ATKINS' PHYSICAL CHEMISTRY-ACCESS
ATKINS' PHYSICAL CHEMISTRY-ACCESS
11th Edition
ISBN: 9780198834700
Author: ATKINS
Publisher: OXF
Question
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Chapter 12, Problem 12.4IA
Interpretation Introduction

Interpretation:

The plot of [I]0 against (δν)1 is a straight line with the slope [E]0Δν and y-intercept K1.

Concept introduction:

Many nuclei and electrons have spin, due to this spin magnetic moment arises.  The energy of this magnetic moment depends on the orientation of the applied magnetic field.  In NMR spectroscopy, every nucleus has a spin.  There is an angular momentum related to the spin.  The difference between its resonance frequency and that of the reference standard is known as the chemical shift of a nucleus.  Tetramethylsilane (TMS) is taken as reference.

Expert Solution & Answer
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Answer to Problem 12.4IA

The plot of [I]0 against (δν)1 is a straight line with the slope [E]0Δν and y-intercept K1 has been rightfully shown.

Explanation of Solution

If the plot of [I]0 against (δν)1 is a straight line with the slope [E]0Δν and y-intercept K1, then the relation given below will hold true.

    [I]0=[E]0ΔνδνK1

The dissociation reaction of the enzyme-inhibitor complex is shown below.

    EIE+I

The expression for the resonance frequency (ν) of the single peak in the NMR spectrum is shown below.

    ν=fIνI+fEIνEI        (1)

Where,

  • νI is the frequency of free inhibitor, I.
  • νEI is the frequency of bound inhibitor, EI.
  • fI is the frequency of free inhibitor, I.
  • fEI is the frequency of bound inhibitor, I.

The expression for fI is given below.

    fI=[I][I]+[EI]

The expression for fEI is given below.

    fI=[EI][I]+[EI]

Where,

  • [I] is the concentration of free inhibitor, I.
  • [EI] is the concentration of bound inhibitor, EI.

Substitute the value of fI and fEI in equation (1).

    ν=[I][I]+[EI]νI+[EI][I]+[EI]νEI        (2)

The expression for [I] is given below.

    [I]0=[I]+[EI][I]=[I]0[EI]

Where,

  • [I]0 is the initial concentration of the inhibitor, I.

Substitute the value of [I]+[EI] and [I] in equation (2).

    ν=[I]0[EI][I]0νI+[EI][I]0νEI        (3)

Equation (3) was solved further as shown below.

    ν=[I]0νI[EI]νI+[EI]νEI[I]0ν[I]0=[I]0νI+[EI](νEIνI)ν[I]0[I]0νIνEIνI=[EI][EI]=[I]0(ννI)νEIνI        (4)

The dissociation constant (K1) is given by the expression shown below.

    K1=[E][I][EI]        (5)

Where,

  • [E] is the concentration of the enzyme.

The expression for [E] is given below.

    [E]=[E]0[EI]

Substitute the value of [E] and [I] in equation (5).

    K1=([E]0[EI])([I]0[EI])[EI]        (6)

As the initial concentration of the inhibitor is greater than the concentration of the bound inhibitor, the equation (6) can be written as shown below.

    K1=([E]0[EI])[I]0[EI]K1[EI]=[E]0[I]0[EI][I]0[E]0[I]0=K1[EI]+[EI][I]0[E]0=[EI](K1+[I]0)[I]0        (7)

Substitute the value of [EI] from equation (4) in equation (7).

    [E]0=[I]0(ννI)νEIνI×(K1+[I]0)[I]0=(K1+[I]0)ννIνEIνI        (8)

The expression for ννI is given below.

    ννI=δν

The expression for νEIνI is given below.

    νEIνI=Δν

Substitute the value of ννI and νEIνI in equation (8).

    [E]0=(K1+[I]0)δνΔν[E]0Δν=K1δν+[I]0δν[I]0δν=[E]0ΔνK1δν[I]0=[E]0ΔνδνK1        (9)

The equation (9) is in the form of an equation for straight line.  Therefore, the plot of [I]0 against (δν)1 is a straight line with the slope [E]0Δν and y-intercept of K1.

