Calculations

The fHo of Ethanol (CH3CH2OH) (l)), Water (H2O (l)) and carbon Dioxide (CO2(g)) at 298K are respectively; -277.1 kJ mol-1, -285.8 kJ mol-1 and –393.5 kJ mol-1. Given that the latent heat of fusion of water is 6.02 kJ mol-1 and the average heat capacity of water is 75.3 J K-1 mol-1 how much ice (in grams) would be needed to put in a measure of whisky (25ml, 40% ethanol by volume) such that all the enthalpy of combustion of the ethanol is taken up raising the temperature of the ice to body temperature (35C). Density of ethanol = 0.789 g cm-3 (16 marks)

Answers
Your first stage in answering this question is calculating the enthalpy of combustion for ethanol using the enthalpy of formation data.
science2chemistrytutor
05 April 2012
I need help with the same type of question. What do you do after this?
emmalouram
12 April 2012
For the question above: Next you need to calculate enthalpy of combustion for the amount of ethanol in a measure of whisky.
science2chemistrytutor
13 April 2012
Ok, I've done this. Im confused at what to do next
emmalouram
13 April 2012
Once you have calculated the enthalpy of combustion of the ethanol in a measure of whisky. The next steps involve calculating the mass of ice you would need so when the ethanol is burnt in complete combustion the ice melts and the temperature of water raises to 35C
science2chemistrytutor
19 April 2012
If the value of Enthalpy of combustion for 10 ml ethanol has equal to the total energy required to melt the ice and heat the ice to 35C. You can rewrite this as an equation like so....................... Enthalpy of combustion for 10 ml ethanol. = (Enthalpy of fusion (melting) of x mol of water) + (heat energy to raise x mol of water by 35C). The only problem being you don't know x number of moles need.
science2chemistrytutor
19 April 2012
To calculate the x mol of ice needed you can work out the energy require to melt and heat 1 mole of ice to 35C (= (latent heat of fusion 1 mol of water) + (heat energy required to heat 1 mol water to 35C). To calculate the actual number of mole of ice require take enthalpy of combustion of ethanol in a measure of whisky / energy require to melt and heat 1 mole of ice to 35C. Once you have the number of mole you can then calculate the mass.
science2chemistrytutor
19 April 2012
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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?