Module 5: Enthalpy and Entropy V1 Flashcards
Define the term “first ionisation energy”. Include the symbol in your answer and predict whether the enthalpy change will be exothermic (negative) or endothermic (positive). Explain why.
ΔH sign: Positive. ✓
Energy required to break electrostatic attraction between negative electron and positive protons. ✓
Define the term “second ionisation energy”. Include the symbol in your answer and predict whether the enthalpy change will be exothermic (negative) or endothermic (positive). Explain why.
ΔH sign: Positive. ✓
Energy required to break electrostatic attraction between negative electron and positive protons. ✓
Define the term “first electron affinity”. Include the symbol in your answer and predict whether the enthalpy change will be exothermic (negative) or endothermic (positive). Explain why.
ΔH sign: Negative. ✓
Bond formed between negative electron and positive protons in nucleus. ✓
Define the term “second electron affinity”. Include the symbol in your answer and predict whether the enthalpy change will be exothermic (negative) or endothermic (positive). Explain why.
ΔH sign: Positive. ✓
Energy required to overcome repulsion between negative ion and negative electron. ✓
Explain why electron affinity decreases down the group.
Atomic radius increases. ✓
Number of shells increases, and so shielding increases. ✓
Nuclear charge increases, but this is outweighed by increase in shielding and atomic radius. ✓
Less nuclear attraction, so more difficult for larger atoms to attract electrons needed to form an ion. ✓
Remember, gaining or losing electrons is always explaining using NANCARS
Define the term “enthalpy change of atomisation”. Include the symbol in your answer and predict whether the enthalpy change will be exothermic (negative) or endothermic (positive). Explain why.
ΔH sign: Positive. ✓
Energy is required to break a bond. ✓
Define the term “lattice enthalpy”. Include the symbol in your answer and predict whether the enthalpy change will be exothermic (negative) or endothermic (positive). Explain why.
ΔH sign: Negative. ✓
Energy is released when forming a bond. ✓
Define the term “enthalpy change of hydration”. Include the symbol in your answer and predict whether the enthalpy change will be exothermic (negative) or endothermic (positive). Explain why.
Negative. ✓
Bond formation between ions and water molecules releases energy. ✓
Define the term “enthalpy change of solution”. Include the symbol in your answer and predict whether the enthalpy change will be exothermic (negative) or endothermic (positive). Explain why.
ΔH sign: Positive or negative. ✓
Depends on the balance between bond breaking for the lattice ✓ and bond making between the ions and water molecules. ✓
Determine the type of enthalpy change for each of the following equations.
Write equations and determine whether the enthalpy change is positive/ negative for the following:
a. The first electron affinity of sulfur
b. The second electron affinity of sulfur
c. The atomisation of bromine
d. The atomisation of oxygen
Write equations and determine whether the enthalpy change is positive/ negative for the following:
a. The first ionisation of aluminium
b. The formation of propan-1-ol
c. The lattice enthalpy of potassium chloride
d. The lattice enthalpy of magnesium chloride
e. The lattice enthalpy of aluminium oxide
Which enthalpy change(s) is/are exothermic?
1 The third electron affinity of nitrogen
2 The bond enthalpy of the C–H bond
3 The standard enthalpy change of formation of Cl2(g)
A 1, 2 and 3
B Neither
C Only 1 and 2
D Only 2 and 3
B, Niether are correct ✓
Which equation represents the change that accompanies the standard enthalpy change of atomisation of bromine?
D ✓
Bromine is a liquid under standard states and conditions
How does lattice enthalpy link to ionic bonding?
Lattice enthalpy is a measure of the strength of an ionic bond. ✓
How does average bond enthalpies link to covalent bonding?
average bond enthalpy is a measure of the strength of a covalent bond. ✓
What does the size of lattice enthalpy indicate?
A large exothermic value for lattice enthalpy means there are very strong electrostatic forces of attraction ✓
between the oppositely charged ions ✓
and a stronger ionic bond. ✓
Write the equation for the lattice enthalpy of sodium chloride. Include state symbols in your answer.
Explain why NaCl has a more exothermic lattice enthalpy than CsCl?
