# Problem: The aqueous reaction of KOH with HBr is:KOH(aq) + HBr(aq) → KBr(aq) + H2O(l)At the start of such a reaction, the concentration of KOH(aq) is 0.275M. After 3.05 seconds the concentration is observed to be 0.0557M KOH(aq).What is the rate of the reaction?

###### FREE Expert Solution
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###### FREE Expert Solution

KOH(aq) + HBr(aq) → KBr(aq) + H2O(l)

Recall that for a reaction aA  bB, the rate of a reaction is given by:

$\overline{){\mathbf{Rate}}{\mathbf{=}}{\mathbf{-}}\frac{\mathbf{1}}{\mathbf{a}}\frac{\mathbf{\Delta }\mathbf{\left[}\mathbf{A}\mathbf{\right]}}{\mathbf{\Delta t}}{\mathbf{=}}\frac{\mathbf{1}}{\mathbf{b}}\frac{\mathbf{\Delta }\mathbf{\left[}\mathbf{B}\mathbf{\right]}}{\mathbf{\Delta t}}}$

where:

Δ[A] = change in concentration of reactants or products (in mol/L or M), [A]final – [A]initial

Δt = change in time, tfinal – tinitial

Since KOH is a reactant, the rate with respect to KOH is negative (–) since we’re losing reactants.

The rate of the reaction is:

94% (152 ratings)
###### Problem Details

The aqueous reaction of KOH with HBr is:

KOH(aq) + HBr(aq) → KBr(aq) + H2O(l)

At the start of such a reaction, the concentration of KOH(aq) is 0.275M. After 3.05 seconds the concentration is observed to be 0.0557M KOH(aq).

What is the rate of the reaction?