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ddd [48]
3 years ago
8

On the following chemical equation, label the acid, the base, the

Chemistry
1 answer:
beks73 [17]3 years ago
3 0

Answer:

HX (aq)    +     H₂O (l)  ⇄      H₃O⁺ (aq)    +     X⁻ (aq)  

Acid                  Base          Conj. acid          Conj. base  

Explanation:

The equation is:

HX (aq) + H₂O (l)  ⇄   H₃O⁺ (aq) + X⁻ (aq)

This is the typical equilibrium for a weak acid. It would complete if we notice the Ka.

HX (aq) + H₂O (l)  ⇄   H₃O⁺ (aq) + X⁻ (aq)    Ka

1 mol of hypothetic HX acid react to 1 mol of water in order to release a proton and make hydronium and generate the X⁻ anion.

HX will be the acid, in this case a weak one and water will be the base. Water is able to accept a proton to make itslef hydronium

Hydronium is the conjugate acid.

The X⁻ will be the conjugate strong base.

This ion can generate the acid form again, that's why it is strong, because it can make hydrolisis.

X⁻ (aq)  +  H₂O (l)  ⇄  HX (aq) +  OH⁻(aq)     Kb

In this case, the anion will be the conjugate base which it takes a proton from water (acid form) to make a conjugate acid, the HX and a conjugate base, OH⁻

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The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol. Part A If an enzyme increases the
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Answer:

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

Explanation:

From the given information:

The activation barrier for the hydrolysis of sucrose into glucose and fructose is 108 kJ/mol.

In this  same concentration for the glucose and fructose; the reaction rate can be calculated by the rate factor which can be illustrated from the Arrhenius equation;

Rate factor in the absence of catalyst:

k_1= A*e^{^{^{ \dfrac {- Ea_1}{RT}}

Rate factor in the presence of catalyst:

k_2= A*e^{^{^{ \dfrac {- Ea_2}{RT}}

Assuming the catalyzed reaction and the uncatalyzed reaction are  taking place at the same temperature :

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the ratio of the rate factors can be expressed as:

\dfrac{k_2}{k_1}={  \dfrac {e^{ \dfrac {- Ea_2}{RT} }} { e^{ \dfrac {- Ea_1}{RT} }}

\dfrac{k_2}{k_1}={  \dfrac {e^{[  Ea_1 - Ea_2 ] }}{RT} }}

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Ea_1-Ea_2 = RT In \dfrac{k_2}{k_1}

Let say the assumed temperature = 25° C

= (25+ 273)K

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Ea_1-Ea_2 = 8.314 \  J/mol/K * 298 \ K *  In (10^6)

Ea_1-Ea_2 = 34228.92 \ J/mol

\mathbf{Ea_1-Ea_2 = 34.23 \ kJ/mol}

The barrier has to be 34.23 kJ/mol lower when the sucrose is in the active site of the enzyme

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The bond angle between the atoms of phosphine is 93°. It has one lone pair. The central atom is covered with 4 atoms.

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