suppose-a-reaction-has-the-equilibrium-constant-k-17-times-10-8-at-a-particular-temperature-will-the

Question

Suppose a reaction has the equilibrium constant $K=$ $1.7 	imes 10^{-8}$ at a particular temperature. Will there be a large or small amount of unreacted starting material present when this reaction reaches equilibrium? Is this reaction likely to be a good source of products at this temperature?

Ifeoluwa Bolujo · Accepted Answer

so looking at this, where we still going to interpret what and its equilibrium? Constant Told you. If you&#x27;re caving in constant, which is K is greater than one, then your equation will probably yield mostly products. But if your equilibrium constants less than one, then your reaction will. Civilians reacted. So given, in example, we have K and the value of K and 1.7 times tender than it did meet. This is significantly less than one, so they yield mostly reactive. So with this, the reaction would probably be a good source for reactivates and not rocks.

Ronald Prasad · Answer

if a reaction has an equilibrium constant that is equal to 1.3 times 10 to the eighth. Uh, that suggests, uh, that the concentration of the products will be favored the definition of an equilibrium constant. It&#x27;s equal to the product concentration over the reacting concentration on if this fraction is greater than one that would favor the products. And since in this question here were definitely greater than one, we&#x27;re favoring the products, and this would be definitely a good source for the products if we

David Collins · Answer

the degree to which a reaction occurs can be expressed by its equilibrium. Constant assuming in equilibrium occurs equilibrium constant is equal to the concentration of the products over the concentration of the react ends. Now this is only true so long as the products I&#x27;m sorry, the products and the reactant are either gasses or salutes in solution. We will assume that that is the case. So if we have a very large equilibrium constant 1.3 times 10 to the eight, then that means we must have a large amount of product and a small amount of reactant. In orderto have this ratio be a large number. Therefore, this reaction that has a very large equilibrium constant producing a large amount of product relative to a small amount of reactant would definitely be a good source of product. So long as we could make sure that the equilibrium or the conditions that established this equilibrium are met. You will find out later that this equilibrium constant can change is a function of temperature. For example,

Ifeoluwa Bolujo · Answer

so he at this. It shows the equilibrium constant. Basically, you have a value for that constant. What doesn&#x27;t need so sorry? Law, Your equilibrium constant is greater than one. You&#x27;ll probably have mostly products in reactors, but then, if it&#x27;s less than one that you have mostly reactors. So given, for example, in equilibrium constant ah, one quote degree turns tend to the E. This value is significantly higher than one. So if that&#x27;s significantly higher than one, that it&#x27;s actually going to be mostly products. And this reaction with likely be a good source for if you if you want more products.

Ronald Prasad · Answer

if a reaction has an equilibrium constant that is equal to 1.3 times 10 to the eighth. Uh, that suggests, uh, that the concentration of the products will be favored the definition of an equilibrium constant. It&#x27;s equal to the product concentration over the reacting concentration on if this fraction is greater than one that would favor the products. And since in this question here, we&#x27;re definitely greater than one, we&#x27;re favoring the products, and this would be definitely a good source for the products if we

Shanie Holman · Answer

in this video, we&#x27;re going to consider what it is that a particular value of K is telling us. And in this case, we&#x27;re talking about 4.5 times 10 to the negative six. The actual value here doesn&#x27;t really matter. The point is that because this is negative six here. We&#x27;re talking about a very, very, very small. Okay, So what does a small K tell us to consider that we need to think about what the equilibrium expression is. And remember that its products concentration of products over the concentration of the reactant since and because our what it works out to be is very small. That means that this number has to be fairly low. And this number has to be fairly high toe low and high because that&#x27;s just how division works, right? If you&#x27;ve got one pie and you need to divide it among a 1,000,000 people, nobody is going to get a very big piece, which is what this is telling us up here. 4.5 Times Center. Think of six. It&#x27;s a very small result. So this this particular, uh, equilibrium constant would not be useful to make product because you get so little, um, so little product relative to the amount of reactant is you need to use. So to be able to make this useful, you&#x27;re going to have to find a way to shift that equilibrium right towards products. Now, one way you can do that, of course, is to increase reactant. But you may not want to do that necessarily. You may be able to manipulate, um, pressure or temperature, depending on what your reaction is. Oops. Pressure. We&#x27;re chunk something that will let you shift that equilibrium to the right if you want to produce more product.

Problem

For the reaction $$\mathrm{Br}_{2}(g)+5 \mathrm{…

Problem 44 Easy Difficulty

Suppose a reaction has the equilibrium constant $K=$ $1.7 \times 10^{-8}$ at a particular temperature. Will there be a large or small amount of unreacted starting material present when this reaction reaches equilibrium? Is this reaction likely to be a good source of products at this temperature?

Answer

No.

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