THERMODYNAMICS (General questions)

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Q: 1 2 3 4 5 6 7 8 9 10 11 12

1

Calculate ΔH° (kilojoule mole-1) for the reaction:

2H2S(g) + SO2(g) → 2H2O(l) + 3S(s)

2

Carbon monoxide can be burned in oxygen to form carbon dioxide.

(1.) Calculate the standard enthalpy change for the oxidation of 1.00 mole of carbon monoxide.

(2.) Is the reaction exothermic or endothermic?

(3.) Calculate the heat released when 2.00 litre of carbon monoxide is combusted at 298 K and 1.00 atmosphere pressure.

3

Calculate the molar enthalpy of formation of methane under standard conditions using the following data:

C(s) + O2(g) → CO2(g) (ΔH° = -393 kJ mol-1 )

H2(g) + O2(g) → H2O(l) ( ΔH° = -285 kJ mol-1 )

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) (ΔH° = -890 kJ mol-1 )

4

CalculateΔH° for the reaction:

4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l)

5

Ethene undergoes an addition reaction with hydrogen forming ethane according to the following equation:

C2H4(g) + H2(g) → C2H6(g)

30.0 litre of ethene and 40.0 litre of hydrogen each at 298 K and 1.50 atmosphere pressure are mixed in the presence of finely divided platinum catalyst. After the formation of ethane has ceased the physical conditions are adjusted to those originally prevailing. Calculate the enthalpy change involved.

6

Calculate the standard enthalpy change for each of these reactions:

(1.) CaO(s) + H2O(l) → Ca(OH)2(s)

(2.) SO2(g) + O2(g) → SO3(g)

(3.) 2NO(g) + O2(g) → 2NO2(g)

(4.) CH4(g) + 2O2(g) → CO2(g) + 2H2O(g)

(5.) C(s) + H2O(l) → CO(g) + H2(g)

(6.) 2Na(s) + 2H2O(l) → 2NaOH(s) + H2(g)

(7.) 2H2S(g) + SO2(g) → 3S(s) + 2H2O(g)

(8.) The decomposition of solid KClO3 into solid KCl and gaseous oxygen.

7

 Water-gas, a mixture of H2 and CO, is an important industrial fuel produced by the reaction of steam with red hot coke (regarded as pure carbon):

C(s) + H2O(g) → CO(g) + H2(g)

Assuming that coke has the same enthalpy of formation as graphite, calculate ΔH°298 for this reaction.

8

By referring to the Chemical Data Book determine the enthalpy change in each of the following processes. Write suitable equations to represent each process and state whether the process is exothermic or endothermic.

(1.) The vaporisation of one mole of lead at its normal boiling point (latent heat of vaporisation).

(2.) The conversion of one mole of potassium vapour to potassium ions at 298 K.

(3.) The condensation of one mole of carbon tetrachloride vapour to liquid carbon tetrachloride at 298 K.

(4.) The dissociation of gaseous bromine molecules into one mole of gaseous bromine atoms.

(5.) The conversion of one mole of gaseous fluorine atoms to gaseous fluoride ions.

(6.) The combustion of one mole of liquid hexane to give liquid water and gaseous carbon dioxide at 298 K and 1.00 atmosphere pressure.

9

Calculate the enthalpy change involved in burning 1.00 mole of ethane, C2H6, in excess oxygen when reactants and products are in their standard states.

10 The bond enthalpy is the enthalpy change for the breaking of one mole of the bonds. Bond enthalpy values can be used to approximate the enthalpy of formation of substances, and to find approximate values for enthalpies of reaction. Estimate the standard enthalpy of formation for one mole of hydrazine, H2NNH2, from the average bond enthalpies.
11

Calculate the average bond enthalpy for the C-H bond in methane, (CH4(g)), using the following data:

Equation 1: C(s) + 2H2(g) → CH4(g) (ΔH = -75 kJ mol-1)

Equation 2: H2(g) → H(g) (ΔH = +218 kJ mol-1)

Equation 3: C(s) → C(g) (ΔH = +717 kJ mol-1)

12

Using the equations below, calculate the energy of the hydrogen/chlorine bond in hydrogen chloride:

(1.) As energy per mole of bonds

(2.) As energy per individual HCl bond

Equation 1: H2(g) + Cl2(g) → HCl(g) (ΔH° = -92 kJ mol-1)

Equation 2: H2(g) → H(g) (ΔH° = +218 kJ mol-1)

Equation 3: Cl2(g) → Cl(g) (ΔH° = +121 kJ mol-1)

 
THERMODYNAMICS (General answers)  

1

Enthalpy is a measure of the heat change during a reaction at constant pressure, and is the difference between the final and initial enthalpies of the molecules in the reaction: ΔH = Hfinal - Hinitial. The enthalpy change accompanying a reaction can be measured experimentally using a calorimeter (essentially just an insulated, open container) and a thermometer.

