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A2 chemistry 4.5
Edexcel chemistry - equilibria
| Question | Answer |
|---|---|
| Define reversible reaction, equilibrium, dynamic equilibrium | Reversible reaction goes in both directions @ same temperature so doesn't go to completion. Equilibrium - [reactants/products] remain constant. Dyanmic equilibrium - rates of forward + backward reactions same so no further change in [reactants/products] |
| What is the equilibrium yield? | The % of reactant that is converted into product in a reversible reaction |
| What is Le Chatelier's principle? | When the external conditions of a system in equilibrium are changed, the position of equilibrium shift to nullify the change |
| Write the Kc expressions for the following reactions: 1) MgCl2 + H2O -> Mg2+(aq) + 2Cl-(aq). 2) HCOOH + CH3OH -> HCOOCH3 + H2O | 1) Kc = [Mg2+][Cl-]2. 2) Kc = [HCOOCH3][H2O]/ [HCOOH][CH3OH] |
| Describe the relationship between Kc and Q | Kc is equilibrium constant, Q is reaction quotient. Q = Kc at equilibrium. Q> Kc means position of equilibrium lies to right, system converts more products into reactants. Q < Kc means equilibrium lies to left, system converts more reactants to product |
| Describe the steps for finding Kc | Balanced equation, table: initial mol, delta mol, eq mol, concentration (mol/dm3) - use x for delta mol if not given, use v if volume isn't given. Write out correct Kc expression, sub in concentrations, find units for Kc (if none, state this) |
| Which of the following species would you exclude from a Kc calculation and why: CO2, H2O, NaCl, Br2? | H2O if it is the solvent - solvent concentration remains relatively constant. NaCl unless if it is dissolved in water - solids are excluded. Br2 - liquids are excluded, no concentration |
| Find Kc for hydrolysis of ethyl ethanoate (acidic) -200g of ethyl ethanoate, 7g water, @ equilibrium 0.25 mol ethanoic acid | 0.1905 |
| Define partial pressure and mol fraction | Partial pressure = pressure exerted by 1 gas in a gaseous mixture if it alone occupied the same volume under the same conditions, partial pressure = mol fraction x total pressure. Mol fraction = amount of gas/ total amount |
| Describe the steps for determining Kp | BAlanced equation, table: intial mol, delta mol, eq mol, mol fraction, partial pressure (units), write out correct Kp equation, sub in partial pressures, find units of Kp |
| 2NO2 (g) <-> 2NO (g) + O2 (g). Eq moles: 0.96 NO2, 0.04 NO, 0.02 O2. Kp = 6.8 x 10^-6 atm, what must total pressure have been? | |
| How does ΔS total affect value of K | ΔStotal = RlnK. As ΔStotal becomes more positive, K value increases. Endothermic: increase T = increase K. Exothermic: increase T = decrease K. |
| If 0<ΔStotal<200J/K/mol, then where does the position of equilibrium lie/ value of K? | 1< K < 10^10. Position of equilibrium lies on the right, so reaction favours products |
| If ΔStotal < -200, what is value of K/where does equilibrium lie? | K<10^-10. Position of equilibrium lies FAR to the left, reaction doesn't go |
| How does temperature affect K + position of equilibrium? | Endothermic: increase T, K increases, Q<K, system converts more reactants to products (shift to right) so Q=K. Exothermic: increase T, K decreases, Q>K, system converts more products to reactants (shifts to left) so Q=K. Le Chatelier |
| How does temperature affect rate of reaction? | Increase T increases rate of reaction (collision theory), rate increase more in endothermic direction due to high Ea |
| How does pressure affect K + position of equilibrium? | Le Chatelier's principle: shifts position of equilibrium to side with fewer gas moles THEREFORE no change in value of K. |
| How does pressure affect rate of reaction? | Increases rate of reaction (collision theory) equally in both directions |
| How does concentration affect K + position of equilibrium? | No effect on value of K, but it changes the reaction quotient so Q doesn't equal K, the position of equilibrium shifts (Le Chatelier's principle: to the side that reverses the change) so Q=K again |
| How does concentration affect rate of reaction? | Increases rate of reaction (collision theory) |
| How does catalysis affect K, position of equilibrium and rate of reaction? | No effect on value of K or Q (position of equilibrium), but it increases rate of reaction equally in both directions so dynamic equilibrium is reached more quickly, although rate is limited by solid catalyst's active site availability |
| What are the desirable features of industrial processes and how are they achieved? | High equilibrium yield (continually remove products), low cost (heat exchanger/catalysis/compromise pressure), high rate, high atom economy (recycle unreacted reactants) |
| State Haber process equation + explain conditions | N2 + 3H2 -> 2NH3, exothermic. Compromise temperature, 400 degrees c - reasonable yield at an economically acceptable rate. Compromise pressure. Iron catalyst. Unreacted reactants pass through catalyst many times so atom economy is 98% |
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