A mixture of 0.2000 mol of CO2, 0.1000 mol of H2, and
0.1600 mol ...

Question

A mixture of 0.2000 mol of CO2, 0.1000 mol of H2, and
0.1600 mol of H2O is placed in a 2.000-L vessel. The following
equilibrium is established at 500 K:
CO21g2 + H21g2 Δ CO1g2 + H2O1g2
(d) Calculate Kc for the reaction.

Accepted Answer

hi everyone for this problem. It reads in a 1.50 liter vessel 0.150 moles of hydrogen gas 0.190 moles of nitric oxide and 0. to zero moles of nitrogen gas were initially mixed. The equilibrium for the reaction was reached at 435 kelvin. What is the K. C. Or the equilibrium constant for the reaction? If the partial pressure of of nitrogen gas is 3.81 atmospheric pressure at equilibrium. Okay, so our goal for this question is to answer, What is the value of Casey? Okay. And K. C is our equilibrium constant. And it's defined by molar concentrations. And here we're dealing with gasses. Um so there is an equation that relates R K C two K P where K P is our equilibrium constant as well. But it's defined by partial pressures of the gasses. And as you can see here, they gave us the partial pressure of nitrogen gas at equilibrium. So that equation that relates K P and K C. Is K. C. R, K p is equal to K. C. Times are t raised to delta N. And we'll define these variables right now. So our goal is to find the value of K C. So that means we're going to need to rearrange this equation so that we can isolate our K. C. So when we rearrange it, we get K C is equal to K P over R T raised to the delta N. So K P we said, is our Equilibrium constant. That is defined by partial pressures. This is a value we don't have. And so we're going to need to solve for it, our is our universal gas constant. And that is the number we should have memorized. That is 0. leaders times atmospheric pressure over mole times kelvin, delta N. Is equal to our change and moles. So our moles of product our final moles minus our initial moles. So looking at our reaction up above, we need to take a look at how many moles we have in total. So on our product side we have three moles total. And on our reactant side we have four moles total. So we're going to take our final minus our initial. So we have three minus four equals negative one. So that is our delta N. And we said, K. P is a question mark. We don't know what that value is. So let's go ahead and solve for K. P. The way that we can solve for K. P. Is by setting up an ice table. Because we're given the equilibrium partial pressure for our nitrogen gas. We can set up an ice table to figure out what is our equilibrium partial pressures for everything. So, let's start off our ice table by rewriting out our reaction. So let's take a minute to do that. Okay. Mhm. All right, So, we can move this up a little bit and we're going to create an ice table. Like we said, we'll ride out ice on the right. And I like to draw this line here to separate the products from the reactant. So the ira of our ice table represents our initial partial pressures. Okay. And so since we're trying to solve for K P. Here, our initial partial pressures are going to go in this place, but we don't know what those are. So we're going to need to solve for it based off of what was given in the problem. And the equation that we can use to find out our individual partial pressures is our ideal gas law and that's PV equals N R T. Okay, so, we want to solve for P R pressure. So if we isolate this, we get P is equal to N R T over V. And we have all of this information because in the problem we are given the amount of moles, we know what our our gas constant is. T represents our temperature in kelvin and V represents our volume. So we know what these values are. So let's go ahead and solve for our individual pressure. So we can put them into our iroh of our ice table. Okay, so let's go ahead and start off with hydrogen gas. So our partial pressure of hydrogen gas is going to be equal. We said N R T over V. So let's just plug in. Our values were told, we have 0.150 mol RR is 0. 206 leader times atmosphere over More Times Kelvin and our temperature is 435 Kelvin. So once we solve this, we get a partial pressure for hydrogen gas equal to 3.56961. Okay, so let's go ahead and plug that into the table for hydrogen gas. And we're going to do this for every each one. So we'll plug in as we go. So we have 3.56961. Okay, so let's move forward with our partial pressure for our nitric oxide. Okay, so are partial pressure. All right to here because this is the second one we're solving for. So our partial pressure of nitric oxide is equal to NRT over V. Okay, so we're told the molds for nitric oxide is 0.190 moles. And everything else is going to be the same. Okay, this is all over For the first one. We should have this all over our volume of 1.50 L. Okay, And we can go ahead and make sure our units cancel properly. So the partial pressure for hydrogen gasses is correct. But let's just make sure our units cancel our moles cancel here. Are leaders cancel on our kelvin's cancel. So you see our unit is atmospheric pressure. Okay, So for nitric oxide, we have divided by our volume so we get a partial pressure of nitric oxide equal to 4.5-15 atmospheric pressure. And again our units will cancel the same way they canceled in solving for hydrogen gas. Okay, Alright, so we have that value. So let's go ahead and plug it into our ice table. So we have 4.5215. And the last value they gave us is our partial pressure of nitrogen gas. Okay, so let's go ahead and plug in for that. Okay, so we're told the number of moles is 0.1-0 for that and everything else will be the same. So only our amount of moles changed. Okay, Alright, so now we have a partial pressure for nitrogen gas equal to 2. atmospheric pressure. Okay, so let's go ahead and plug that value into our nice table. So we have 2.8557. And this for our water, this is going to be zero. Okay, so now our next row of our equilibrium or our ice table is our change row. And so we need to pay attention to our stoke e a