Standard enthapies are enthalpies measured at standard conditions, viz at a pressure of one atmosphere (strictly at 100kPa) and, if solutions are involved, at concentrations of each solute = one molar, and are indicated by a superscript on the H thus, H° .

ΔH° for a reaction can also be calculated from tables of standard enthalpies of formation data, ΔHf°, of the individual reactants and products. The standard enthalpy of formation of any compound is the enthalpy change for the formation of one mole of that compound in its standard state from its component elements in their standard states - i.e. the most stable form of each element at standard conditions. Because absolute enthalpy values cannot be measured, the standard enthalpy of formation of any element is defined to be zero. As an example, the standard enthalpy of formation of carbon dioxide gas would be the enthalpy change measured for the reaction of one mole of carbon (as graphite, its most stable form) with one mole of oxygen gas to produce one mole of carbon dioxide gas, each partial pressure of the gaseous components being maintained at one atmosphere.

[The temperature at which the standard enthalpy data applies must also be specified and is usually at 298 K, but this is not part of the standard state definition.]

Enthalpy is expressed in units of kJ mol-1. The mol-1 component which is part of this unit does not mean that the value necessarily applies to one mole of any particular substance but merely that the units used for quantities are moles as distinct from say, atoms or molecules. [This will become more apparent in the following question where the enthalpy calculated is for a reaction equation in which only one of the components (the SO2(g)) has a stoichiometric coefficient = 1]. The enthalpy change applies to the equation as written with the number of moles of each of the reactants and products being their stoichiometric coefficients. Thus in this question we intend to calculateΔH° for the reaction of 2 mole of H2S(g) with 1 mole of SO2(g) to form 2 moles of H2O(l) and 3 moles of S(s).

For the reaction 2H2S(g) + SO2(g) → 2H2O(l) + 3S(s):

Using the tabulated data,

sum of enthalpies of formation of reactants, ΣΔHf° (reactants)

= 2 x ΔHf°[H2S(g)] + ΔHf°[SO2(g)]

= (2 x -21 kJ mol-1) + (-297 kJ mol-1) = -339 kJ mol-1.

Sum of enthalpies of formation of products, ΣΔHf° (products)

= 2 x ΔHf°[H2O(l)] + 3 x ΔHf°[S(s)]

= (2 x -285 kJ mol-1) + (3 x 0 kJ mol-1) = -570 kJ mol-1.

∴ ΔH° = ΣΔHf° (products) - ΣΔHf° (reactants)

= (-570 kJ mol-1) - (-339 kJ mol-1) = -231 kJ mol-1.

This is a negative value, which means that the reaction is exothermic (ie the products have a lower enthalpy than the reactants, with the balance released as thermal energy during reaction).

 

2

(1.) Balanced equation:

CO(g) + O2(g) → CO2(g)

Sum of enthalpies of formation of reactants,

ΣΔHf° (reactants) = ΔHf°[CO(g)] + 0.5 x ΔHf°[O2(g)]

= (-111 kJ mol-1) + (0.5 x 0 kJ mol-1) = -111 kJ mol-1.

Sum of enthalpies of formation of products,

ΣΔHf° (products) = ΔHf°[CO2(g)] = (-393 kJ mol-1)

= -393 kJ mol-1.

∴ ΔH° = ΣΔHf° (products) - ΣΔHf° (reactants)

= (-393 kJ mol-1) - (-111 kJ mol-1) = -282 kJ mol-1.

(2.) This is a negative value, therefore the reaction is exothermic.

(3.) No. mol CO (at 298 K, 1 atm) = volume CO / molar volume CO

= 2.00 L / 24.5 L mol-1 = 0.0816 mol.

Enthalpy change for the combustion of 0.0816 mol CO

= no. mol CO x enthalpy change for 1.00 mol CO

= 0.0816 mol x -282 kJ mol-1 = -23.0 kJ.

