For the following reaction: CH3OH(aq) + H^+(aq) + Cl^-(aq) yields CH3Cl(aq) + H2

Question

For the following reaction:

CH3OH(aq) + H^+(aq) + Cl^-(aq)  yields CH3Cl(aq) + H20(l)

t(s): 0, 34.0, 60.0, 125.0

[H^+] (M): 2.20, 1.98, 1.86, 1.69

the value of [H^+] was measured over the a period of time. Given the data, find the average rate of disspearance of H^+ (aq) for the time interval between each measurement.

intervals:a) 0 s to 34.0 s, b)34.0 s to c) 60.0 s, 60.0 s to 125.0 s. Answers must be in M/s

what is the average rate of appearance of CH3Cl(aq) for the same intervals?

interval:d) 0 s to 34.0 s, e)34.0 s to 60.0 s, f)60.0s to 125.0 s. answers must be in M/s

Explanation / Answer

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


(a) From t = 0 s to 34.0 s

Rate = (2.20 - 1.98)/(34.0 - 0) = 0.00647 M/s = 6.47 x 10^(-3) M/s


(b) From t = 34.0 s to 60.0 s

Rate = (1.98 - 1.86)/(60.0 - 34.0) = 0.00462 M/s = 4.62 x 10^(-3) M/s


(c) From t = 60.0 s to 125.0 s

Rate = (1.86 - 1.69)/(125.0 - 60.0) = 0.00262 M/s = 2.62 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 34.0 s

Rate = 0.00647 M/s = 6.47 x 10^(-3) M/s


(e) From t = 34.0 s to 60.0 s

Rate = 0.00462 M/s = 4.62 x 10^(-3) M/s


(f) From t = 60.0 s to 125.0 s

Rate = 0.00262 M/s = 2.62 x 10^(-3) M/s

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