For the following reaction t (s) [H] (M) CH3OH(aq) H+(aq) CI-(aq) CH3C1(aq) H20(

Question

For the following reaction t (s) [H] (M) CH3OH(aq) H+(aq) CI-(aq) CH3C1(aq) H20(1) - 238 + + + 29.0 2.16 the value of [H'] was measured over a period of time. Given the data, find the average 65.0 2.04 rate of disappearance of H'(aq) for the time interval between each measurement. 119.0 1.87 Interval: o s to 29.0s 29.0 s to 65.0s 65.0s to 119.0s Number Number Reaction rate MIs M/s What is the average rate of appearance of CH3Cl(aq) for the same time intervals? Interval: 0 s to 29.0s 29.0 s to 65.0s 65.0 s to 119.0s Number Reaction rate M/s M/s

Explanation / Answer

Average rate of disappearance of H+ = (initial concentration - final concentration)/(final time - initial time)


(a) From t = 0 s to 29.0 s

Rate = (2.38 – 2.16)/(29.0 - 0) = 0.00759 M/s = 7.59 x 10^(-3) M/s


(b) From t = 29.0 s to 65.0 s

Rate = (2.16 – 2.04)/(65.0 - 29.0) = 0.00333 M/s = 3.33 x 10^(-3) M/s


(c) From t = 65.0 s to 119.0 s

Rate = (2.04-1.87)/(119.0 - 65.0) = 0.00315 M/s = 3.15 x 10^(-3) M/s


CH3OH(aq) + H^+(aq) + Cl^-(aq) => CH3Cl(aq) + H2O(l)

1 mole of H+ gives 1 mole of CH3Cl

Average rate of appearance of CH3Cl = average rate of disappearance of H+


(d) From t = 0 s to 29.0 s

Rate = 0.00759 M/s = 7.59 x 10^(-3) M/s


(e) From t = 29.0 s to 65.0 s

Rate = 0.00333 M/s = 3.33 x 10^(-3) M/s


(f) From t = 65.0 s to 119.0 s

Rate = 0.00315 M/s = 3.15 x 10^(-3) M/s

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