Which of the following expressions is the correct equilibrium-constant expression for the equilibrium between dinitrogen tetroxide and nitrogen dioxide?N2O4(g) = 2N...

Which of the following correctly relates the equilibrium constants for the two reactions shown? $A+B \rightleftarrows 2 C \quad K_{1}$ $2 \mathrm{A}+2 \mathrm{B} \rightleftarrows 4 \mathrm{C} \quad K_{2}$ (a) $K_{2}=2 K_{1}$ (c) $K_{2}=1 / K_{1}$ (b) $K_{2}=K_{1}^{2}$ (d) $K_{2}=1 / K_{1}^{2}$

For equilibrium, heterogeneous equilibrium that include a pure liquid or pure solid. The pure liquids and pure solids are excluded from the equilibrium expression. So for the first reaction, we do see that N B. R. Two is a solid, so it should be excluded from the expression, although it is included in the expression in the question. So we need to remove it in order to have the correct equilibrium expression, remembering that pure solids are excluded for the next one we have See you, oh, solid, that should be excluded and then we have a product. See you liquid that should be excluded. However, in the equilibrium expression they do include to see you liquid. But pure liquids should be excluded. So the correct equilibrium expression would be this with the removal of the copper liquid, because pure liquids and pierre solids should be excluded from the equilibrium expression. The last one is very interesting because we only have one product and it is a solid. So what are we going to include in the numerator? If there is only one product and it does appear solid, we can't include that concentration, and we can't put zero because an equilibrium. Constant is never zero. So what we do is we put one. If there is nothing to include, we put one. So the only thing to include as a reactant which would be down in the denominator, is the oxygen gas that we would raise to the third power.

Okay, so we have the following equilibrium expressions and they want us to relate K one k two. So let's talk about the equations first. So we see that K one is literally k tube of flipped. Right? So we see that no CEO is the reactant and K one, but we see that NLC Elysa product in K two and we see that N o N c l two is the product and Tae won an n o N c l two is You reacted in k to buy. So what we can dio Here's looking flip k one around, right? So if we were to flip it around, we will get n o gas plus 1/2 CEO to gas. We'll get it an o c l gas. Right? But if we flip an equation around disputes that we had to take the inverse of the K right, So this means that the new K of this expression is one over K one, right? The inverse is just one over devalue. Now let's compare that. We don't care about this one anymore. Now let's compare the two equations we have, right? We see that K two is literally double K one. Right. So what we can do is weaken double dis expression down here. Let's double it is Multiply everything by two. So you have. So we'll get to an O plus two times. 1/2 is you go to one. So we get C l to write gas and then we get to and O c l s right. When we double an equation, DK value is squared. Right? So let's say we square this whole expression right here. We get one square over K one square, one squares, just one. So we have one over K one square, and then let's look at these. We see that these two are literally the same exact equations. So how did a K values relate to each other? But we see that K to physical 21 over K one square, right? And I believe that is choice. See? So this is our answer right here.

Okay, So for this question, they want us to relate the equilibrium constants of thes two equations. Right? So the first thing that we should do is we should write the equally room constant the expression for the equilibrium constant for both of the equations first. So it's to K one first so we know that equilibrium costing is the product over the reactant. Right? So we have the concentration of the product, which is just see, but there's a two in France. We must square it over the concentration of the reactant because we have to. Now it's like a two. We have the products which is see to the fourth power because of the four when we have both A and B to D second power because to to Now let's compare to to right. So if we say we were to square K one right, the whole entire expression were to square K one. Would that be ok? One square will be C two squares for and then this square right? And we would have be square too. But we see that this expression is exact same thing sk to this means of the relationship between K one and K two is that K one is a square Ruoff K two or K? Too busy square of K one. Right? And if we look at the choicest, it looks like that is choice. Be choice. Be right. So that is the answer.

Okay, So this question we have to chemical equations. I've rewritten the have used different variables for this question. But the variable like a or acts does This isn't really matter. So let's find the equilibrium. Constant equation for K one. So we know that the equilibrium, uh, constant is calculated by taking the product, the Constitution of the product over the concentration of the reactions. Right. So this w here has a two in France. We must square it and then we have concentration of x times concentration of why? Now it's two k two. So we have to product over the reactant. So we have product, which is W to the fourth power because of the four over X Square over times of my wife square. Right now, let's compare the two. We see that if we were just square, let's say we square K one right. We square this whole expression. What would it be? Right. So we have appeared w square and every square That'll be W to the 4th May. We went to square X. It'll be X square, and we would scare Why will be we have by square life. Now let's look atyou we said these two are the exact same things, right? So this means that K one square is equal to K two or K one is to go to the square root of K two, right? Cause square root and squaring are the inverse of each other. So this is our answer.