Therfore, heat released = 23.0 kJ.

 

3

This question differs from previous ones by giving the change in enthalpy of several reactions, and asking to solve for the molar enthalpy of formation of one of the reactants.

Note also, that reference to data book values for ΔHf° of CO2(g) and H2O(l) is unnecessary, as the first two equations describe the formation of CO2(g) and H2O(l) from their respective elements. Therefore:

ΔHf°[CO2(g)] = -393 kJ mol-1

ΔHf°[H2O(l)] = -285 kJ mol-1

Since O2(g) is an element under standard conditions, its ΔHf° will be 0 kJ mol-1. We now know the ΔHf° values for all reactants and products (except methane), and ΔH°rxn, given as -890 kJ mol-1 for the reaction:

CH4(g) + 2O2(g) → CO2(g) + 2H2O(l) (ΔH° = -890 kJ mol-1)

Sum of enthalpies of formation of reactants,

SDHf° (reactants) = ΔHf°[CH4(g)] + 2 x ΔHf°[O2(g)]

= ΔHf°[CH4(g)] + 2 x 0 kJ mol-1 = ΔHf°[CH4(g)]

Sum of enthalpies of formation of products,

ΣΔHf° (products) = ΔHf°[CO2(g)] + 2 x ΔHf°[H2O(l)]

= (-393 kJ mol-1) + (2 x -285 kJ mol-1)= -963 kJ mol-1.

∴ ΔH° = ΣΔHf° (products) - ΣΔHf° (reactants)

∴ -890 kJ mol-1 = -963 kJ mol-1 - ΔHf°[CH4(g)]

Solving for ΔHf°[CH4(g)]:

ΔHf°[CH4(g)] = -73 kJ mol-1

4

For the reaction 4NH3(g) + 5O2(g) → 4NO(g) + 6H2O(l):

Sum of enthalpies of formation of reactants,

ΣΔHf° (reactants) = 4 x ΔHf°[NH3(g)] + 5 x ΔHf°[O2(g)]

= (4 x -46 kJ mol-1) + (5 x 0 kJ mol-1) = -184 kJ mol-1.

Sum of enthalpies of formation of products,

ΣΔHf° (products) = 4 x ΔHf°[NO(g)] + 6 x ΔHf°[H2O(l)]

= (4 x +90 kJ mol-1) + (6 x -285 kJ mol-1) = -1350 kJ mol-1.

∴ ΔH° = ΣΔHf° (products) - ΣΔHf° (reactants)

= (-184 kJ mol-1) - (-1350 kJ mol-1) = -1166 kJ mol-1.

5

STEP 1: calculate the enthalpy change for 1 mole of ethene reacting.

From the balanced equation and tabulated ΔHf° values, ΔH° of the reaction can be calculated for the reaction C2H4(g) + H2(g) → C2H6(g):

Sum of enthalpies of formation of reactants,

ΣΔHf° (reactants) = ΔHf°[C2H4(g)] + ΔHf°[H2(g)]

= (+52 kJ mol-1) + (0 kJ mol-1) = +52 kJ mol-1.

Sum of enthalpies of formation of products,

ΣΔHf° (products) = ΔHf°[C2H6(g)] = -85 kJ mol-1.

∴ ΔH° = ΣΔHf° (products) - ΣΔHf° (reactants)

= (-85 kJ mol-1) - (+52 kJ mol-1) = -137 kJ mol-1.

STEP 2 - calculate the moles of ethene and hydrogen available

moles C2H4 (298K, 1.5 atm) = 1.5 x (volume C2H4 / molar volume)

= 1.5 x (30.0 L / 24.5 L mol-1) = 1.84 mol.

(The molar volume of 24.5L per mole is correct for gas at 1 atm, 298K. The factor of 1.5 introduced is to correct for the 1.5 atm pressure of the gases in this case.)

moles H2 (at 298 K, 1.5 atm) = 1.5 x (volume H2 / molar volume)

= 1.5 x (40.0 L / 24.5 L mol-1) = 2.45 mol.

From the equation, one mole of H2 reacts with one mole of C2H4, so C2H4 will be the limiting reagent in this reaction. [Alternatively, as gases combine in simple whole-number ratios by volumes provided they are at the same conditions, then the 30.0 L of ethene would react exactly with 30.0 L of hydrogen (as their combining ratio from the equation is 1:1), so the hydrogen is in excess.]

