Halogen lamps are ordinary tungsten filament lamps in which the lamp bulb contains a small amount of a halogen (often bromine). At the high temperatures of the lamp, the halogens dissociate and exist as single atoms.
(c) When the WBr41g2 diffuses back toward the filament, it decomposes, depositing tungsten back onto the fila- ment. Show quantitatively that the pressure of WBr4 from part (a) will cause the reaction in part (a) to go in reverse direction at 2800 K. [The pressure of Br1g2 is still 0.010 atm.] Thus, tungsten is continually recycled from the walls of the bulb back to the filament, allow-ing the bulb to last longer and burn brighter.

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Hi everyone for this problem it reads metal Halide lamps are a type of gas discharge lamp that contains tungsten electrodes doped with a metal M. And an additional hey lied material, usually sodium iodide the working temperature of the lamp varies between 1000 to 3000 degrees Celsius. At the working temperature of the lamp, sodium iodide dissociates to produce sodium and iodine atoms in gaseous form. The iodine atoms react with the metal em to produce gaseous M. I. Four. The equilibrium pressure of iodine gas at 12, degrees Celsius within the lamp is 40.24. Atmospheres. Show quantitatively that the pressure of M. I. Four at 12,000 degrees Celsius will cause the reaction to go in the reverse direction at 2500 degrees Celsius. The pressure of the iodine gas is the same. Assume that the assume that for the theoretical medal, the equilibrium constant for the reaction is 210 at 12, degrees Celsius in 15 at 2500 degrees Celsius. Okay, so the question that we want to answer here is or what we want to show is we want to show quantitatively that the pressure of M. I. Four at 1200 degrees Celsius will cause the reaction to go in the reverse direction. So what we're dealing here, what we're dealing with here is lush outliers principle. And what we're going to need to do is we're going to need to compare K two Q. To determine which direction the reaction is going. So in the problem, we're told that we're told that the k p. Okay, the K P. For the reaction is 210 at 12,000 degrees Celsius and 15 at 2500 degrees Celsius. Okay. So what we'll need to remember is when Q is equal to K. That means the reaction is at equilibrium when Q is less than K. That means the reaction shifts in the forward direction to reach equilibrium. So if it's shifting in the forward direction, it's going to shift right? And if Q is greater than K, that means the reaction shifts in the reverse direction to reach equilibrium. If it shifts in the reverse direction then it shifts left. Okay, so let's go ahead and write out our equilibrium constant expression. Okay, and that for this problem is going to be K. P is equal to the concentration of products over the concentration of react ints. Okay, and so here remember for we're looking at our reaction here and we don't include solids in our equilibrium expression. Okay. And so when we write our equilibrium expression, what it becomes is the pressure of the product which is M. I four over the pressure of the reactant, which is I. And we're going to raise this to its Tokyo metric coefficient. So we have four moles of that. So the four now becomes an exponent. Okay, and so that's our equilibrium constant expression and we know that the K P. Is 210 at 1200 degrees Celsius. So let's just go ahead and plug that value in. So we have K. P. Which is 2 10 is going to equal the pressure of AM I four over the pressure of reacting. And we're told in the problem what the pressure is? We're told that the pressure, we're told that the pressure is 0.0 to 40 atmospheres. So that is right here. All right. So we're going to plug that in. So this is going to be over 0. to 40 race to the fourth. Okay, so let's go ahead and simplify this. Okay, so now we have 2 10 is equal to the pressure of our product over. Let's simplify our reactant pressure. This becomes 3. 1776 times 10 to the negative seven. So, we're going to multiply both sides. We're going to isolate Our numerator. OK, so we're going to isolate our numerator by taking And multiplying it by 3. times 10 to the -7. And this is going to equal the partial pressure of AM I four. Okay, so now what we're going to do is solve so what we're gonna get is the partial pressure of M I four is going to equal 6. times 10 to the negative five atmospheres. Okay, So what that means then is at 2500°C. This is what's going to take place. So we're right here at 2500°C. If K is equal to Q. The the reaction is at equilibrium if Q is less than K, it's going to shift left and if Q is greater than K, it's going to shift right. So let's go ahead and solve for Well this was the pressure, this was at 1200°C. Okay, so disregard what I just said up here. Alright, so this should be right here where I just put it. So at 1200°C, the pressure is equal to 6.97 times 10 to the negative fifth atmosphere. So now we're going to calculate it at 2500°C. Okay, so our Q. Then becomes Q. is equal to are our product over our reactant. So we just calculated the partial pressure of our product. So we're just plugging it in here. Okay, So we have 6.97 times 10 to the negative five atmospheres over Our reactant. zero to the fourth. So, Q. When we solve is equal to 210. All right, so as we'll see here, Okay, Q. At 210 is greater than the K. P. at 15. So, what this means is the reaction shifts in the reverse direction. So it's going to shift to the left to reach equilibrium and this is going to be our final answer. Okay. And this is the reasoning behind it, Since Q is greater than K, the reaction shifts in the reverse direction to reach equilibrium. That is it. For this problem, I hope this was helpful.

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Key Concepts

Chemical Equilibrium

Gibbs Free Energy

Partial Pressure and Dalton's Law