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:


Rate=-1aΔ[A]Δt=1bΔ[B]Δt


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?

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Based on our data, we think this problem is relevant for Professor Savrov's class at UNM.