 [A] - Incorrect. Submissions to practice problems are not permanently recorded. Tries 2/45 Previous Tries

Answer 1

- Acans Yoll M B con 6.71M Initial Final Yolltu 6.71-X sa= sgo +RT InQ Q: (B3 I CA] 4.93 kJ +(8,314 x 205 x 10-3)/n(1.6326) 621 = 1.6326 4.11 251 = = 4.93 KJ +1.0229 kJ = 5.953 kJ From After 11.1 hours, It means value of a reverse direetion Ah = 2011 kJ decreases, reaction goes into .. Ah= sGo + RT in 6.71-7 Yolltu 2011 - 4.93 + 8.314 x 251 x10^3 in 6.71 –71 4.11+ . 66.71-1 2011.4.93 = -1.3513 4.11tu 2.0868 - -1.3513 = 0.26 ', 671-3 Yollax 6.71-75 1.0640 + 0.26 2 n = 6.71- 1.064. = 4.481 1-26 :- (A) = 4.11+4.48 = 8.59 M1