CHEM 104 1st Edition Lecture 20- Example (from previous chemical equilibrium notes): butane isobutene, kc = [isobutene]/[butane] = 2.5 at 298 KIf concentration of butane is 0.97 M and concentration of isobutene is 2.18 M, is system at equilibrium? Q = Cisobutane/Cbutane = 2.18 M/0.97 M = 2.25System is not at equilibrium, Q<kc, reaction will proceed towards the right- If you double the volume of a liquid product, the concentration of each of the other reactants/products will halve (since c = n/v); the resulting Q value will depend on the stoichiometry of the equation (there could be no shift, or equilibrium could shift to the right or to the left) - Decreasing volume (or increasing pressure) will shift equilibrium towards a smaller number of particles o Example: 2HI(g) H2(g) + I2(g)If we double pressure, all concentrations double Q = kc, there is no shift in equilibriumo Changing pressure (by changing the volume) only shifts equilibrium is gaseous systems where change in n isn’t equal to 0 - Changes in temperature o Endothermic reactions: increase in temperature will determine an increase in kc – reaction is more product favored at higher temperaturesExothermic reactions: increase in temperature will determine a decrease in kc – reaction is more reactant favored at higher temperaturesThese notes represent a detailed interpretation of the professor’s lecture. GradeBuddy is best used as a supplement to your own notes, not as a
View Full Document
Unlocking...