Chemical Equilibrium Chapter 14.2 & 14.3

Dynamic Equilibrium Is the point at which the forward and reverse rates of a reaction are equal to each other Is represented in equations by a double arrow in a reaction Ex. 2H2+O2 ⇌ 2H2O In addition to equal rates concentrations become stable on both sides of the equation Although rates are equal concentrations at equilibrium are not equal

How to look at a Reaction Considering Dynamic Equilibrium Beginning Only reactant molecules exist so only reactant molecules collide, there is only the forward reaction happening Middle Concentration of the product(s) is increasing so collisions increase causing reverse reaction Equilibrium Collisions are happening at the same rate on both sides

What Does This Look Like Graphically?

Equilibrium Constant (K) The equilibrium constant (K) is the way to quantify the concentrations at equilibrium and how far the reaction proceeds after equilibrium K>>1 means the forward reaction is favored and there are high concentration for products and low concentration for reactants, forward reaction proceeds fully K<<1 means the reverse reaction is favored and there are low concentrations for products and high concentration for reactants, forward reaction doesn’t proceed far K≈1 means neither is favored and the reaction proceeds about half way

How to find K To find K we use the Law of Mass Action Using the base equation aA+bB⇌cC+dD where lowercase represent coefficients we divide the concentration of the products by that of the reactants while all values are raised to their respective coefficients as following Note that this formula is in your reference tables at the top of the section labeled Equilibrium

Changes to the Equilibrium Constant When the base equation you are working with needs to be modified in some way the equilibrium constant must be change to account for the modifications There are three common modifications Reversing the equation Adjusting coefficients Adding multiple equations together

Adjusting the Constant When reversing the equation invert the equilibrium constant When adjusting coefficients (multiplying the coefficients by a factor) raise the equilibrium constant to the same factor When combining equations find the individual constants then multiply them together to obtain an overall equilibrium constant for the combined equation

What the college board says

Find the Equilibrium constant for the following equation 2H2S(g)⇌2H2(g) + S2(g) K=[H2]² [S2] / [H2S]²

Consider the reaction A(g)⇌ B(g) where K=10 If initially [A]=1.1M and [B]=0.0M which of the following would be the result at equilibrium? [A]=1.0M and [B]=0.1M 2. [A]=0.1M and [B]=1.0M 3. Concentrations will be equal

Consider the following reaction and solve for K 2A + B ⇌ C + 2D Where the concentrations at equilibrium are 0.100M , 0.600M, 0.200M and 0.400M respectively. [0.200][0.400]²/[0.100]²[0.600] = K K= 5.33 And if we reverse the reaction? [0.100]²[0.600]/[0.200][0.400]² K=.1875