metric coefficients here and our reaction is going to be moving towards the right to establish equilibrium because as you can see we have a zero there for water. And so that means if if our system is shifting in the forward direction or to the right, that means our concentration of reactant is going to decrease and our concentration of products will increase. So for our reactant swill have a negative sign. And we're taking a look at the number of moles are spokeo metric coefficient. So for hydrogen gas is going to be minus two X. And for a nitric oxide it's going to be minus two X. Because our products are increasing, we're going to have a positive over here. So this is going to be plus two X and this is going to be X. Okay, let me write this two X. Just a little clearer. So that is a two here and one key piece of information we were told. And the problem is that the The partial pressure for nitrogen gas at Equilibrium is 3.81. So our hero of our ice table represents our equilibrium row. And so because we're told it's 3.81, we can just go ahead and plug that 3.81 here. Okay, so that's what this that means now, is we can solve for X. Because we know what the equilibrium the equilibrium partial pressure is for nitrogen gas. So we can set our nitrogen gas row of our ice table equal to each other to solve for X. Okay, so our partial pressure Of nitrogen gas is equal to we're looking at our nitrogen gas uh column of our ice table. So we have 2. 2.857. Excuse me. 2.8557 Plus X is equal to 3.81. Okay, so that means we have what we need to solve for X. So that means X is equal to 0.954312. Okay. And this is all an atmospheric pressure. So now that we know what the value is for X. We can solve for the equilibrium row of our ice table for everything. Okay, so let's just bring down everything for our equilibrium rose. So it's just a combination of the first two rows. So for our hydrogen or for our water that's in its gas form, our equilibrium rose two X for our nitric oxide, it's going to be 4.5 to 15 minus two X. And then four hydrogen gas, it's 3.56961 minus two X. So like we said, we know what the value of X is. So we'll just plug it in for everything in our equilibrium realm. And that will allow us to solve for R. K. P. Which is our missing value that we're looking for to plug into our equation up above. Okay, so R K P is going to have an expression where we can plug in those values from our equilibrium row. And that expression is R K P is equal to the partial pressure of our products over the partial pressure of our reactant. Each race to its Tokyo metric coefficient. So we have two products. So we have our partial pressure of hydrogen or of water and gas form. And it has an exponent or it has a coefficient of two. So that means it gets an exponent of two. And this is going to be multiplied by the partial pressure of our nitrogen gas. There's only one mole of that. So we raise it to the power of one and we can not write that in here. It's not necessary. And then we're going to that's going to be over our partial pressure of our reactant. Okay, so we have our partial pressure of hydrogen gas squared because it's Tokyo metric coefficient is two times our partial pressure of nitric oxide squared. Okay, so let's go ahead and do this. All right. So for a partial pressure of water and its gas form, our equilibrium of the ice table tells us it's two X. Okay, so, we know what X is. So we know it's that's what we saw. 490.954312. So, this gives us a partial pressure of 1.9086248 atmospheric pressure. And remember in the problem they gave us the partial pressure of nitrogen gas equilibrium already, that was 3.81. Okay. Our partial pressure for nitric oxide based off the equilibrium row of our ice table is equal to 4.5 to 15 atmospheric pressure minus. Or we can Yeah. So it's 4.5 to 15 minus two X. Okay, so let's go ahead and plug in X. Here. So we have 4.5 to 15 minus two times 0.954312. Okay, so this gives us a partial pressure of 2.612876 atmospheric pressure for nitric oxide. And lastly we have our hydrogen gas. So the equilibrium rover ice table tells us it's 3.56961 minus two X. So we'll just go ahead and do the same thing and plug in X. Okay, so we get a value of 1.6609868 p. M. So this is what we're going to plug into this K. P. Expression here. So let's go ahead and do that. So we get and remember we need to raise everything to the correct exponent and power. So we're just plugging in and this is squared times 3.81. Okay, over our pressure pressure of our reactant so our hydrogen gas is squared and our nitric oxide is also squared. Okay, so let's go ahead and salt finish this off so we can get our value for K P. So we get K P. Is equal to 0.736879. Okay, so this was the value we were missing to plug into our equation above that relates K P. Two K C. Okay, so now that we have our value for K P we can finish this off because we already know what are our is we know what our change in moles is and we just solved for K P. Okay, so we can erase this question mark and put K. P. Based off of what we just saw for is equal to 0.736879. Alright, so we're in the home stretch here. We just need to plug in and solve. Alright, so we have K P. Is equal to 0. divided by r gas constant. R So 0.8206 Leaders times atmospheric pressure over multi times kelvin. This is going to be multiplied by our temperature in kelvin which was given in the problem. Okay. And raise to our change in moles are delta N. And we said that value is negative one. Alright, so let's go ahead and solve. And when we do we get K C is equal to 26.3037. So let's go ahead and write that over here. So KC is equal to 26.3. This is our final answer. And that is the end of this problem. I hope this was helpful

A mixture of 0.2000 mol of CO2, 0.1000 mol of H2, and 0.1600 mol of H2O is placed in a 2.000-L vessel. The following equilibrium is established at 500 K: CO21g2 + H21g2 Δ CO1g2 + H2O1g2 (d) Calculate Kc for the reaction.

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