Elements Of Physical Chemistry
Elements Of Physical Chemistry
7th Edition
ISBN: 9780198796701
Author: ATKINS, P. W. (peter William), De Paula, Julio
Publisher: Oxford University Press
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Chapter 6, Problem 6.1P

(a)

Interpretation Introduction

Interpretation:

The order of the given reaction with respect to glucose has to be found.

Concept introduction:

Rate of a reaction: It represents the speed at which a chemical reaction runs.  How much concentration of substrates (reactants) consumed and how much concentration of targets (products) formed in a unit of time is said to be rate of reaction.

Rate of reaction depends on time, temperature, pressure, concentration, and pH of the reaction.

Rate of the reaction is the change in the concentration of reactant or a product with time. It can be varied in accordance with temperature, pressure, concentration, presence of catalyst, surface area…

General rate reaction is,

aA+bBcC+dD

Rate=-1aΔ[A]Δt=-1bΔ[B]Δt=1cΔ[C]Δt=1dΔ[D]Δt

The negative sign indicates the reduction of concentration of reactant.

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.

(a)

Expert Solution
Check Mark

Explanation of Solution

Given reaction is,

Enzyme(E)+Glucose(G)Enzymehexokinase(E+G)

Record the given data’s

[C6H12O6]/(mmol dm-3)1.001.543.124.02v0/(mol dm-3s-1)5.007.615.520.0

Let order be (n) with respect to glucose (G) and with respect to enzyme (E).

Rate(r)=k×G/n×Er1/r2=(G1/G2)×nn=log(r1/r2)/log(G1/G2)

So, the rate at,

G1=1.00andG2=1.54,r1=5andr2=7.6n1=log(5/7.6)/log(1/1.54)=log(0.65)/log(0.64)=(0.187)/(0.193)n1=0.96

Similarly we calculate other data’s

RateatG1=1.54andG2=3.12,r1=7.6andr2=15.5n2=1.01RateatG1=3.12andG2=4.02,r1=15.5andr2=20.0n3=1.03

Over all order is n=n1,n2,n3=1

Let us we can write the rate let

Rate=k×[glucose]xLetustakeRate2/Rate17.6M/s5.0M/s=[1.54mM]x[1.0mM]x1.52=[1.54]x[1.0]x1.52=[1.54]xx1

Therefore, the order of the reaction with respect to glucose is one (1). That means this is first order reaction.

(b)

Interpretation Introduction

Interpretation:

The rate constant of the given reaction has to be found.

Concept introduction:

Rate of a reaction: It represents the speed at which a chemical reaction runs.  How much concentration of substrates (reactants) consumed and how much concentration of targets (products) formed in a unit of time is said to be rate of reaction.

Rate of reaction depends on time, temperature, pressure, concentration, and pH of the reaction.

Rate of the reaction is the change in the concentration of reactant or a product with time. It can be varied in accordance with temperature, pressure, concentration, presence of catalyst, surface area…

General rate reaction is,

aA+bBcC+dD

Rate=-1aΔ[A]Δt=-1bΔ[B]Δt=1cΔ[C]Δt=1dΔ[D]Δt

The negative sign indicates the reduction of concentration of reactant.

(b)

Expert Solution
Check Mark

Explanation of Solution

Given reaction is,

Enzyme(E)+Glucose(G)Enzymehexokinase(E+G)