Na+ is a smaller ion than Cs+ ✓
Na+ ions can pack closer together and have a greater charge density ✓
Na+ ions have a stronger electrostatic force of attraction for Cl- ions ✓
NaCl has the more exothermic lattice enthalpy ✓
Explain why magnesium chloride has a more exothermic lattice enthalpy than sodium chloride.
Mg2+ has a greater charge compared to Na+ ✓
Mg2+ ions can pack closer together and have a greater charge density ✓
Mg2+ ions have a stronger electrostatic force of attraction for Cl- ions ✓
MgCl2 has the more exothermic lattice enthalpy ✓
Describe how the lattice enthalpy of MgO differs from the lattice enthalpy of BaO.
Mg2+ is a smaller ion than Ba2+ ✓
Mg2+ ions can pack closer together and have a greater charge density ✓
Mg2+ ions have a stronger electrostatic force of attraction for O2- ions ✓
MgO has the more exothermic lattice enthalpy ✓
Explain why magnesium oxide has a more exothermic lattice enthalpy than sodium oxide.
Mg2+ has a greater ionic charge than Na+ ✓
Mg2+ ions can pack closer together and have a greater charge density ✓
Mg2+ ions have a stronger electrostatic force of attraction for O2- ions ✓
MgO has the more exothermic lattice enthalpy ✓
Why is it difficult to predict whether the lattice enthalpy for magnesium sulfide is more or less exothermic than the lattice enthalpy of sodium oxide.
Mg2+ smaller and has a greater charge compared to Na+ ✓
therefore Mg2+ stronger attraction ✓
O2- smaller than S2- ✓
O2- stronger attraction ✓
Hard to predict which is more exothermic ✓
Using a Born-Haber Cycle and the data provided, calculate the lattice enthalpy
Draw a Born-Haber cycle for BaO.
Using the following data, draw a dissolving cycle and calculate a value for the standard enthalpy change of solution of KCl
Calculate a value for the standard enthalpy change of hydration for chloride ions using the cycle below
Using the following data, draw a dissolving cycle and calculate a value for the standard enthalpy change of solution of aluminium chloride
Explain the difference in hydration enthalpies between Cl- and Br -
Cl- is a smaller ion than Br - ✓
Cl- has the same ionic charge as Br - but Cl- has a greater charge density ✓
Cl- has stronger electrostatic attractions for water molecules ✓
Cl- has the more exothermic hydration enthalpy ✓
Explain the difference in hydration enthalpies between Na+ and Mg2+
Mg2+ is a smaller ion than Na+ ✓
Mg2+ a greater ionic charge and greater charge density than Na+ ✓
Mg2+ has stronger electrostatic attractions for water molecules ✓
Mg2+ has the more exothermic hydration enthalpy ✓
Use the results below to determine the enthalpy change of solution for KCl
Define what is meant, entropy, S
is a measure of the degree of disorder in a system. ✓
Explain why entropy, S is always positive.
All substances have some degree of disorder ✓
Particles are in constant motion ✓
Explain what a +ΔS shows
A positive ΔS value means the system is becoming more disordered. ✓
Explain what a -ΔS shows
A negative ΔS value means the system is becoming more ordered. ✓
State whether the entropy change will be a positive or negative value for the following. Explain your answer.
Positive because entropy increases ✓
Liquid becomes gas and there is more disorder ✓
State whether the entropy change will be a positive or negative value for the following. Explain your answer.
Negative because entropy decreases ✓
3 moles of gas to 2 moles of gast herefore there is less moles of gas product.✓
therefore more ordered ✓
State whether the entropy change will be a positive or negative value for the following. Explain your answer.
Positive because entropy increases ✓
Solid becomes aqueous, aqueous particles are more disordered than solid particles ✓
State whether the entropy change will be a positive or negative value for the following. Explain your answer.
Negative because entropy decreases ✓
4 moles of gas to 2 moles of gas therefore there is less moles of gas product. ✓
therefore more ordered ✓
State whether the entropy change will be a positive or negative value for the following. Explain your answer.
Positive because entropy increases ✓
1 moles of gas to 2 moles of gas therefore there is more moles of gas product.✓
therefore more disorder ✓
State whether the entropy change will be a positive or negative value for the following. Explain your answer.
Positive because entropy increases ✓
Solid becomes gas and liquid and there is more disorder ✓
State whether the entropy change will be a positive or negative value for the following. Explain your answer.
