Consider the following equation :If volume of the container was decreased, what would happen tothe equilibrium? What would happen to each of the reactants andproduc...

Each reaction is allowed to come to equilibrium, and then the
volume is changed as indicated. Predict the effect (shift right,
shift left, or no effect) of the indicated volume change.
a. I2(g)2 I(g) (volume is increased)
b. 2 H2S(g)2 H2(g) + S2(g) (volume is decreased)
c. I2(g) + Cl2(g)2 ICl(g) (volume is decreased)

They were going to be discussing the shuttlers principle again in the context of chemical equilibrium. So this is a principle that tells us when a chemical system is my guitar equilibrium on it is then disturbed. The system will shift to a direction in order to minimize that disturbance on maintain that position of equilibrium prior to the disturbance. So the equation we're looking at here we have CEO at H 20 is an equally opium with CO two on H two, where H. DeWitt was in steam for me, concedes in the gas state. So we will consider it as part more equilibrium, whereas if it was a liquid that we would neglect it. So when the reaction volume is increased, depression decreases on nasty number of reactions and products of the same. This does not have an effect on the reaction. That is because we have the same number of moles on both sides of the starting materials on the products we have. Imaginary ones are straight geometric coefficients before every single material in that equation. So our next example we have the reaction volume is decreased in the scenarios of the pressure increases as a result, however, again, because the total number of the acting on products are the same. In this given reaction, we have two moles on the product size two moles on the react inside. It will have no effect on the reaction, nor will it have an effect on the equilibrium position.

They were going to be discussing the shuttlers principle again in the context of chemical equilibrium. So this is a principle that tells us when a chemical system is my guitar equilibrium on it is then disturbed. The system will shift to a direction in order to minimize that disturbance on maintain that position of equilibrium prior to the disturbance. So the equation we're looking at here we have CEO at H 20 is an equally opium with CO two on H two, where H. DeWitt was in steam for me, concedes in the gas state. So we will consider it as part more equilibrium, whereas if it was a liquid that we would neglect it. So when the reaction volume is increased, depression decreases on nasty number of reactions and products of the same. This does not have an effect on the reaction. That is because we have the same number of moles on both sides of the starting materials on the products we have. Imaginary ones are straight geometric coefficients before every single material in that equation. So our next example we have the reaction volume is decreased in the scenarios of the pressure increases as a result, however, again, because the total number of the acting on products are the same. In this given reaction, we have two moles on the product size two moles on the react inside. It will have no effect on the reaction, nor will it have an effect on the equilibrium position.

Just looking again at Le Shatner's principle. So this is a principle that tells us when a chemical system is measured to be out its equilibrium, it can be disturbed. And then so our system will then act in order to shift the direction in order to minimize the disturbance on retained that same equilibrium position that was possessed prior to the disturbance. So we've got a new example equation here. I too fancy l two. There's an equilibrium with I c E o. Well, we have to for the value of the story key metric coefficients. So when the reaction volume is increased, the pressure decreases on the total number off. Reactant on products are the same. So this has no effect on the reaction. So when the reaction volume has decreased, the pressure increases at the total number off reactant and products are the same again. It has no effect on the equilibrium position for this reaction. So that is because we have two miles on the left inside for starting materials, and then we have two moles on the right hand side for products

A single shot list principle again in the context of chemical equilibrium. So this is a principle that tells us when a chemical system is measured at to be an equilibrium is in fact disturbed by an opposing outside force. The system will then shift to a direction and or destroying minimize the disturbance on maintained that same previous under starved equilibrium. So the reaction we're taking a look at here was to age to ask, is an equilibrium with 28 just out as to so the chemical equation has three moles of gas on the right and two moles on the left. So increasingly volume off reaction mixture decreases the pressure, which causes the reaction to shift to the right just again to note that we have three moles of gas and other left two moles of gas on the right. By this, I mean we can count of Boston geometric value. So here we have two over on the left. Then we have to, and that is an imaginary one, and hair that makes three so decreasing the volume off reaction mixture increases the pressure, which causes a reaction equilibrium to shift to the left hand side.