Help
Options
Didn't know it?
click below
click below
Knew it?
click below
click below
Don't Know
Remaining cards (0)
Know
retry
shuffle
restart
0:00
Embed Code - If you would like this activity on your web page, copy the script below and paste it into your web page.
Normal Size Small Size show me how
Normal Size Small Size show me how
AQA A2Thermodynamics
AQA A2 Thermodynamics
| Question | Answer |
|---|---|
| Define Enthalpy change of Formation | The enthalpy change when 1 mole of a compound is formed from its elements in their standard states under standard conditions. |
| Define Bond dissociation enthalpy | The enthalpy change when all the bonds of the same type in one mole of gaseous molecules are broken. |
| Define Enthalpy change of atomisation of an element | The enthalpy change when 1 mole of gaseous atoms is formed from an element in its standard state. |
| Define First ionisation enthalpy | The enthalpy change when one mole of gaseous 1+ ions is formed from one mole of gaseous atoms. |
| Define Second ionisation enthalpy | The enthalpy change when one mole of gaseous 2+ ions is formed from one mole of gaseous 1+ ions. |
| Define First electron affinity | The enthalpy change when one mole of gaseous 1– ions is formed from one mole of gaseous atoms. |
| Define Second electron affinity | The enthalpy change when one mole of gaseous 2– ions is formed from one mole of gaseous 1– ions. |
| Define enthalpy change of hydration | The enthalpy change when one mole of aqueous ions is formed from one mole of gaseous ions. |
| Define enthalpy change of solution | The enthalpy change when one mole of solute is dissolved in sufficient solvent that no further enthalpy change occurs on further dilution. |
| Explain why the second ionisation energy is more endothermic than the first ionisation energy | Removing an electron from a positive ion will be require more energy (not simply “difficult”) because there are more p+ than e- so the e- are more attracted to the nucleus |
| Explain why the first electron affinity is always exothermic | Energy is given out when an electron JOINs onto an atom |
| Explain why the second Electron Affinity is endothermic | Energy is required to add an electron to an already negative ion and to overcome the forces of repulsion |
| Explain why enthalpy change of hydration is exothermic | because a bond forms: an attraction between an ion and a polarised molecule |
| Define Standard conditions | 100kPa and 298 K |
| Define Pure ionic model | Ions are spherical and have charge evenly spread |
| Define Partial covalent bonding | The electrons in the negative ions are pulled towards the positive ion. |
| Explain why there is a large difference in the experimental and theoretical values? | the compound is not purely ionic, there is a covalent character |
| In which compound is there more a covalent character? NaF or AlCl3? | AlCl3 because the Al+3 is highly charged and will polarise the electron cloud of the large Cl- ion (larger than F- ion); KEYWORDS: CHARGE, SIZE, POLARISATION, ELECTRON CLOUD |
| Order the compound in increasing lattice enthalpy: LiF or NaCl or MgO or BaS? | NaCl |
| What is the Enthalpy change of solution equal to? | Lattice dissociation enthalpy + Enthalpy of hydration (of each ion) (mnemotechnic:hydration=lucozade sport) |
| What is entropy? | Entropy is the measure of the number of ways that particles can be arranged. |
| How does changing state affect entropy? | Melting or boiling increases entropy, condensing or solidifying decreases entropy. |
| How does dissolving a solid affect entropy? | Causes an increase. |
| How does number of particles affect entropy? | More particles lead to an increased entropy. |
| What is a spontaneous reaction | A reaction that does not require any external energy to occur; deltaG<0 |
| How do you work out Free energy change | ΔG = ΔH – T ΔS (Remember to convert all units to J or kJ) |
| What are the traps on a Born Haber cycle? | remember moles, moles, moles; direction of arrows |
| How do you work out the minimum temperature for a reaction to be feasible? | deltaG MUST be at least equal to 0, so T=deltaH/deltas |
| How does FreeGibbs energy fit the equation of a straight line? | deltaG=deltaH – T deltas or y =a –bx where a=deltaH and b=deltaS=gradient |
| Given the entropy of individual reactants and products, how do you work out the entropy change? | deltaS=(molesxentropy of each products – molesxentropy of reactants); REMEMBER MOLES MOLES MOLES |
| How do you work out the enthalpy change of solution in a calorimetry experiment? | deltaH=-Q/moles; ie deltaH=-m c deltaT/moles |
| Explain why. the enthalpy change of hydration of Magnesium is more exothermic than the enthalpy change of hydration of Calcium. | Both have the same charge; Magnesium is a smaller ion so will attract more the water molecule KEYWORDS: CHARGE, SIZE, ATTRACTION |
| Explain why it is difficult to predict which of Na2S or CaCl2 would have the most exothermic Lattice Enthalpy | Ca+2 is larger ion than Na+ so will attract less the anion but Ca+2 is more charged than Na+ so will attract more the anion; similarly S-2 is more charged but larger than Cl- |
| Explain why the enthalpy change of atomisation of Chlorine is more endothermic than the enthalpy change of atomisation of Bromine | Cl-Cl bond needs to be broken: as the 2 atoms of Cl are smaller so closer than the 2 atoms of Br, the bonding pair between Cl atoms is more attracted to muclei and will require more energy to be broken |
| Describe the interaction between water and a Sodium ion | the O-H bond in water is polarised as O is more electronegative than H; Sodium ion attract the partially negative oxygen atom; It is an ION-DIPOLE interaction |
| Explain why a magnesium appears to be stable in air at room temperature, despite a negative value for the Free Gibbs energy for the reaction between Magnesium and Oxygen | High activation energy (ie the reaction maybe THERMODYNAMICALLY feasible but not KINETICALLY) |
| In a graph deltaG on y-axis vs Temperature on x-axis, what does the gradient of the straight line represent? | deltaS |
| The enthalpy of solution for KCl is +17.2 kJmol–1. 5.00 g of KCl are dissolved in 20.0 g of water. Initial temperature of the water is 298 K. Calculate the final temperature of the solution. Specific heat capacity of water is 4.18 J K–1 g–1. | moles KCl = 5/74.6 = 0.067(0) mol; Heat absorbed = 17.2 × 0.0670 = 1.153 kJ Heat absorbed = mass × c × ΔT; (1.153 × 1000)= 20 × 4.18 × ΔT ; ΔT = 1.153 × 1000 / (20 × 4.18) = 13.8 K; T = 298 – 13.8 = 284(.2) K |
| What enthalpy change does this equation represent? Cl2(g)--> 2Cl(g)? | TWICE the enthalpy change of atomisation |
Created by:
UrsulineChem
Popular Chemistry sets