Were given this reaction for the conversion of the hydrated form of any to eso for into its anhydrous form. We're also told the value of the equilibrium constant K p at a temperature of 25 degrees Celsius at the same temperature in part a. We wanted a German with the vapor pressure of water is since we have a value for KP, we can write out the equilibrium expression and we see that H 20 vapor is the only gashes species on the product side of this reaction. So that means that KP is equal to the partial pressure of H 20 to the power of 10. And now, since we know the value of K P, we can solve for the partial pressure of H 20 in the system, which will be equal to the vapour pressure at the temperature at which we know the value of K P. So the partial pressure of H 20 is equal to the 10th root of KP, or K P, to the power of 1/10 and that is equal to the 10th root of the given value of K P 4.8 times 10 to the negative 25. And that comes out to a vapor pressure of water in this system, a 25 degree Celsius to be about three 0.64 times 10 do the negative third atmospheres. And now in part B, we want to use the chandeliers principle in order to determine whether each one of the given disturbances to the system will increase, decrease or have no effect on the ratio of the hydrated form to the anhydrous form of any two s 04 So this is the hydrated form and this is the anhydrous form because the hydrated form has those 10 water is bonded to it. So we start with number one and part B. We're increasing the concentration of any to S 04 solid. This is our equilibrium expression. And we know that KP is only dependent on the partial pressure of of each 20 and so that is. That means that the reaction quotient is also only dependent upon that. So the equilibrium will only shift if we change The partial pressure of H 20 since in a two s set of four is a solid. If we add it to the system that will know it, shift the equilibrium and therefore will have no change on that ratio of the hydrated form to the anhydrous form. Number two, we decrease the volume. On the reactant side of this equation, we have zero moles of gas and on the product side, we have one mole of gas. If we decrease the volume, we know that that means that the pressure was increased and so the system will respond by decreasing the pressure. The way that we do that is by shifting to the left to decrease the total number of moles of gas. And so that means that more reactant swill be formed them products and the reactive has the hydrated form and the product side has the anhydrous form. So that means that the ratio of the hydrated form to the anhydrous form will increase. So we increase at ratio when we decrease. Of all you number three, we increase the amount of H 20 gas. So that means that we increase the partial pressure of of H 20 And so when we do that, when we increase that value and research to the power of 10 then the result for the reaction quotient is greater than the equilibrium. Constant. So cute is greater than K, and so that means that the reaction will respond by shifting to the left to reestablish equilibrium. So when that's the case again, just like in number two for Part B, we have more reactant than products, so we have more of the hydrated form of any to S 04 than the anhydrous forms. So that ratio increases again. And number four, we add, and to gas and to does not appear anywhere in a reaction system. So it is an inert gas. Since it is a gas, it will increase the pressure of the system. But to the equilibrium is only affected by the partial pressure of of water. And we did not change that by by adding an inert gas. We only increased the total pressure, so that will have no change on that ratio of the hydrated form to the anhydrous form, because we did not change the equilibrium by adding an inert