Course Home Shapter 14 Homework Applying Le Chätelier\'s Principle Applying Le C
ID: 557939 • Letter: C
Question
Course Home Shapter 14 Homework Applying Le Chätelier's Principle Applying Le Châtelier's Principle Part A When a chemical reacion is at equilbrium, Q (the Consider the following system at equilbrium reaction quobient) is equal to K (the equilbrium constat). lf a stress is applied to the mixture Classify eachofthe following actions by whether it canes a leeward n the direction of the net reaction changes the value of Q, then the system a rightward shift, or no shim no longer at equilibrium. To regain equiibrium, the reaction will either proceed forward or in reverse Drag the appropriate items to their respective bins unti Q is equal to K once again. Allernatively equilibrium can be disrupted by a change in Hints emperature, which changes the value of K. The result however is the same, and the reaction will proceed forward or in reverse unsil Q is equal to he new k Le Chátelier's principle summarizes this idea: If a stress is applied to a reaction mture at equil brium, a net reaction occurs in the direction relieves the tRightward shit No shit Leftward shnt Increase [C Decreate [A increase IAl Double JA] and My Answers Give Up PS BJExplanation / Answer
First, let us define the equilibrium constant for any species:
The equilibrium constant will relate product and reactants distribution. It is similar to a ratio
The equilibrium is given by
rReactants -> pProducts
Keq = [products]^p / [reactants]^r
For a specific case:
aA + bB = cC + dD
Keq = [C]^c * [D]^d / ([A]^a * [B]^b)
Where Keq is constant at a given temperature, i.e. it is dependant on Temperature only
If Keq > 1, this favours products, since this relates to a higher amount of C + D
If Keq < 1, this favours reactants, since this relates to a higher amount of A + B
If Keq = 1, this is in equilibrium, therefore, none is favoured, both are in similar ratios
Note that the concentrations MUST be in equilibrium. If these are not in equilibrium, then the reaction will take place until there is equilibrium achieved.
For this, we use Q, the reaction quotient of products/reactants, it allows us to understand the ratio distribution and the direction/shit of equilibrium
Q is defined as:
Q = [C]^c * [D]^d / ([A]^a * [B]^b)
In this Case, the concentrations are NOT in equilibrium
Therefore:
If Q < Keq; this has much more reactants than products, therefore expect reactants to form more product in order to achieve equilibrium
If Q > Keq; this has much more products than reactants, therefore expect products to form more reactants in order to achieve equilibrium
If Q = Keq; this has the same ratio in equilibrium for reactants and products. Expect no reaction. It is safe to assume this is already in equilibrium.
Special notes:
Typically; we use aqueous and gas phases. (Gases and Aqueous Concentrations can be related via PV = nRT, since M = n/V as well)
Solids and Liquids, i..e. (s) and (l) have activity of 1. Therefore, they must not be considered in the ratios.
from the list:
a)
K = C^5 /((A^2)(B^2))
a)
leftward shift will favour A and B... this will cuase this:
increase of C, since shift must go reverse
Decrease in either A or B, since C must form them again
double of [B] and [C] will favour reactants, since C^5 and B^2 only
b)
right shift favours products, C
this is done via:
icnrease of A and B
decrase of C, since the reaciton must proceed forward
c)
No effect will be doubling [A] but halving [B], since 2x * 1/2 = 1