Consider the reaction:
SO2Cl2(g) ⇌ SO2(g) + Cl2(g)
Kc = 2.99 * 10...

Question

Consider the reaction:
SO2Cl2(g) ⇌ SO2(g) + Cl2(g)
Kc = 2.99 * 10-7 at227°C
If a reaction mixture initially contains 0.175 M SO2Cl2, what is the equilibrium concentration of Cl2 at 227 °C?

Accepted Answer

hi everyone for this problem, we're told to consider the following equilibrium reaction at 373 kelvin calculate the equilibrium concentration of C O. If a reaction vessel initially contains 15. moller C O C L two. Okay, so our mission here is to calculate the equilibrium concentration of C. O. Okay. And we can calculate equilibrium concentrations if we're given initial concentrations. And so what we can do now is set up an ice table to do that. So we're going to rewrite out our reaction And that is COCL two. Is that equilibrium? And then we have C. O. Gas plus cl two, gas. Okay, I am going to draw a line here to separate our products from our reactant. Is and I'm going to write out ice on the side. Alright, so our eye represents our initial concentration. So in the problem we're told we have an initial concentration of 15.4 for COCL two. Okay. And we don't know what these are. So we're going to say it is zero. And if that is the case, that means our reaction is going to shift towards the right and it's going to shift towards the right to establish equilibrium. So that means our concentration of products is going to decrease and our concentration of reactant is going to increase for our second row of our ice table. The C represents that change in concentration. So we have our concentration of reactant is going to decrease. So that is represented by the minus and we have one mole of each thing. So this is just going to be minus X. Now our concentration of products is going to increase so we have plus X and plus X. Alright, so R E row represents our equilibrium concentrations, so we're just going to add our first two rows together So we have 15.4 minus X. And then we just have X. Here. So our ice table is complete. Now we can write out our equilibrium expression and that is our K. C is going to equal our concentration of products Over our concentration of reactant. Each race to its Tokyo metric coefficient. So in the problem we're given our KC value right here. So we're going to plug that into the left side of our equation. So our KC is 2.19 times 10 to the negative 10. And our products, we have two products, we have C. O and C. L two. So we're going to look at our equilibrium role for this part. Our concentration of products over reactant. So we have our products is X times X and it's raised to its Tokyo metric coefficient. We only have one of each so we don't put anything And our reactant is COCL two and that is 15. minus X. Okay, so we just transferred what's over in our equilibrium role to this right side of our equation. So now what we can do is we can check to see if this X right here is negligible. If it's negligible, we can go ahead and eliminate it from our equation. But if it's not we need to leave it there. So the way that we check if X is negligible is by dividing our initial concentration by our K. C value. So our initial concentration is 15.4. That was our initial concentration for COCL two and R K C we said is 2.19 times 10 to the negative 10. If this value is greater than 500 that means X is negligible. And if it's not that means X is not negligible. So when we do this, we see that the value is going to be greater than 500. So that means that this X right here is negligible. So we're going to ignore it. All right, So let's write out our equation with that X not there. And we'll simplify. So we have 2.19 times to the negative 10 is equal to X squared over 15.4. So, our goal here is to solve for this X. And when we solve for that X, we can plug it into our equilibrium row to solve for our concentration of C. O. Which is what the problem is asking us for. So let's go ahead and solve and isolate our X squared. So we'll multiply both sides of our equation by 15.4. So we'll get X squared is equal to 2. times 10 to the negative 10 times 15.4. So we'll simplify our right side. And when we do we get 3. times 10 to the negative nine. We want to get rid of that square. So we're going to take the square root of both sides. When we take the square root of both sides are X squared is going to eliminate. So we get X is equal to 5.8074 times 10 to the negative five. So now that we know what X is, we can plug X into our equilibrium row to figure out the concentration of everything in our reaction. But the question is asking us specifically for the equilibrium concentration of C. O. And as you can see here and our table C. O. The equilibrium concentration is represented by X. And so that means that our equilibrium concentration of C. O. Is equal to X. And we said that X is equal to 5.807, four times 10 to the negative five molar. So this is going to be our final answer for the equilibrium concentration of C. O. That's the end of this problem. I hope this was helpful

Consider the reaction: SO2Cl2(g) ⇌ SO2(g) + Cl2(g) Kc = 2.99 * 10-7 at227°C If a reaction mixture initially contains 0.175 M SO2Cl2, what is the equilibrium concentration of Cl2 at 227 °C?

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