Summary CHAPTER 14 CHEMICAL KINETICS PRACTICE EXAMPLES ( NEW UPDATE 2021)

CHAPTER 14 CHEMICAL KINETICS PRACTICE EXAMPLES 1A (E) The rate of consumption for a reactant is expressed as the negative of the change in molarity divided by the time interval. The rate of reaction is expressed as the rate of consumption of a reactant or production of a product divided by its stoichiometric coefficient. rate of consumption of A     A 5 1 0.3187 M 0.3629 M 1min = = = 8.93 10 M s t 8.25 min 60 s         rate of reaction = rate of consumption of A2 = 5 1 8.93 10 M s 5 1 4.46 10 M s 2        1B (E) We use the rate of reaction of A to determine the rate of formation of B, noting from the balanced equation that 3 moles of B form (+3 moles B) when 2 moles of A react (–2 moles A). (Recall that “M” means “moles per liter.”) 0.5522M A 0.5684M A 3moles B 4 1 rateof B formation= 1.62 10 M s 60s 2moles A 2.50 min 1min         2A (M) (a) The 2400-s tangent line intersects the 1200-s vertical line at 0.75 M and reaches 0 M at 3500 s. The slope of that tangent line is thus 0 M 0.75 M 4 1 slope = = 3.3 10 M s = instantaneous rate 3500 s 1200 s       of reaction The instantaneous rate of 4 1 reaction = 3.3 10 M s    . (b) At 2400 s, H O = 0.39 M. At 2450 s, H O = 0.39 M + rate  2 2   2 2   t   4 1 1 At 2450 s, H O = 0.39 M + 3.3 10 mol H O L s 50s 2 2 2 2 = 0.39 M 0.017 M = 0.37 M            2B (M) With only the data of Table 14.2 we can use only the reaction rate during the first 400 s,   4 1 2 2 H O / = 15.0 10 M s t      , and the initial concentration,  2 2 0 H O = 2.32 M. We calculate the change in  HO2 2  and add it to  2 2 0 HO to determine 2 2 100 HO .   4 1 H O = rate of reaction of H O = 15.0 10 M s 100 s = 0.15 M 2 2 2 2 t          22 22 22        100 0 H O = H O + H O = 2.32 M + 0.15 M = 2.17 M   This value differs from the value of 2.15 M determined in text Example 14-2b because the text used the initial rate of reaction   4 1 17.1 10 M s    , which is a bit faster than the average rate over the first 400 seconds.