Question

The reaction between ethyl bromide (C2H5Br) and hydroxide ion in ethyl alcohol at 330 K, C2H5BR(alc) + OH-(alc) --> C2H5OH(l) + Br-(alc), is first order each in ethyl bromide and hydroxide ion. When [C2H5Br] is 0.0477 M and [OH-] is 0.100 M, the rate of disappearance of ethyl bromide is 1.7 x 10^-7 M/s.
What is the value of the rate constant?
k=?

Answer 1

Answer: $3.5\times 10^{-5}mol^{-1}Lsec^{-1}$

Explanation:

Rate law says that rate of a reaction is directly proportional to the concentration of the reactants each raised to a stoichiometric coefficient determined experimentally called as order.

$C_2H_5Br(alc)+OH^-(alc)\rightarrow C_2H_5OH(l)+Br^-(alc)$

Given: Order with respect to $C_2H_5Br$ = 1

Order with respect to $OH^-$ = 1

Thus rate law is:

$Rate=k[C_2H_5Br]^1[OH^-]^1$

k= rate constant

The rate in terms of reactants is given as negative as the concentration of reactants is decreasing with time whereas the rate in terms of products is given as positive as the concentration of products is increasing with time.

$Rate=-\frac{1d[C_2H_5Br]}{dt}=k[C_2H_5Br]^1[OH^-]^1$

Given: $\frac{d[C_2H_5]}{dt}]=1.7\times 10^{-7}$

Putting in the values we get:

$Rate=1.7\times 10^{-7}=k[0.0477]^1[0.100]^1$

$k=3.5\times 10^{-5}mol^{-1}Lsec^{-1}$

Thus the value of the rate constant is $3.5\times 10^{-5}mol^{-1}Lsec^{-1}$