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

ch3oh(aq)+hcl(aq)→ch3cl(aq)+h2o(l)< br />
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

(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