Chemistry

What happens to stearic acid and paraffin wax at a molecular level and the velocity of molecules when it solidifies and melts?

Answers
Hey Johnny,When a substance is melted, its molecules gain energy (due to the increase in heat) and due to this energy the molecules move faster (increased velocity). This is why solid paraffin wax becomes liquid - the molecules have more kinetic energy and are thus moving around more. When cooled, the molecules lose energy and move slower - the wax becomes solid. This is a very brief answer to your question but its an overview of the kinetic theory of particles. Please message me if you want a detailed explanation and we can talk it through!
Layla G.
14 September 2017
Both molecules have van der Waals forces of attraction acting between them. However, the stearic acid molecules (C17H35COOH) also have hydrogen bonds between them. These are sometimes referred to as simply "weak bonds", but are better called "intermolecular forces of attractions" - forces of attraction between small discrete molecules.The higher the temperature - the higher the KE of these molecules - and the faster they move around. It makes sense that as the molecules move slower (at lower temperatures) these forces of attraction are more successful at forming actual bonds between them. The attractions don't change - but the bonds do. In the gaseous state there are no bonds. In the liquid state bonds constantly form and break (allowing the molecules to slide past one another). In the solid state the bonds between the molecules are permanent.
franklychemistry
01 October 2017
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Catalyst equation help

A catalyst consisting of palladium on an α-Al2O3 support, Pd/α-Al2O3, has been used for the oxidation of CO at room temperature:

Equation 1 CO(g) + ½O2(g) = CO2(g)

Under certain conditions, the oxidation reaction was found to involve competitive adsorption of the reactants with CO being non-dissociatively adsorbed and oxygen undergoing dual-site adsorption:

Equation 2 ka CO(g) + * ↔ CO(ad) Kd

Equation 3 ka’ O2(g) + 2* ↔ O(ad) + O(ad) kd’

The rate-limiting step is then a bimolecular reaction between CO(ad) and O(ad):

Equation 4 CO(ad) + O(ad) → CO (ad)

Carbon dioxide can be assume to be weakly adsorbed and, as a consequence, desorbs as quickly as it is formed.

(i) For the competitive adsorption of CO show that the following expression can be derived by equating the rates of adsorption and desorption of CO:

Equation 5 θCO = bCOpCO(1 – θCO – θO)

Where θCO and θO are the fractional surface coverages of CO and O, respectively. The quantity bCO (= ka/kd) is the adsorption coefficient for CO and pCO is the partial pressure of CO. (Hint In your working you should represent the total number of adsorption sites by N and provide expressions for both the rate of adsorption and the rate of desorption of CO.)

(ii) Equation 5 can be used in a more detailed analysis of the mechanism to derive the following two expressions:

Equation 6 θCO = (bCOpCO) / (1 + bCOpCO + (bO2pO2)1/2)

and

Equation 7 θO = θCO((bO2pO2)1/2 / bCOpCO)

where b02(= ka’ / kd’) and pO2 are, respectively, the adsorption coefficient and partial pressure of O2.

Given Equations 6 and 7, in conjunction with the information about the rate-limiting step at the start of this question, show that the theoretical rate equation takes the form:

Equation 8 r = (kθbCOpCO(bO2pO2)1/2) / {1 + bCOpCO + (bO2pO2)1/2}^2

(iii) The experimental rate equation for the CO oxidation reaction, under conditions for which the mechanism given in part (ii) is valid, takes the form:

Equation 9 r = (kR(pO2)1/2) / pCO

How can this result be rationalised in terms of the theoretical rate equation (Equation 8) that has been derived for the mechanism?

Catalyst equation help (redone)

A catalyst consisting of palladium on an α-Al2O3 support, Pd/α-Al2O3, has been used for the oxidation of CO at room temperature:

Equation 1 CO(g) + ½O2(g) = CO2(g)

Under certain conditions, the oxidation reaction was found to involve competitive adsorption of the reactants with CO being non-dissociatively adsorbed and oxygen undergoing dual-site adsorption:

Equation 2 ka CO(g) + * ↔ CO(ad) kd

Equation 3 ka’ O2(g) + 2* ↔ O(ad) + O(ad) kd’

The rate-limiting step is then a bimolecular reaction between CO(ad) and O(ad):

Equation 4 CO(ad) + O(ad) → CO (ad)

Carbon dioxide can be assume to be weakly adsorbed and, as a consequence, desorbs as quickly as it is formed.

(i) For the competitive adsorption of CO show that the following expression can be derived by equating the rates of adsorption and desorption of CO:

Equation 5 θCO = bCOpCO(1 – θCO – θO)

Where θCO and θO are the fractional surface coverages of CO and O, respectively. The quantity bCO (= ka/kd) is the adsorption coefficient for CO and pCO is the partial pressure of CO. (Hint In your working you should represent the total number of adsorption sites by N and provide expressions for both the rate of adsorption and the rate of desorption of CO.)

(ii) Equation 5 can be used in a more detailed analysis of the mechanism to derive the following two expressions:

Equation 6 θCO = (bCOpCO) / (1 + bCOpCO + (bO2pO2)1/2)

and

Equation 7 θO = θCO((bO2pO2)1/2 / bCOpCO)

where b02(= ka’ / kd’) and pO2 are, respectively, the adsorption coefficient and partial pressure of O2.

Given Equations 6 and 7, in conjunction with the information about the rate-limiting step at the start of this question, show that the theoretical rate equation takes the form:

Equation 8 r = (kθbCOpCO(bO2pO2)1/2) / {1 + bCOpCO + (bO2pO2)1/2}^2

(iii) The experimental rate equation for the CO oxidation reaction, under conditions for which the mechanism given in part (ii) is valid, takes the form:

Equation 9 r = (kR(pO2)1/2) / pCO

How can this result be rationalised in terms of the theoretical rate equation (Equation 8) that has been derived for the mechanism?