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Chapter 12 Solutions

ATKINS' PHYSICAL CHEMISTRY-ACCESS

Ch. 12 - Prob. 12A.2BECh. 12 - Prob. 12A.3AECh. 12 - Prob. 12A.3BECh. 12 - Prob. 12A.4AECh. 12 - Prob. 12A.4BECh. 12 - Prob. 12A.5AECh. 12 - Prob. 12A.5BECh. 12 - Prob. 12A.6AECh. 12 - Prob. 12A.6BECh. 12 - Prob. 12A.7AECh. 12 - Prob. 12A.7BECh. 12 - Prob. 12A.8AECh. 12 - Prob. 12A.8BECh. 12 - Prob. 12A.9AECh. 12 - Prob. 12A.9BECh. 12 - Prob. 12A.1PCh. 12 - Prob. 12A.3PCh. 12 - Prob. 12B.1DQCh. 12 - Prob. 12B.2DQCh. 12 - Prob. 12B.3DQCh. 12 - Prob. 12B.4DQCh. 12 - Prob. 12B.5DQCh. 12 - Prob. 12B.1AECh. 12 - Prob. 12B.1BECh. 12 - Prob. 12B.2AECh. 12 - Prob. 12B.2BECh. 12 - Prob. 12B.3AECh. 12 - Prob. 12B.3BECh. 12 - Prob. 12B.4AECh. 12 - Prob. 12B.4BECh. 12 - Prob. 12B.5AECh. 12 - Prob. 12B.5BECh. 12 - Prob. 12B.6AECh. 12 - Prob. 12B.6BECh. 12 - Prob. 12B.7AECh. 12 - Prob. 12B.7BECh. 12 - Prob. 12B.8AECh. 12 - Prob. 12B.8BECh. 12 - Prob. 12B.9AECh. 12 - Prob. 12B.9BECh. 12 - Prob. 12B.10AECh. 12 - Prob. 12B.10BECh. 12 - Prob. 12B.11AECh. 12 - Prob. 12B.11BECh. 12 - Prob. 12B.12AECh. 12 - Prob. 12B.12BECh. 12 - Prob. 12B.13AECh. 12 - Prob. 12B.13BECh. 12 - Prob. 12B.14AECh. 12 - Prob. 12B.14BECh. 12 - Prob. 12B.1PCh. 12 - Prob. 12B.2PCh. 12 - Prob. 12B.3PCh. 12 - Prob. 12B.5PCh. 12 - Prob. 12B.6PCh. 12 - Prob. 12B.7PCh. 12 - Prob. 12B.8PCh. 12 - Prob. 12B.9PCh. 12 - Prob. 12C.1DQCh. 12 - Prob. 12C.2DQCh. 12 - Prob. 12C.3DQCh. 12 - Prob. 12C.4DQCh. 12 - Prob. 12C.5DQCh. 12 - Prob. 12C.1AECh. 12 - Prob. 12C.1BECh. 12 - Prob. 12C.2AECh. 12 - Prob. 12C.2BECh. 12 - Prob. 12C.3AECh. 12 - Prob. 12C.3BECh. 12 - Prob. 12C.4AECh. 12 - Prob. 12C.4BECh. 12 - Prob. 12C.5AECh. 12 - Prob. 12C.5BECh. 12 - Prob. 12C.4PCh. 12 - Prob. 12C.5PCh. 12 - Prob. 12C.6PCh. 12 - Prob. 12C.10PCh. 12 - Prob. 12D.1DQCh. 12 - Prob. 12D.2DQCh. 12 - Prob. 12D.1AECh. 12 - Prob. 12D.1BECh. 12 - Prob. 12D.2AECh. 12 - Prob. 12D.2BECh. 12 - Prob. 12D.3AECh. 12 - Prob. 12D.3BECh. 12 - Prob. 12D.4AECh. 12 - Prob. 12D.4BECh. 12 - Prob. 12D.5AECh. 12 - Prob. 12D.5BECh. 12 - Prob. 12D.6AECh. 12 - Prob. 12D.6BECh. 12 - Prob. 12D.1PCh. 12 - Prob. 12D.2PCh. 12 - Prob. 12D.4PCh. 12 - Prob. 12D.5PCh. 12 - Prob. 12D.6PCh. 12 - Prob. 12D.7PCh. 12 - Prob. 12D.8PCh. 12 - Prob. 12.3IACh. 12 - Prob. 12.4IA
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