ThereforeΔH° for the reaction of 1.84 mol C2H4 = 1.84 x -137 kJ

= -252 kJ mol-1.

6

Problems (2.) - (8.) are all solved using the same general method outlined in the worked solution for part (1.).

(1.) For the reaction CaO(s) + H2O(l) → Ca(OH)2(s)

Sum of enthalpies of formation of reactants,

SDHf° (reactants) = ΔHf°[CaO(s)] + ΔHf°[H2O(l)]

= (-635 kJ mol-1) + (-285 kJ mol-1) = -920 kJ mol-1.

Sum of enthalpies of formation of products,

ΣΔHf° (products) = ΔHf°[Ca(OH)2] = -986 kJ mol-1.

∴ ΔH° = ΣΔHf° (products) - ΣΔHf° (reactants)

= (-986 kJ mol-1) - (-920 kJ mol-1) = -66 kJ mol-1.

(2.) -99 kJ

(3.) -114 kJ

(4.) -802 kJ

(5.) +174 kJ

(6.) -284 kJ

(7.) -145 kJ

(8.) -45 kJ mol-1 (balanced equation: KClO3(s) → KCl(s) + 1.5O2(g))

7

For the reaction C(s) + H2O(g) → CO(g) + H2(g):

Sum of enthalpies of formation of reactants,

ΣΔHf° (reactants) = ΔHf°[C(s)] + ΔHf°[H2O(g)]

= (0 kJ mol-1) + (-242 kJ mol-1) = -242 kJ mol-1.

Sum of enthalpies of formation of products,

ΣΔHf° (products) = ΔHf°[CO(g)] + ΔHf°[H2(g)]

= (-111 kJ mol-1) + (0 kJ mol-1) = -111 kJ mol-1.

∴ ΔH° = ΣΔHf° (products) - ΣΔHf° (reactants)

= (-111 kJ mol-1) - (-242 kJ mol-1) = +131 kJ mol-1.

8

(1.) The enthalpy change of this process is described by the enthalpy of vaporisation, ΔHv, of lead, which specifically describes the change from liquid (at a substance's boiling point temperature) to vapour (at this same temperature). From the data sheet, ΔHv[Pb] = +190 kJ mol-1, the positive value indicating that the process is endothermic.

(2.) K(g) → K+(g) + e-. This is the process of ionisation, and the change in enthalpy is found from the first ionisation energy for potassium (the first ionisation is the the removal of the first electron from a neutral atom). From the data sheet, this value is +425 kJ mol-1, an endothermic process.

(3.) The process of condensation is the reverse of vaporisation, ie ΔH°(condensation) = -ΔH°(vaporisation). From the data sheet, ΔH°v[CCl4] = +33 kJ mol-1. Therefore the enthalpy of condensation = -33 kJ mol-1, an exothermic process.

(4.) Br2(g) → Br(g). The change in enthalpy associated with converting molecules to atoms is the heat of atomisation. Note the question specifies one mole of bromine atoms formed. From the data sheet the heat of atomisation for the formation of two mol Br atoms is +193 kJ mol-1, ∴ heat of atomisation of one mol Br atoms is (193 / 2) kJ mol-1 = +96.5 kJ mol-1, an endothermic process.

(5.) F(g) + e- → F-(g). The enthalpy change associated with this process is given by the first electron affinity of F(g), from the data sheet this value is -334 kJ mol-1, so the reaction is exothermic.

(6.) C6H14(l) + 9.5O2(g) → 7H2O(l) + 6CO2(g) is a combustion reaction, and the enthalpy change is given by the heat of combustion which is often listed in data books for organic compounds. For C6H14, ΔHc° = -4163 kJ mol-1, an exothermic reaction.

9

Balanced equation:

C2H6(g) + 3.5O2(g) → 2CO2(g) + 3H2O(l)

Sum of enthalpies of formation of reactants,

ΣΔHf° (reactants) = ΔHf°[C2H6(g)] + 3.5 x ΔHf°[O2(g)]

= (-85 kJ mol-1) + (3.5 x 0 kJ mol-1) = -85 kJ mol-1.

Sum of enthalpies of formation of products,

ΣΔHf° (products) = 2 x ΔHf°[CO2(g)] + 3 x ΔHf°[H2O(l)]

= (2 x -393 kJ mol-1) + (3 x -285 kJ mol-1) = -1641 kJ mol-1.

