Energy Cycles in Reactions (DP IB Chemistry)

The equation that would be used to calculate the enthalpy change for the conversion of graphite to diamond is ΔH_r = ΔH₁ - ΔH₂

False.

Hess's Law is used to calculate enthalpy changes which cannot be found experimentally using calorimetry.

The two common methods for solving Hess's Law problems are:

• Using Hess's law cycles

• Using equations

False.

When using cycles to solve Hess's Law problems, if you follow the direction of the arrow you add the quantity.

True.

In Hess's Law calculations, you always need to adjust for different molar amounts.

The general equation for solving Hess's Law problems using equations is:

ΔH(reaction) = Σ(ΔH products) - Σ(ΔH reactants)

The enthalpy change for the reaction is:

ΔH_r = - ( - 544 x 2) + (- 1648) = - 560 kJ

The standard enthalpy of formation is the enthalpy change when one mole of a compound is formed from its elements under standard conditions.

False.

In enthalpy of formation cycles, arrows always point upwards because the definition of enthalpy of formation must go from elements to compounds.

The equation used to calculate enthalpy changes from enthalpy of formation data is:

ΔH = ΣΔH_f^Ꝋ(products) - ΣΔH_f^Ꝋ(reactants)

The enthalpy of combustion is:

• ΔH = ΔH_{f products}  - ΔH_{f reactants}

• ΔH = - 1996 - (+ 31.4) = -2027.4 kJ

A Hess's law cycle that would allow the determination of overall enthalpy change for the following reaction using enthalpy of formation data is:

The enthalpy change for this reaction using the enthalpy of formation data given is:

ΔH_r^Ꝋ = +365 - 484 - 197 = -316 kJ mol^-1

The enthalpy of formation of elements in their standard states is always ‎zero.

False.

The enthalpy of formation can be a positive or negative value.

Standard enthalpy of combustion is the enthalpy change that occurs when one mole of a substance burns completely under standard conditions.

False.

In enthalpy of combustion cycles, arrows should be pointing downwards.

The general equation for calculating ΔH using enthalpy of combustion data is:

ΔH = ΣΔH_c^Ꝋ(reactants) - ΣΔH_c^Ꝋ(products)

True.

The enthalpy of combustion is always an exothermic process.

A Hess's law cycle that could be used to determine the overall enthalpy change using enthalpy of combustion data is:

The enthalpy change of formation of propanone using the Hess's law cycle is:

ΔH_f^Ꝋ = -1182 - 858 + 1821 = -219 kJ mol^-1

The The sign for enthalpy change of combustion values will always be negative.

True.

Enthalpy of combustion can be used to calculate enthalpy of formation using Hess's Law.

The enthalpy of formation of propane is:

• ΔH = ΣΔH_c^Ꝋ(reactants) - ΣΔH_c^Ꝋ(products)

• ΔH = (3x -393)+ (4 x -286) - (-2220)

• ΔH = -103 kJ mol^-1

Enthalpy of combustion is significant in practical applications as it helps determine the energy content of fuels and food.

The symbol to represent the enthalpy change of combustion under standard conditions is ‎ ΔH_c^Ꝋ.‎

A Born-Haber cycle is a specific application of Hess's Law for ionic compounds that enables the calculation of lattice enthalpy.

False.

The first ionisation energy is always an endothermic process.

Lattice enthalpy is the energy required to separate one mole of an ionic compound into its constituent gaseous ions.

The enthalpy of atomisation is the energy required to convert one mole of an element in its standard state to gaseous atoms.

The energy change when one mole of gaseous atoms gains one mole of electrons to form one mole of gaseous anions is known as electron affinity.

False.

While electron affinity is often exothermic, it can be endothermic for some elements.

In a Born-Haber cycle for NaCl, the arrow from Na (s) to Na (g) represents the enthalpy of atomisation of sodium.

True.

Arrows pointing upwards in a Born-Haber cycles represent endothermic reactions.

An equation to represent the first electron affinity of chlorine is:

Cl (g) + e^- → Cl^- (g)

An equation to represent the second ionisation energy of magnesium:

Mg⁺ (g) → Mg²⁺ (g) + e^-

The enthalpy change is shown in Step 1 is the enthalpy change of formation.

The enthalpy change is shown in Step 2 is the enthalpy of atomisation of potassium.

The enthalpy change is shown in Step 3 is the enthalpy of atomisation of fluorine.

The enthalpy change is shown in Step 4 is the first ionisation energy of potassium.

The enthalpy change is shown in Step 5 is the first electron affinity of fluorine.

The enthalpy change is shown in Step 6 is the lattice enthalpy of formation.

Th equations for the three missing steps in the Born-Haber cycle are:

To calculate lattice enthalpy for KF, ΔH^θ_lattice KF:

ΔH^θ_lattice KF = -563-90-79-419+328

ΔH^θ_lattice KF = -823 (kJ mol⁻¹)

To calculate the enthalpy of formation LiF. ΔH^θ_f LiF_:

ΔH^θ_f LiF = 216 + 79 + 520 + (-348) + (-1061)

ΔH^θ_f LiF = -594 (kJ mol^-1)

The enthalpy of atomisation of chlorine is 121 kJ mol^-1:

ΔH^θ_at Cl₂ = 242 / 2 = 121 kJ mol^-1

To calculate the enthalpy change for the following conversions:

Ca (s) → Ca²⁺ (g) + 2e^-

ΔH^θ_IE1 Ca = 590 kJ mol^-1   

ΔH^θ_IE2 Ca = 1145 kJ mol^-1                       

ΔH^θ_at Ca = 178 kJ mol^-1

• ΔH^θ_at Ca + ΔH^θ_IE1 Ca + ΔH^θ_IE2 Ca = 178 + 590 + 1145 = (+)1913 (kJ mol^-1)

True.

When performing a Born-Haber cycle calculation for MgCl₂, the value for the first electron affinity of chlorine would need to be doubled.

To calculate the enthalpy change for the following conversion:

Cl₂ (g) + 2e^- → 2Cl^- (g)

ΔH^θ_BE Cl₂ = 242 kJ mol^-1

ΔH^θ_EA Cl = -349 kJ mol^-1

• ΔH^θ_BE Cl₂ + (2 x ΔH^θ_EA Cl) = 242 + (2 x -349) = -456 (kJ mol^-1)

To calculate lattice enthalpy,  ΔH^ꝋ_lat:

ΔH^ꝋ_f = ΔH^ꝋ_at + ΔH^ꝋ_at + ΔH^ꝋ_IE + ΔH^ꝋ_EA - ΔH^ꝋ_lat

ΔH^ꝋ_lat = - ΔH^ꝋ_f + ΔH^ꝋ_at + ΔH^ꝋ_at + ΔH^ꝋ_IE + ΔH^ꝋ_EA

To calculate the lattice enthalpy of CaCl₂, ΔH^θ_lattice CaCl₂.

ΔH^θ_lattice CaCl₂ = -814 - 178 - 590 - 1145 - 242 + (349 x 2)

ΔH^θ_lattice CaCl₂ = -2271 (kJ mol⁻¹)