When 2.00 mol of SO₂Cl₂ is placed in a 2.00-L flask at 303 K, 56% of the SO₂Cl₂ decomposes to SO₂ and Cl₂: SO₂Cl₂(𝑔) ⇌ SO₂(𝑔) + Cl₂(𝑔) (a) Calculate 𝐾_𝑐 for this reaction at this temperature.

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Hi everyone for this problem. It reads in a 1.50 liter flask, 1.67 moles of hydrogen peroxide is placed at 100 and 56 kelvin. If 78% of hydrogen peroxide decomposes to hydrogen gas and oxygen gas. What is the value of K C. For the reaction? Okay. So what we want to answer it here is the value of K. C. And K C represents our equilibrium constant. And what we express K C in is it's defined by molar concentrations. And so in order to solve this problem, let's start off by writing out our equilibrium expression. Okay, and what that looks like is K C is equal to our molar concentration of products over our molar concentration of reactant. Since and when we take a look at our reaction, we can go ahead and substitute our products and reactant into this. So our concentration of products, we have two products, We have hydrogen gas and we have oxygen gas and we want to make sure that we pay attention to the stoke eo metric coefficient. Because whatever that's Tokyo metric coefficient is, it's going to become an exponent in our equilibrium expression. So here we only have one mole of everything. So we're just going to raise everything to the power of one. And here we won't write that in. Okay, over our concentration of reactant. So our reactant is hydrogen peroxide. Okay, so now that we have our case equilibrium expression written out, we need to figure out what is our value of K C. And the way that we're able to do that is we need to find out our equilibrium concentrations and that requires us to make an ice table. Okay, so by making an ice table we're going to be able to figure out our equilibrium concentrations. So we need to rewrite our reaction. Okay, we're going to draw a line here and then write ice on the side. Our eye represents our initial concentration. Our initial concentration needs to be in malaria T or in moles over leader. Okay. And so in the problem we're told we have 1.67 moles of hydrogen peroxide and we're given the volume of the of the flask. So we can solve for the initial concentration of hydrogen peroxide. So we can input that into our rice table. So let's go ahead and do that. So our our initial concentration of hydrogen peroxide is equal to and remember concentration is moles over leader. So in the problem we're told we have 1.67 moles of hydrogen peroxide over 1. liters. So this gives us an initial concentration of 1.1133 moller of hydrogen peroxide. So this is what we're going to put into our ice table under our initial concentration four hydrogen peroxide. Okay, so we have one point let's do it in a different color. So we have 1.1133. Okay. And we have zero for hydrogen gas and zero for oxygen gas, which means our reaction is going to shift to the left to establish equilibrium. So that means our concentration of reactant is going to decrease and our concentration of products is going to increase. Okay, so R C represents our c row of our ice table represents that change. Okay. And we can figure out what our change in hydrogen peroxide is because we were told in the problem that 78% of the hydrogen peroxide decomposes to hydrogen gas and oxygen gas. Okay. So we can also solve for r change or r zero of the equilibrium of our ice table by doing the following. Okay, so our change in hydrogen peroxide is going to equal Armel arat e our initial concentration. So we said that's 1.1133 times 78%. And so when we convert that to decimal form it becomes 0.78. So that means our change is 0.8684. Okay, so once we do that we can plug now we can plug that in. So we have minus because we're shifting to the right here so our concentration of reactant is decreased so that's represented by minus and our change is 0.8684. Okay, now our change for our hydrogen gas and oxygen gas is going to be positive because we're creating product, our reaction is shifting to the right, so this is going to be plus 0.8684. And our hero of our equilibrium table is going to be the sum of the first two rows. So we have 1. 3 -0.8684. And then 0.8684 for our hydrogen gas and oxygen gas. So now that we know what our equilibrium values are because we just did an ice table, we can plug those values into our K. C. Expression or our equilibrium constant expression up here to solve for the value of K. C. So let's go ahead and plug in. So based off of our table, we set our equilibrium concentration for hydrogen gas is 0.8684. This is multiplied by our equilibrium concentration of oxygen gas which is the same thing. 0.8684. And this is divided by our equilibrium concentration of hydrogen peroxide. So we said if we look at our table it's 1. minus 0.8684. And that value is 0.2449. Okay, so now all we have to do is solve and when we do we get K C Is equal to 3.08. So the value of our equilibrium constant for this reaction is 3.08. That's the end of this problem. I hope this was helpful

When 2.00 mol of SO₂Cl₂ is placed in a 2.00-L flask at 303 K, 56% of the SO₂Cl₂ decomposes to SO₂ and Cl₂: SO₂Cl₂(𝑔) ⇌ SO₂(𝑔) + Cl₂(𝑔) (a) Calculate 𝐾_𝑐 for this reaction at this temperature.

Verified Solution

Key Concepts

Chemical Equilibrium

Equilibrium Constant (Kc)

Molarity and Concentration Calculations

When 2.00 mol of SO2Cl2 is placed in a 2.00-L flask at 303 K, 56% of the SO2Cl2 decomposes to SO2 and Cl2: SO2Cl2(𝑔) ⇌ SO2(𝑔) + Cl2(𝑔) (a) Calculate 𝐾𝑐 for this reaction at this temperature.

Verified Solution

Key Concepts

Chemical Equilibrium

Equilibrium Constant (Kc)

Molarity and Concentration Calculations

When 2.00 mol of SO₂Cl₂ is placed in a 2.00-L flask at 303 K, 56% of the SO₂Cl₂ decomposes to SO₂ and Cl₂: SO₂Cl₂(𝑔) ⇌ SO₂(𝑔) + Cl₂(𝑔) (a) Calculate 𝐾_𝑐 for this reaction at this temperature.