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Chemistry 6
| Question | Answer |
|---|---|
| how to write expressions for the Keq (constant) | {products}/[reactants] nA + mB <-->xC + yD [C]^x[D]y/[A]n[B]m no solids or liquids, only gases and aqeous CONSTANT IS DEPENDENT ON TEMP include charges |
| when you change the concentration of an equalibrium... | add a chemical, shift to the other side remove a chemical - shift to the same side |
| if you change the temperature of an equalibrium... | write heat into the rxn (- goes on right and + goes on left) then same rules as concentration |
| when you change the pressure of an equalibrium | increasing the P = shift to the side with fewest moles of gas decreasing the P = shift to the side with most moles of gas |
| how volume can determine the direction a reaction can shift | volume is increased (pressure decreases) → equilibrium shifts toward the side with more moles of GAS(to increase pressure). volume is decreased (pressure increases) → equilibrium shifts toward the side with fewer moles of GAS(to reduce pressure). |
| equalibriumin saturated solutions | Solute (solid)⇌Solute (dissolved in solution) |
| what direction would equalibrium shift if solid were added to a saturaed solution | No change (the solid would just settle to the bottom, the amount of dissolved solute would remain the same) |
| what is equal in equalibrium | the RATES of forward and reverse reactions |
| how does a catalyst affect equalibrium | it gets to equilibrium faster but does not change the position of equilibrium (equilibrium constant stays the same) 'speeds up both the forward and reverse reactions equally |
| Kc is dependent on __________ | temperature (so if the temp. is changed then the equilibrium constant will change, think of Le Chatelier) Endothermic reactions → K increases with temperature Exothermic reactions → K decreases with temperature |
| Le Chatelier | If a system at equilibrium is disturbed by a change in concentration, pressure, volume, or temperature, the system will shift in the direction that counteracts the disturbance. |
| why dont concentration, volume, or preassure affect the Kc | |
| if one person uses whole numbers to balance the equation, and another uses fractions, will the Kc value change? | Yes, equilibrium constants are written for specific coefficients. If you multiply all coefficients by 𝑛, the new equil. constant is related to the original by: 𝐾new = 𝐾old^𝑛 If you divide all coefficients by 𝑛, then: 𝐾new = 𝐾old^1/𝑛 |
| why does using a fraction as coefficients instead of whole numbers change the value of Kc | |
| what does the K value mean | large --> many products are formed small --> few products are formed (more reactant) |
| given an amount of product & reactant, determine the direction of the rxn or if it is already at equalibrium (if either the products or reactants start at zero, this should be obvious) -given inital amount, the balanced rxn, and the K value | write the Q expression substitute in the initial concentrations given solve for Q if Q < K, the rxn proceeds in the forward direction (there is not enough product, think prod/react) if Q > K, then the rxn proceeds in the reverse direction |
| difference between K (equilibrium constant) and Q (reaction quotient) | -both are prod./reactants -K is only @ equil. but Q is at a random place (can still be @ equilibrium) -use [] for K and () for Q |
| how Le Chatelier's princple applied to K and Q | some stresses, (changes in {}) cause a change in Q ONLY. A change in T causes a change in K. In either case, species redistribute (state to react more agressively in one direction) to bring Q and K back into equality |
| if two gases are in a 4L container, what is the volume of each gas | 4L Both gases are mixed together, they don't split or separate the volume. Therefore, both gases have the same volume |
| the different equalibtium constatnts and what they mean | Kc = concentration (ONLY IN MOLARITY) Kp = pressure (gases) Ka = acids Kb = bases Kw = water Ksp = solubility product (solutions - allow us to calculate where a solution will be saturated) |
| how to solve for unknown concentrations given the balanced equation and the K | write k expression make ICE chart (using x for amount of change) subsitute K & the final values from ICE chart into expression use quadratic formula & solve for x plug xs in, tif one produces a neg. end [] it's wrong & use other X to find ans. |
| what units can you use in an ICE chart | moles (under any conditions) M (if the V is constant) atm (if the V and T are constant) and NEVER gram (ICE is based on balanced rxns & stoiciometry but every substance has a different molar mass) |
| WARNING {2x]^2 = 4x^2 | |
| *Kp is asked to be in atm but we are given pressures in mmHg* you can solve for Kp in mmhg & convert back to atm at the end. Look at the K expression and cancle out units. If only one mmHg on the denominatior is left, then it's a straight conversion | However, if, for example, the units were mmHg^2, then divide by 760^2 to convert to atm^2 (leave atm^2, in this senario it is the correct unit). Therefore, its better to convert evey value into atm in the beggining |
| how to convert Kc into Kp | take Kc & assume --> original value moles over 1 L plug these values into ideal gas law P(1) = (og moles)[(0.00821)(298) ]^Δn Kp=Kc(RT)^Δn Δn=∑coeffic. of gaseous prod.−∑coefficients of gaseous react. -change in gaseous moles |
| Kp | (Pb)^b/(Pa)^a dependent only on terperature |
| how to convert Kc into Kp (second flashcard) | Kp = Kc (RT)^Δng Vatiation of ideal gas law: PV = nRT P = n/v R T P = M RT Kp = P and Kc = M |
| what does it mean if Δn is zero | Kp = Kc the moles of gaseous product and moles of gaseous reactant are the same (doesnt mean their molarities are the same) Δn₍g₎ = 0 → same number of gas moles before and after Kp=Kc(RT)^0 → anything to the zero power is 1 |
| the Kc of a rxn shows that he equation likes to progress forward. However, when I convert this K into Kp, the reaction prefers to progress in the reverse. why is this | But concentration and pressure are not proportional unless the number of gas moles is the same on both sides. Pressure changes depend on number of gas particles |
| when solving for K by maipulating rxns, the rules are different from hess law ... | reverse a rxn --> 1/k (instead of changing the sign) Halve =coefficients in an equation --> take the square root of K double the coefficients -> K^2 |
| when finding the K and given miltple K expressions, | multiply the k values of each instead of adding like Hess' multiply all the K expressions and cancle out common compounds that are on both the bottom and the top (cross cancle) |
| if you have a lot of either a reactant or product, why does the rxn that consumes this chemical occur faster | if there are many A mcs around, then when A needs to react, it has a higher chance of colliding with other mcs. So A is being consumed faster |
| if a chemical in a reaction has a higher pressure, why will the reaction that consumes it occure faster | 1. higher P = more of that substance (GAS) is present becuase more gas particles are colliding with the edges of the container 2. more of a substance means more frequent collisions (collision theory) |
| if the volume is decreased, why does the rxn shift to the side with fewer moles of gas | a smaller volume means that the pressure is increased fewer moles of gas occupy less space and thereofre exert less pressure shifting to the side with fewer moles of gas relieves the stress of increased pressure |
| why does a smaller volume means that the pressure is increased | 1. each particles has a smaller distance to travel before colliding with a wall 2. particles are closer together (increase in []) (when particles collide they transfer kinetic E, ensures they're changing direction & speed --> more wall collisions) |
| how does T affect equilirbium | if heat is on the reactant side, then adding heat will increase the amount of product meaning that K is bigger you can figure out the rest |
| what does the value of Ksp tell you | molar solubility (how moles of solid dissolve per leter) |
| in dissociation equations, what is alawys the intial concentrations of the ions that can form from dissociation | intiial solid (before being dissolved) - 0 becuase no ions can form from the solid -because (curretly) the solution is alawys made using pure water, the concetration of ion in the water is zero (0 for each ion) |
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Blessed Rectangle 10