∴ ΔH° = ΣΔHf° (products) - ΣΔHf° (reactants)

= (-1641 kJ mol-1) - (-85 kJ mol-1) = -1556 kJ mol-1.

10

The enthalpy of formation can be approximated by using average bond enthalpies:

ΔHf° = ΔH°element bonds broken + ΔH°product bonds formed

Nitrogen and hydrogen, in their standard states, contain N≡N and H-H bonds, respectively. The first step is to consider the enthalpy change involved in breaking these bonds. From the data sheet, the average bond enthalpy of the N≡N bond is 945 kJ mol-1, and the H-H bond is 436 kJ mol-1:

1 x N≡N → 2N (ΔH° = +945 kJ mol-1)

2 x H-H → 4H (ΔH° = 2 x +436 kJ mol-1 = +872 kJ mol-1)

ΣΔH°element bonds broken = 945 kJ mol-1 + 872 kJ mol-1

= +1817 kJ mol-1

Note that we now have the 2 N atoms and 4 H atoms required for the atoms present in the hydrazine. Note also that the enthalpy changes for these bond breaking steps are taken as being positive since breaking bonds is an endothermic process.

The second step is to consider the bonds present in the hydrazine molecule, and the enthalpy change associated with their creation. The hydrazine molecule contains 1 N-N bond and 4 N-H bonds. From the data sheet the average bond enthalpy of the N-N bond is 163 kJ mol-1, and the N-H bond is 391 kJ mol-1. The enthalpies of the bonds being formed are:

1 x N-N (ΔH° = -163 kJ mol-1)

4 x N-H (ΔH° = 4 x -391 kJ mol-1 = -1564 kJ mol-1)

ΣΔH°product bonds formed = -163 kJ mol-1 + (-1564 kJ mol-1)

= -1727 kJ mol-1

Using Hess' law, ΔH for on overall process equals the sum of ΔH values for individual steps.

∴ ΔHf° = ΔH°element bonds broken + ΔH°product bonds formed

= 1817 kJ mol-1 + (-1727 kJ mol-1) = +90 kJ mol-1.

Note that since the bond enthalpies in data sheets are average values (determined over a wide range of compounds), the results produced for a particular compound will only be an approximation of the actual bond strengths in the compound.

Note also that bond enthalpy values apply only when all atoms and molecules are in the gas phase - when using bond enthalpy values to obtain the enthalpy change for a reaction involving liquids or solids, the relevant enthalpies for the phase changes must be included.

11

We require ΔH° for the reaction C(g) + 4H(g) → CH4(g). As all components are in the gas phase, the ΔH° for this reaction would be the bond enthalpy for 4 C-H bonds, so dividing it by 4 will give the C-H bond enthalpy.

Rearranging the 3 reaction equations given:

equation 1 + 4 x reverse of equation 2 + reverse of equation 3 provides

C(s) + 2H2(g) + 4H(g) + C(g) → CH4(g) + 2H2(g) + C(s)

which when common terms are cancelled, becomes

C(g) + 4H(g) → CH4(g).

The component ΔH° values are

(-75) + 4 x (-218) + (-717) ΔH° = kJ mol-1

= -1664 kJ mol-1

Note that reversing the equations 2 and 3 requires that the sign of the accompanying ΔH° be reversed but the magnitude remains unaltered.

ΔH°[C-H] = -1664 kJ mol-1 / 4 = -416 kJ mol-1.

The bond enthalpy value is normally tabulated for a mole of bonds breaking, so would have a + sign. For a mole of bonds forming, the value would be the same but a - sign would be attached.

Note that this value differs somewhat from the tabulated average C-H bond value of 413 kJ mol-1.

12

(1.) We require ΔH° for the reaction H(g) + Cl(g) → HCl(g).

Combine reaction equation 1 with the reverse of reaction equation 2 and the reverse of reaction equation3:

H2(g) + Cl2(g) + H(g) +Cl(g) → HCl(g) + H2(g) + Cl2(g)

which when common terms are cancelled, becomes

H(g) + Cl(g) → HCl(g).

The component ΔH° values are

(-92) + (-218) + (-121) kJ mol-1

= -431 kJ mol-1

and the C-H bond enthalpy = 431 kJ mol-1.

(2.) The energy of a single H-Cl bond = ΔH°[H-Cl] / NA

= 431 kJ mol-1 / 6.022 x 1023 mol-1 = 7.16 x 10-22 kJ.