Negative because entropy decreases ✓
2 moles of gas to 1 moles of gas therefore there is less moles of gas product.✓
therefore more ordered ✓
State whether the entropy change will be a positive or negative value for the atomisation of iodine. Explain your answer.
Positive because entropy increases ✓
Solid becomes gas so there is more disorder ✓
State whether the entropy change will be a positive or negative value for the lattice enthalpy of calcium bromide. Explain your answer.
Negative because entropy decreases ✓
Gases become solid, order increases ✓
Give the units for entropy, S and entropy change, ΔS
Explain why this exothermic reaction is spontaneous at low temperatures but does not occur at very high temperatures.
∆H is negative (exothermic reaction) ✓
∆S is negative (3 moles of gas gives 2 moles of gas) so T∆S is negative ✓
At higher temperatures, T∆S becomes more negative ✓
At higher temperatures, T∆S becomes more negative than ∆H and outweighs ∆H
✓
At higher temperatures, ∆G becomes more positive, and the reaction becomes unfeasible ✓
Complete the following table
Write the equation for Gibbs Free energy in terms of the equation of a line (y=mx+c) and state how you can find each term graphically.
Plot graph of ∆G on the y-axis against temperature on the x-axis gives a straight with gradient - ∆S. ✓
y-Intercept: ∆H. ✓
Gradient: -∆S. ✓
a) No, as ∆G is positive ✓
b) 760 – 273 = 487 degrees celcius ✓
c) 160 kJ mol-1 ✓
d) -∆S = -0.25 kJ K-1 mol-1 ✓
so ∆S = 0.25 x 1000 = 250 J K-1 mol-1 ✓
This question is about free energy changes, ∆G, enthalpy changes, ∆H, and temperature, T.
The Gibbs’ equation is shown below.
∆G = ∆H − T∆S
A chemist investigates a reaction to determine how ∆G varies with T.
The results are shown
What is significant about the gradient of the line and the values P and Q shown
Explain your reasoning.
ΔG = −ΔST + ΔH ✓
Gradient = −ΔS ✓
P: ΔH enthalpy change ✓
Q: Temperature for a reaction to be feasible ✓
Using the graph attached, determine the value for ∆S, ∆H and ∆G for lines A, B, C, D
a. ∆S = -ve
∆H = +ve
∆G = +ve. ✓
b. ∆S = -ve
∆H = -ve
∆G = +ve at high temperatures only. ✓
c. ∆S = +ve
∆H = +ve
∆G = +ve at low temperatures only. ✓
d. ∆S = +ve
∆H = -ve
∆G = -ve. ✓
∆H is negative (exothermic reaction) ✓ ∆S is positive so T∆S is positive ✓
At higher temperatures, T∆S becomes more positive ✓
At higher temperatures, ∆G becomes more negative and the reaction becomes more feasible ✓
∆G is always going to be negative. ✓
∆H is negative (exothermic reaction) ✓
∆S is negative (4 moles of gas gives 2 moles of gas) so T∆S is negative ✓
At higher temperatures, T∆S becomes more negative ✓
At higher temperatures, T∆S becomes more negative than ∆H and outweighs ∆H
✓
At higher temperatures, ∆G becomes more positive and the reaction becomes unfeasible ✓
∆H is positive (endothermic reaction) ✓
∆S is positive so T∆S is positive ✓
At higher temperatures, T∆S becomes more positive ✓
At higher temperatures, T∆S becomes more positive than ∆H and outweighs ∆H ✓
At higher temperatures, ∆G becomes more negative and the reaction becomes feasible ✓
∆S for an endothermic reaction is negative. Will the reaction ever be feasible? Explain your answer using both low and high temperature.
∆H is positive (endothermic reaction) ✓
∆S is negative so T∆S is negative ✓
At higher temperatures, T∆S becomes more negative ✓
Negative value is subtracted from positive ∆H, so ∆G will always be positive and reaction will always be not feasible. ✓
∆S for an exothermic reaction is positive. Will the reaction ever be feasible? Explain your answer using both low and high temperature.
∆H is negative (exothermic reaction) ✓
∆S is positive so T∆S is postive ✓
At higher temperatures, T∆S becomes more positive ✓
Positive value is subtracted from negative ∆H, so ∆G will always be negative and reaction will always be feasible. ✓
