For the equilibrium
PH3BCl31s2 Δ PH31g2 + BCl31g2
Kp = 0.052 at 6...

Question

For the equilibrium
PH3BCl31s2 Δ PH31g2 + BCl31g2
Kp = 0.052 at 60 C. 
(b) After 3.00 g of
solid PH3BCl3 is added to a closed 1.500-L vessel at 60 C,
the vessel is charged with 0.0500 g of BCl31g2. What is the
equilibrium concentration of PH3?

Accepted Answer

Hello everyone today. We are being given the following problem and asked to solve for it. So nitrogen nitrogen dioxide decomposed into nitric oxide and oxygen gas. And the following chemical reaction is given to us of that process a 0.6 most sample of nitrogen dioxide was placed in a one liter flask and allowed to reach equilibrium. What is the equilibrium concentration of oxygen gas? If our equilibrium constant is equal to 1.2 times 10 to the negative fifth. So the first thing we wanna do is we want to calculate our case the expression and that is going to have our products, the concentration of our products in the numerator over our concentration of our reactant and the denominator. With our case expressions, we with our Casey expressions, we only pay attention to the gas just forms of these substances no solves or liquids and with that we see that we have two products. We have nitric oxide. We're gonna write in o in brackets there and since we have a two coefficient for that, we're going to make that expanded to, we're going to multiply the other product molecular oxygen or oxygen gas. And then for our reactant we only have one, it's going to be nitrogen dioxide and we're gonna write that in brackets. And so that's our K. C. Expression. Next. We're going to construct our ice chart. So before we do that we just want to write our chemical equation over again and then we're going to write I C. E. The I stands for the initial concentrations. The C stands for the change in concentrations and the is just the equilibrium. So our initial concentration of our nitrogen dioxide as given to us in the question stem is going to be 0.6 molar. And so how did we get that? So malaria t is simply moles over liters. The question stem gave us 0.6 moles and it also gave us one leader And that would give us our .6 moles. Of course our initial concentrations for the products will be zero each. Since we are taking away from our reactive we're gonna denote that by the change by being negative X. And positive X for the products. And as noted before for equilibrium, we're going to just take I and subtract it by our seed, the equilibrium. So we're gonna have 0.6 minus X. For the first one, nitrogen dioxide for our first product, we're gonna have to X since we have two of those nitric oxide molecules. And we're simply just gonna have an X. For oxygen. And so working this into our expression, we want to check and see if this X value is negligible or not. And so to do that, we're gonna take our Casey expression which would be equal to our 0.6 Divided by our 1.2 Times 10 to the -5. And it wasn't mentioned before, but we have our concentration of nitrogen dioxide over RKC RKc constant. And so what we saw for that, we see that we have a magnitude of 55,000 to 500. And therefore, since the value is so insignificant, we can say that X is ignored. So when we write out our K. C equation, we take our K. C value, which is 1.2 times 10 to the negative fifth. And we're going to equal that too. Our concentrations of our variables that we just found. And so in the numerator, we're going to have to X squared times X for our products here. And they were going to place that over our 0.6 for our reacted there. This equation can simplify itself to 7.2 times 10 to the negative six is equal to four X. Cute solving for X, which is also the concentration for our oxygen at equilibrium, we get a value of 0.10 moller as our answer. I hope this helped. And until next time.

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