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1

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

2H₂S(g) + SO₂(g) → 2H₂O(l) + 3S(s)

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

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) + ½O₂(g) → H₂O(l) ( ΔH° = -285 kJ mol^-1 )

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

CalculateΔH° for the reaction:

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.

 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.

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.

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.

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: ½H₂(g) → H(g) (ΔH = +218 kJ mol^-1)

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

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

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 = H_final - H_initial. 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, ΔH_f°, 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, ΣΔH_f° (reactants)

= 2 x ΔH_f°[H₂S(g)] + ΔH_f°[SO₂(g)]

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

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

= 2 x ΔH_f°[H₂O(l)] + 3 x ΔH_f°[S(s)]

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

∴ ΔH° = ΣΔH_f° (products) - ΣΔH_f° (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).

CO(g) + ½O₂(g) → CO₂(g)

Sum of enthalpies of formation of reactants,

ΣΔH_f° (reactants) = ΔH_f°[CO(g)] + 0.5 x ΔH_f°[O₂(g)]

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

Sum of enthalpies of formation of products,

ΣΔH_f° (products) = ΔH_f°[CO₂(g)] = (-393 kJ mol^-1)

= -393 kJ mol^-1.

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

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

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 ΔH_f° 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:

ΔH_f°[CO₂(g)] = -393 kJ mol^-1

ΔH_f°[H₂O(l)] = -285 kJ mol^-1

Since O₂(g) is an element under standard conditions, its ΔH_f° will be 0 kJ mol^-1. We now know the ΔH_f° 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,

SDH_f° (reactants) = ΔH_f°[CH₄(g)] + 2 x ΔH_f°[O₂(g)]

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

Sum of enthalpies of formation of products,

ΣΔH_f° (products) = ΔH_f°[CO₂(g)] + 2 x ΔH_f°[H₂O(l)]

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

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

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

Solving for ΔH_f°[CH₄(g)]:

ΔH_f°[CH₄(g)] = -73 kJ mol^-1

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

Sum of enthalpies of formation of reactants,

ΣΔH_f° (reactants) = 4 x ΔH_f°[NH₃(g)] + 5 x ΔH_f°[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,

ΣΔH_f° (products) = 4 x ΔH_f°[NO(g)] + 6 x ΔH_f°[H₂O(l)]

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

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

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

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

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

Sum of enthalpies of formation of reactants,

ΣΔH_f° (reactants) = ΔH_f°[C₂H₄(g)] + ΔH_f°[H₂(g)]

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

Sum of enthalpies of formation of products,

ΣΔH_f° (products) = ΔH_f°[C₂H₆(g)] = -85 kJ mol^-1.

∴ ΔH° = ΣΔH_f° (products) - ΣΔH_f° (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)

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

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.

Sum of enthalpies of formation of reactants,

SDH_f° (reactants) = ΔH_f°[CaO(s)] + ΔH_f°[H₂O(l)]

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

Sum of enthalpies of formation of products,

ΣΔH_f° (products) = ΔH_f°[Ca(OH)₂] = -986 kJ mol^-1.

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

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

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

Sum of enthalpies of formation of reactants,

ΣΔH_f° (reactants) = ΔH_f°[C(s)] + ΔH_f°[H₂O(g)]

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

Sum of enthalpies of formation of products,

ΣΔH_f° (products) = ΔH_f°[CO(g)] + ΔH_f°[H₂(g)]

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

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

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

Balanced equation:

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

Sum of enthalpies of formation of reactants,

ΣΔH_f° (reactants) = ΔH_f°[C₂H₆(g)] + 3.5 x ΔH_f°[O₂(g)]

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

Sum of enthalpies of formation of products,

ΣΔH_f° (products) = 2 x ΔH_f°[CO₂(g)] + 3 x ΔH_f°[H₂O(l)]

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

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

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

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

ΔH_f° = 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.

∴ ΔH_f° = 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.

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