The following initial rate data are for the reaction of tertiary butyl bromide w

Question

The following initial rate data are for the reaction of tertiary butyl bromide with hydroxide ion at 55 °C (CH3)3C Br OHT (CH3)3CoH Br ICCH303CBrlo, M OH1er M Initial Rate, Ms Experiment 7.14x10-3 0.632 0.422 7.14x10-3 0.844 0.632 1.42x10-2 1.26 0.422 4 0.844 1.42x10-2 1.26 Complete the rate law for this reaction in the box below Use the form klAlm[B]n, where '1' is understood for m or n and concentrations taken to the zero power do not appear. Don't enter 1 for m or n Rate From these data, the rate constant is

Explanation / Answer

Answer - We are given, concentrations of reactants and rate of formation

First we need to calculate the order of reaction

Rate = k [(CH3)3CBr]m [OH-]n

We assume m and n order with respect to (CH3)3CBr and OH-

Rate1 = k [(CH3)3CBr]1m [OH-]1n

Rate2 = k [(CH3)3CBr]2m [OH-]2n

Rate3 = k [(CH3)3CBr]3m [OH-]3n

Rate4 = k [(CH3)3CBr]4m [OH-]4n

Now first we need to calculate order of (CH3)3CBr

So,

Rate3/ Rate1 = k [(CH3)3CBr]3m [OH-]3n / k [(CH3)3CBr]1m [OH-]1n

1.42*10-2 / 7.14*10-3 = (1.26)m /(0.632)m * (0.422)n / (0.422)n

2 = (2)m

So, m = 1

Now order of OH-

Rate2/Rate1 = k [(CH3)3CBr]2m [OH-]2n / k [(CH3)3CBr]1m [OH-]1n

7.14*10-3 / 7.14*10-3 = (0.632)m /( 0.632)m *(0.844)n /(0.422)n

1 = (2)n

So, n = 0

So order with respect OH- is zero

So overall order = 1+ 0 = 1. Overall order is first order.

So rate law, Rate = k [(CH3)3CBr]

Rate constant-

we know rate law

Rate = k [(CH3)3CBr]

so, 7.14*10-3 = k (0.632)

k = 1.13*10-2 s-1

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