sapling learning The following reaction is a single-step, bimolecular reaction:

Question

sapling learning The following reaction is a single-step, bimolecular reaction: CH,Brit NaoH CH,oH+ NaBr When the concentrations of chaBr and NaoH are both o.160 M, the rate of the reaction is o 0050 M/s. (a) What is the rate of the reaction if the concentration ofCH3Br is doubled? Number M/s (b) What is the rate of the reaction if the concentration of NaoH is halved? Number (c) What is the rate of the reaction if the concentrations of CHaBr and NaoH are both increased by a factor of six? Number MIs Previous Give Up & View Solution check Answer Next Exit Hint

Explanation / Answer

For a single step, bimolecular reaction, the rate law can be written as:

rate = k[CH3Br][NaOH]

From the given information, we can determine the rate constant k:

k = rate / ( [CH3Br][NaOH] ) = ( 0.0050 M/s ) / ((0.160 M)(0.160 M)) = 0.195 M-1 s-1.

Now we can determine the rates if we change the concentrations:

Part (a)

Doubling CH3Br gives a concentration of 0.320 M, so:

rate = (0.195 M-1 s-1)(0.320 M)(0.160 M) = 0.00998 M/s

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Part (b)

Halving the NaOH concentration gives a new concentration of 0.08 M, so:

rate = (0.195 M-1 s-1)(0.160 M)(0.08 M) = 0.002496 M/s

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Part (C)

Increasing the concentration of both reactants by a factor of six gives a new concentration of 0.96 for each reactant, so:

rate = (0.195 M-1 s-1)(0.96 M)(0.96 M) = 0.180 M/s

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