Record the given data’s

[C6H12O6]/(mmol dm-3)1.001.543.124.02v0/(mol dm-3s-1)5.007.615.520.0

Hence rate law can be written as

Rate=k×[Glucose]Taketherateconstantvaluesfromexpriment1

Rate=5.0M/s[Glucose]=1.0mM=1.0×103Mk=5.0M/s1.0×103Mk=5.0×103s1

Therefore, the rate constant is k=5.0×103s1_

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Chapter 6 Solutions

Elements Of Physical Chemistry

Ch. 6 - Prob. 6C.4STCh. 6 - Prob. 6D.1STCh. 6 - Prob. 6D.2STCh. 6 - Prob. 6D.3STCh. 6 - Prob. 6E.1STCh. 6 - Prob. 6E.2STCh. 6 - Prob. 6F.1STCh. 6 - Prob. 6F.2STCh. 6 - Prob. 6G.1STCh. 6 - Prob. 6G.2STCh. 6 - Prob. 6G.3STCh. 6 - Prob. 6H.1STCh. 6 - Prob. 6I.1STCh. 6 - Prob. 6I.2STCh. 6 - Prob. 6A.1ECh. 6 - Prob. 6A.2ECh. 6 - Prob. 6A.3ECh. 6 - Prob. 6B.1ECh. 6 - Prob. 6B.2ECh. 6 - Prob. 6B.3ECh. 6 - Prob. 6B.4ECh. 6 - Prob. 6B.5ECh. 6 - Prob. 6C.1ECh. 6 - Prob. 6C.2ECh. 6 - Prob. 6C.3ECh. 6 - Prob. 6C.4ECh. 6 - Prob. 6C.5ECh. 6 - Prob. 6C.6ECh. 6 - Prob. 6C.7ECh. 6 - Prob. 6C.8ECh. 6 - Prob. 6C.9ECh. 6 - Prob. 6C.10ECh. 6 - Prob. 6C.11ECh. 6 - Prob. 6C.12ECh. 6 - Prob. 6D.1ECh. 6 - Prob. 6D.2ECh. 6 - Prob. 6D.3ECh. 6 - Prob. 6D.4ECh. 6 - Prob. 6D.5ECh. 6 - Prob. 6D.6ECh. 6 - Prob. 6D.7ECh. 6 - Prob. 6D.8ECh. 6 - Prob. 6D.9ECh. 6 - Prob. 6D.10ECh. 6 - Prob. 6D.11ECh. 6 - Prob. 6E.1ECh. 6 - Prob. 6E.2ECh. 6 - Prob. 6E.3ECh. 6 - Prob. 6F.1ECh. 6 - Prob. 6F.2ECh. 6 - Prob. 6F.3ECh. 6 - Prob. 6G.1ECh. 6 - Prob. 6G.2ECh. 6 - Prob. 6G.3ECh. 6 - Prob. 6G.5ECh. 6 - Prob. 6H.1ECh. 6 - Prob. 6H.2ECh. 6 - Prob. 6H.3ECh. 6 - Prob. 6H.4ECh. 6 - Prob. 6I.1ECh. 6 - Prob. 6I.2ECh. 6 - Prob. 6I.3ECh. 6 - Prob. 6.1DQCh. 6 - Prob. 6.2DQCh. 6 - Prob. 6.3DQCh. 6 - Prob. 6.4DQCh. 6 - Prob. 6.5DQCh. 6 - Prob. 6.6DQCh. 6 - Prob. 6.7DQCh. 6 - Prob. 6.8DQCh. 6 - Prob. 6.9DQCh. 6 - Prob. 6.10DQCh. 6 - Prob. 6.11DQCh. 6 - Prob. 6.12DQCh. 6 - Prob. 6.13DQCh. 6 - Prob. 6.14DQCh. 6 - Prob. 6.1PCh. 6 - Prob. 6.2PCh. 6 - Prob. 6.3PCh. 6 - Prob. 6.4PCh. 6 - Prob. 6.5PCh. 6 - Prob. 6.6PCh. 6 - Prob. 6.9PCh. 6 - Prob. 6.10PCh. 6 - Prob. 6.11PCh. 6 - Prob. 6.12PCh. 6 - Prob. 6.13PCh. 6 - Prob. 6.15PCh. 6 - Prob. 6.16PCh. 6 - Prob. 6.17PCh. 6 - Prob. 6.19PCh. 6 - Prob. 6.20PCh. 6 - Prob. 6.21PCh. 6 - Prob. 6.22PCh. 6 - Prob. 6.24PCh. 6 - Prob. 6.25PCh. 6 - Prob. 6.26PCh. 6 - Prob. 6.27PCh. 6 - Prob. 6.28PCh. 6 - Prob. 6.29PCh. 6 - Prob. 6.30P
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