When 1.00 g of gaseous I2 is heated to 1000. K in a 1.00 L sealed container, the resulting equilibrium mixture contains 0.830 g of I2. Calculate Kc for the dissociation equilibrium below.
I2 <--> 2 I
What I did was converted the initial I2 into moles then molarity since it is just over 1 L (.00394 M) then I found the molarity of I by subtracting .830 from 1 and converting first to moles (.00134 M of I) I then set up Kc= [.00134 M of I]^2/ [.00394 M of I2] which i found to be .00045...but it's incorrect. I'm not sure what I am missing...
Well, it seems like you're missing something in your calculation. Let's go through it together and see if we can pinpoint the issue.
To calculate the equilibrium constant, Kc, we need to use the concentrations of the reactants and products at equilibrium. You correctly found the concentration of I2 to be 0.00394 M. However, the concentration of I is not simply the difference between 1 g and 0.830 g.
To calculate the concentration of I, you need to consider the stoichiometry of the reaction. From the balanced equation, we can see that 1 mole of I2 dissociates into 2 moles of I. Therefore, the moles of I formed would be twice the moles of I2 consumed.
Let's calculate the moles of I2 first. Since we have 1 g of I2 and its molar mass is 253.8 g/mol, we have:
moles of I2 = mass of I2 / molar mass of I2
= 1 g / 253.8 g/mol
= 0.00394 mol
Now, we can calculate the moles of I formed:
moles of I = 2 x moles of I2
= 2 x 0.00394 mol
= 0.00788 mol
To find the concentration of I, we divide moles by the volume in liters:
concentration of I = moles of I / volume
= 0.00788 mol / 1 L
= 0.00788 M
Now we have all the necessary values to calculate Kc:
Kc = [concentration of I]^2 / [concentration of I2]
= (0.00788 M)^2 / (0.00394 M)
≈ 0.0157
So, the correct value for Kc is approximately 0.0157. Keep in mind that I rounded the answer to two decimal places.
It's always important to double-check calculations and consider the stoichiometry of the reaction. Chemistry can be tricky, but don't worry, I'm here to help!
To solve this problem, you correctly converted the initial amount of I2 into moles and then into molarity. However, there seems to be a mistake in your calculation of the molarity of I.
Let's go through the steps to find the correct value for Kc:
1. Convert the mass of I2 to moles:
moles of I2 = 1.00 g / molar mass of I2
The molar mass of I2 is 2 * atomic mass of I = 2 * 126.90 g/mol = 253.80 g/mol
Therefore, moles of I2 = 1.00 g / 253.80 g/mol
2. Convert moles of I2 to molarity using the given volume of the container:
Molarity of I2 = moles of I2 / volume of container = (1.00 g / 253.80 g/mol) / 1.00 L
3. Calculate the moles of I formed:
moles of I = mass of I2 initially - mass of I2 at equilibrium
= 1.00 g - 0.830 g
4. Convert moles of I to molarity using the given volume of the container:
Molarity of I = moles of I / volume of container = (1.00 g - 0.830 g) / 1.00 L
Now, substitute these values into the expression for Kc:
Kc = [Molarity of I]^2 / [Molarity of I2]
Plug in the calculated values for Molarity of I and Molarity of I2 to find the correct value for Kc.
To solve this problem, you need to consider the stoichiometry of the reaction and use the given information.
First, let's calculate the initial moles of I2. We are given that there is 1.00 g of I2. The molar mass of I2 is 253.80 g/mol, so we can calculate the moles of I2:
moles of I2 = mass of I2 / molar mass of I2 = 1.00 g / 253.80 g/mol = 0.003941 mol
Next, we need to calculate the moles of I in the equilibrium mixture. We are given that there is 0.830 g of I2 in the mixture. The molar mass of I2 is 253.80 g/mol, so we can calculate the moles of I2:
moles of I2 = mass of I2 / molar mass of I2 = 0.830 g / 253.80 g/mol = 0.003269 mol
Now, let's calculate the moles of I in the equilibrium mixture. Since the stoichiometry of the reaction is 1:2 (1 mol of I2 produces 2 moles of I), we can calculate the moles of I:
moles of I = 2 * moles of I2 = 2 * 0.003269 mol = 0.006538 mol
To calculate the molarity, we need to divide the moles of each substance by the volume of the container, which is 1.00 L.
Molarity of I2: moles of I2 / volume of container = 0.003269 mol / 1.00 L = 0.003269 M
Molarity of I: moles of I / volume of container = 0.006538 mol / 1.00 L = 0.006538 M
Now we can calculate the equilibrium constant Kc using these molarities. Remember that Kc is calculated using the concentrations (molarities) of the products over the concentrations of the reactants, each raised to the power of their respective stoichiometric coefficients.
Kc = [I]^2 / [I2] = (0.006538 M)^2 / (0.003269 M) = 0.013180 M^2 / 0.003269 M = 4.03 M
Therefore, the equilibrium constant Kc for the dissociation equilibrium is 4.03 M.