Chemistry Bootcamp Lectures Liquids - Intro A liquid is a nearly incom…. Recommended Videos Problem 2. Problem 3. Problem 4. Problem 5. Problem 6. Problem 7. Problem 8. Problem 9. Problem Video Transcript the only stress that effects the value of the equilibrium.
Numerade Educator. Liquids - Intro A liquid is a nearly incompressible fluid that conforms to the shape of its …. What is an equilibrium constant? What is meant by a stress on a reaction at equilibrium? To what stresses does a chemical equilibrium respond?
What is the equilibrium constant? First, find K p for the target reaction using the Law of Multiple Equilibria. The target reaction has 2 CH 4 g as one of the reactants and 1 C 2 H 2 g as a product. This tells us that the first reaction must have all of the stoichiometric coefficients doubled and that the K p for the given reaction must be squared. The stoichiometric coefficients of the second reaction must all be halved and then the reaction reversed so this changes the K p to the inverse of the square root of the given reaction.
The sum of these two reactions is given below, which conveniently eliminates CO g. K p for the summed reaction is found by multiplying the two equilibrium constants. This is not the target reaction. To reach the target, the third reaction given above must be used with all of its stoichiometric coefficents multiplied by three.
This means that the equilbrium constant must be cubed. Now, adding the last two reactions together gives the target reaction.
K p for the target reaction is found by multiplying equilibrium constants. Since the equilibrium constant data is given in terms of K p for the known reactions, first find K p for the unknown reaction and then, lastly, convert it to K c. Reaction 1 has nitrogen and oxygen as reactants but a wrong product. Reaction 3 can be used to eliminate the nitrogen dioxide product from 1 and also introduces chlorine:. The NO 2 g found on both reactants and products sides has been eliminated.
This has all the correct reactants, but the product is wrong and the stoichiometry is not quite right. Adding 4 and 2 rev gives the desired reaction, both eliminating the NO 2 Cl and correcting the stoichiometry.
The equilibrium concentrations of each component is given, so can be directly put into the mass action expression to give. In a sealed 1. The quantities found at equilibrium are 0. What is the value of K p? To find K c , then, we need to find the equilibrium concentration of each component in the reaction from the given masses, the volume of the container, and the molar mass of each species.
Ammonium hydrogen sulfide dissociates into ammonia gas and hydrogen sulfide gas. Write an equation for the dissociation reaction and determine the value of K p. Equilibrium —x x x. In this case x represents the partial pressure of the gases in the container and the total pressure in the container is equal to the sum of the partial pressures Dalton's Law of Partial Pressures. Thus, 0. Convert the mass of butane abbreviated B to molarity, then set up a table of concentrations to solve for the equilibrium concentration of isobutane abbreviated i -B.
Then convert this molarity back to a mass of isobutane. Initial 6. In general, if a balanced chemical equation contains different numbers of gaseous reactant and product molecules, the equilibrium will be sensitive to changes in volume or pressure. Increasing the pressure on a system or decreasing the volume will favor the side of the reaction that has fewer gaseous molecules and vice versa.
For each equilibrium system, write the reaction quotient for the system if the pressure is decreased by a factor of 2 i. Given: balanced chemical equations. Asked for: direction of reaction if pressure is halved. Two moles of gaseous products are formed from 4 mol of gaseous reactants.
Decreasing the pressure will cause the reaction to shift to the left because that side contains the larger number of moles of gas. Thus the pressure increases, counteracting the stress. When the pressure is decreased by a factor of 2, the concentrations are halved, which means that the new reaction quotient is as follows:.
Two moles of gaseous products form from 2 mol of gaseous reactants. Decreasing the pressure will have no effect on the equilibrium composition because both sides of the balanced chemical equation have the same number of moles of gas.
The new reaction quotient is as follows:. Three moles of gaseous products are formed from 2 mol of gaseous reactants. Decreasing the pressure will favor the side that contains more moles of gas, so the reaction will shift toward the products to increase the pressure. Under the new reaction conditions the reaction quotient is as follows:.
For each equilibrium system, write a new reaction quotient for the system if the pressure is increased by a factor of 2 i. In all the cases we have considered so far, the magnitude of the equilibrium constant, K or K p , was constant. In this case, the system is no longer at equilibrium; the composition of the system will change until Q equals K at the new temperature.
We can express these changes in the following way:. Thus heat can be thought of as a product in an exothermic reaction and as a reactant in an endothermic reaction. Increasing the temperature of a system corresponds to adding heat.
Conversely, an endothermic reaction will shift to the right toward the products if the temperature of the system is increased. This reaction can be written as follows:. Increasing the temperature adding heat to the system is a stress that will drive the reaction to the right, as illustrated in Figure Thus increasing the temperature increases the ratio of NO 2 to N 2 O 4 at equilibrium, which increases K. The effect of increasing the temperature on a system at equilibrium can be summarized as follows: increasing the temperature increases the magnitude of the equilibrium constant for an endothermic reaction, decreases the equilibrium constant for an exothermic reaction, and has no effect on the equilibrium constant for a thermally neutral reaction.
Table The values of both K and K p decrease dramatically with increasing temperature, as predicted for an exothermic reaction. Increasing the temperature causes endothermic reactions to favor products and exothermic reactions to favor reactants.
For each equilibrium reaction, predict the effect of decreasing the temperature:. Asked for: effects of decreasing temperature. The formation of NH 3 is exothermic, so we can view heat as one of the products:. If the temperature of the mixture is decreased, heat one of the products is being removed from the system, which causes the equilibrium to shift to the right. Hence the formation of ammonia is favored at lower temperatures. The decomposition of calcium carbonate is endothermic, so heat can be viewed as one of the reactants:.
If the temperature of the mixture is decreased, heat one of the reactants is being removed from the system, which causes the equilibrium to shift to the left. Hence the thermal decomposition of calcium carbonate is less favored at lower temperatures.
For each equilibrium system, predict the effect of increasing the temperature on the reaction mixture:. Three types of stresses can alter the composition of an equilibrium system: adding or removing reactants or products, changing the total pressure or volume, and changing the temperature of the system. A reaction with an unfavorable equilibrium constant can be driven to completion by continually removing one of the products of the reaction.
Equilibriums that contain different numbers of gaseous reactant and product molecules are sensitive to changes in volume or pressure; higher pressures favor the side with fewer gaseous molecules. Removing heat from an exothermic reaction favors the formation of products, whereas removing heat from an endothermic reaction favors the formation of reactants.
If an equilibrium reaction is endothermic in the forward direction, what is the expected change in the concentration of each component of the system if the temperature of the reaction is increased? If the temperature is decreased? Would you expect the equilibrium to shift toward the products or reactants with an increase in pressure?
What happens to K? If both temperature and pressure are increased? In each system, predict the effect that the indicated change will have on the specified quantity at equilibrium:. H 2 is removed; what is the effect on P I 2? Br 2 is removed; what is the effect on P NOBr? What effect will the indicated change have on the specified quantity at equilibrium? O 2 is added; what is the effect on P H 2?
Cl 2 is removed; what is the effect on P PCl 5? For each equilibrium reaction, describe how Q and K change when the pressure is increased, the temperature is increased, the volume of the system is increased, and the concentration s of the reactant s is increased.
For each equilibrium reaction, describe how Q and K change when the pressure is decreased, the temperature is increased, the volume of the system is decreased, and the concentration s of the reactant s is increased.
What changes in the values of Q and K would you anticipate when a the volume is doubled, b the pressure is increased by a factor of 2, and c COCl 2 is removed from the system? What happens to the values of Q and K if the reaction temperature is increased?
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