The rate of disappearance of HCl was measured for the following reaction:

CH3OH (aq) + HCl (aq) →CH3Cl (aq) + H2O (l)

The following data were collected:

Time (min) [HCl] (M)

0.0 1.85

54.0 1.58

107.0 1.36

215.0 1.02

430.0 0.580 (a) Calculate the average rate of reaction, in M/s, for the time interval between each measurement. (b) Calculate the average rate of reaction for the entire time for the data from t=0.0min to t=430.0min. (c) Which is greater, the average rate between t=54.0 and t=215.0min, or between t=107.0 and t=430.0min

Answer: (a) 0.000083M/s, 0.000069M/s, 0.000052M/s, 0.000034M/s

(b) average reaction rate between t=0.0min to t=430.0min = 0.000049M/s

(c) The average rate between t=54.0 and t=215.0min is greater than the average rate between t=107.0 and t=430.0min

Explanation:

Average reaction rate = change in concentration / time taken

(a) after 54mins, t = 54*60s = 3240s

average reaction rate = (1.58 - 1.85) M / (3240 * 0.0) s

= - 0.27M/3240

= 0.000083M/s

after 107mins, t = 107*60s = 6420s

average reaction rate = (1.36 - 1.58) M / (6420 - 3240) s

= - 0.22M/3180s

= 0.000069M/s

after 215mins, t = 215*60s = 12900s

average reaction rate = (1.02 - 1.36) M / (12900 - 6420) s

= - 0.34M/6480s

= 0.000052M/s

after 430mins, t = 430*60 = 25800s

average reaction rate = (0.580 - 1.02) M / (25800 - 12900) s

= - 0.44M/12900s

= 0.000034M/s

(b) average reaction rate between t=0.0min to t=430.0min

= (0.580 - 1.85) M / (25800 - 0.0) s

= - 1.27M/25800s

=0.000049M/s

(c) average reaction rate between t = 54.0min and t = 215.0min

= (1.02 - 1.58) M / (12900 - 3240) s

= - 0.56M/9660s

= 0.000058M/s

average reaction rate between t=107.0 and t=430.0min

= (0.580 - 1.36) M / (25800 - 6420) s

= 0.78M / 19380s

= 0.000040M/s

Therefore the average rate between t=54.0 and t=215.0min is greater than the average rate between t=107.0 and t=430.0min