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 $\text{pH}$ 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,

$aA+bB\to cC+dD$

$\text{Rate}\text{=}\text{-}\frac{\text{1}}{\text{a}}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=}\text{-}\frac{\text{1}}{\text{b}}\frac{\text{Δ}\left[\text{B}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{c}}\frac{\text{Δ}\left[\text{C}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{d}}\frac{\text{Δ}\left[\text{D}\right]}{\text{Δt}}$

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,

    $\text{aA}\to \text{products}$

Where a= stoichiometric co-efficient of reactant A.

Consider the reaction has first-order rate law,

    $\text{Rate=-}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=k}\left[\text{A}\right]$

The integrated rate law equation can be given as,

    $\text{ln}\frac{{\left[\text{A}\right]}_{\text{t}}}{{\left[\text{A}\right]}_{\text{o}}}\text{=-kt}$

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

Explanation of Solution

Given reaction is,

$Enzyme(E)+Glu\cos e(G)\underset{}{\to }Enzymehexokinase(E+G)$

Record the given data’s

$\begin{array}{ccccc}{\text{[C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}{\text{]/(mmol dm}}^{\text{-3}}\text{)}& 1.00& 1.54& 3.12& 4.02\\ {\text{v}}_{\text{0}}{\text{/(mol dm}}^{\text{-3}}{\text{s}}^{\text{-1}}\text{)}& 5.00& 7.6& 15.5& 20.0\end{array}$

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

$\begin{array}{l}Rate\left(r\right)=-k\times G/n\times E\\ r_1/r_2=\left(G_1/G_2\right)\times n\\ n=\log \left(r_1/r_2\right)/\log \left(G_1/G_2\right)\end{array}$

So, the rate at,

$\begin{array}{c}{G}_1=1.00and{G}_2=1.54,r_1=5and{r}_2=7.6\\ n_1=\log \left(5/7.6\right)/\log \left(1/1.54\right)\\ =\log \left(0.65\right)/\log \left(0.64\right)\\ =\left(-0.187\right)/\left(-0.193\right)\\ n_1=0.96\end{array}$

Similarly we calculate other data’s

$\begin{array}{l}Rateat{G}_1=1.54and{G}_2=3.12,r_1=7.6and{r}_2=15.5\\ n_2=1.01\\ Rateat{G}_1=3.12and{G}_2=4.02,r_1=15.5and{r}_2=20.0\\ n_3=1.03\end{array}$

Over all order is $n=n_1,n_2,n_3=1$

Let us we can write the rate let

$\begin{array}{l}Rate=k\times {\left[\text{glucose}\right]}^x\\ LetustakeRate2/Rate1\\ \frac{7.6M/s}{5.0M/s}=\frac{{\left[1.54mM\right]}^x}{{\left[1.0mM\right]}^x}\\ 1.52=\frac{{\left[1.54\right]}^x}{{\left[1.0\right]}^x}\\ 1.52={\left[1.54\right]}^x\\ x\cong 1\end{array}$

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 $\text{pH}$ 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,

$aA+bB\to cC+dD$

$\text{Rate}\text{=}\text{-}\frac{\text{1}}{\text{a}}\frac{\text{Δ}\left[\text{A}\right]}{\text{Δt}}\text{=}\text{-}\frac{\text{1}}{\text{b}}\frac{\text{Δ}\left[\text{B}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{c}}\frac{\text{Δ}\left[\text{C}\right]}{\text{Δt}}\text{=}\frac{\text{1}}{\text{d}}\frac{\text{Δ}\left[\text{D}\right]}{\text{Δt}}$

The negative sign indicates the reduction of concentration of reactant.

Explanation of Solution

Given reaction is,

$Enzyme(E)+Glu\cos e(G)\underset{}{\to }Enzymehexokinase(E+G)$

Record the given data’s

$\begin{array}{ccccc}{\text{[C}}_{\text{6}}{\text{H}}_{\text{12}}{\text{O}}_{\text{6}}{\text{]/(mmol dm}}^{\text{-3}}\text{)}& 1.00& 1.54& 3.12& 4.02\\ {\text{v}}_{\text{0}}{\text{/(mol dm}}^{\text{-3}}{\text{s}}^{\text{-1}}\text{)}& 5.00& 7.6& 15.5& 20.0\end{array}$

Hence rate law can be written as

$\begin{array}{l}Rate=k\times \left[Glu\cos e\right]\\ Taketherate\text{constant}\text{values}\text{from}expriment1\end{array}$

$\begin{array}{c}Rate=5.0M/s\\ \left[\text{Glucose}\right]=1.0mM\\ =1.0\times {10}^{-3}M\\ k=\frac{5.0M/s}{1.0\times {10}^{-3}M}\\ k=5.0\times {10}^3{s}^{-1}\end{array}$

Therefore, the rate constant is $\underset{\_}{k=5.0\times {10}^3{s}^{-1}}$