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Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074

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BuyFindarrow_forward

Chemistry & Chemical Reactivity

10th Edition
John C. Kotz + 3 others
ISBN: 9781337399074
Textbook Problem

The decomposition of HOF occurs at 25 °C.

HOF(g) → HF(g) + ½ O2(g)

Using the data in the table below, determine the rate law, and then calculate the rate constant.

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Interpretation Introduction

Interpretation:

The rate law for the given chemical equation and the rate constant for the reaction should be determined.

Concept Introduction:

The rate of reaction is the quantity of formation of product or the quantity of reactant used per unit time.  The rate of reaction doesn’t depend on the sum of amount of reaction mixture used.

The raise in molar concentration of product of a reaction per unit time or decrease in molarity of reactant per unit time is called rate of reaction and is expressed in units of mol/(L.s).

Integrated rate law for first order reaction:

Consider A as substance, that gives the product based on the equation,

    aAproducts

Where a= stoichiometric co-efficient of reactant A.

Consider the reaction has first-order rate law,

    Rate=-Δ[A]Δt=k[A]

The integrated rate law equation can be given as,

    ln[A]t[A]o=-kt

The above expression is called integrated rate law for first order reaction.

Explanation

With given concentration of reactant values with respect to time the rate law and the rate constant should be determined.

The given chemical reaction is as follows,

HOF(g)HF(g)+12O2(g)

Analyzing the given chemical reaction it is clear that for given set of reactant concentration to the time taken results in the formation of 12O2(g) and one mole of HF which means 2 moles of HF gives rise to one mole of O2(g) formation.

The given table clearly shows that initial concentration that is in 0 time taken the concentration of reactant is 0.850mol/L.  Using this initial concentration value and the concentration value as the time proceed the rate constant is calculated as follows by first using first order rate equation.

  k=1tln(initial concentration (a)concentration with respect to time taken (ax))

  k1=12sln(0.85mol/L0.81mol/L)=0

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