   Chapter 17, Problem 76E

Chapter
Section
Textbook Problem

# One reaction that occurs in human metabolism is For this reaction ∆G°= 14 kJ at 25°c.a. Calculate K for this reaction at 25°C.b. In a living cell this reaction is coupled with the hydrolysis of ATP. (See Exercise 75.) Calculate ∆G° and K at 25°C for the following reaction: Glutamic acid ( a q ) +   ATP ( a q ) + NH 3 ( a q ) ⇌ Glutamine ( a q ) +   ADP ( a q ) + H 2 PO 4 − ( a q )

(a)

Interpretation Introduction

Interpretation: One of the reactions of human metabolism and its ΔG° value is given. The value of K is to be calculated at the given temperature. The coupling of the stated reaction with the reaction corresponding to the hydrolysis of ATP is given. The value of ΔG° and K for this reaction at given temperature is to be calculated.

Concept introduction: Equilibrium constant K , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, then the free energy change will be,

ΔG=0Q=K

ΔG°=RTln(K)

Explanation

Explanation

Given

The value of ΔG° for the reaction of human metabolism is 14kJ .

Given temperature is 25°C .

The conversion of degree Celsius (°C) into Kelvin (K) is done as,

T(K)=T(°C)+273

Hence,

The conversion of 25°C into Kelvin is,

T(K)=T(°C)+273T(K)=(25+273)K=298K

Formula

ΔG°=RTln(K)<

(b)

Interpretation Introduction

Interpretation: One of the reactions of human metabolism and its ΔG° value is given. The value of K is to be calculated at the given temperature. The coupling of the stated reaction with the reaction corresponding to the hydrolysis of ATP is given. The value of ΔG° and K for this reaction at given temperature is to be calculated.

Concept introduction: Equilibrium constant K , is defined as the ratio of the concentration of products to that of the reactants at equilibrium. If a given reaction is at equilibrium, then the free energy change will be,

ΔG=0Q=K

ΔG°=RTln(K)

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