mod/ibis/wiew.php?id-5857854 Giant Salvma(Sde g Swords, sound effe ? SparkNotes
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mod/ibis/wiew.php?id-5857854 Giant Salvma(Sde g Swords, sound effe ? SparkNotes Toa. D openng ceremonies Feedet Panel, Shee f998 dodge ra ege-CHEM 2423 -Summer18-KOUADIO Activities and Due Dates Ch6 Gradeboo O 7/7/2018 11:30 PM 0 60.8/100 o 7/6/2018 10:46 PM PrintCalculator Periodic Table 13 of 1 Map da Sapling Learning e following reaction is a single-step, bimolecular reaction CH, Br NaHCH,OH NaBr When the concentrations of CH Br and NaOH are both 0.160 M, the rate of the reaction is 0.0020 M/s. (a) What is the rate of the reaction if the concentration of CH Br is doubledt? Number M/s if the concentration of NaOH is halved? (b) What is the rate of the Number M/S (c) What is the rate of the reaction if the concentrations of CH Br and NaOH are both increased by a factor of five? Number 0.2777 M/s Eat O Prevous ®Give up & View Solution. Try Agan?NextExplanation / Answer
Ans. Rate, r = k [CH3Br] [NaOH] ; where, k = rate constant
Or, 0.0020 M s-1 = k (0.160 M) x (0.160 M)
Or, k = 0.0020 M s-1 / 0.0256 M2
Hence, k = 0.0781250 M-1 s-1
#a. Given, [CH3Br] is doubled = 2 x (0.160 M) = 0.320 M
[NaOH] = 0.160 M
Now, putting the values in rate expression-
r = (0.0781250 M-1 s-1) (0.320 M) (0.160 M)
Hence, r = 0.004 M s-1
#b. Given, [NaOH] is halved = 1/2 x (0.160 M) = 0.080 M
[CH3Br] = 0.160 M
Now, putting the values in rate expression-
r = (0.0781250 M-1 s-1) (0.160 M) (0.080 M)
Hence, r = 0.001 M s-1
#c. Given, [NaOH] is increased by a factor of 5 = 5 x (0.160 M) = 0.800 M
[CH3Br] is increased by a factor of 5 = 5 x (0.160 M) = 0.800 M
Now, putting the values in rate expression-
r = (0.0781250 M-1 s-1) (0.800 M) (0.800 M)
Hence, r = 0.050 M s-1
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