At 500 °C, hydrogen iodide decomposes according to
2HI(g)↽−−⇀H2(g)+I2(g)
For HI(g) heated to 500 °C in a 1.00 L reaction vessel, chemical analysis determined these concentrations at equilibrium: [H2]=0.359 M , [I2]=0.359 M , and [HI]=3.03 M . If an additional 1.00 mol of HI(g) is introduced into the reaction vessel, what are the equilibrium concentrations after the new equilibrium has been reached?
[HI]=
M
[H2]=
M
[I2]=
M
[HI]= 2.03 M
[H2]= 0.609 M
[I2]= 0.609 M
Wow, that's a lot of HI! Looks like things got a little crowded in the reaction vessel. Maybe it's time to start handing out name tags.
To solve this problem, we need to use the principles of the law of mass action and reactant stoichiometry. Let's break down the steps to find the equilibrium concentrations after the introduction of 1.00 mol of HI(g).
Step 1: Write the balanced chemical equation for the decomposition of hydrogen iodide:
2HI(g) ⇌ H2(g) + I2(g)
Step 2: Determine the initial moles of HI:
Since the volume and concentration of HI are given, we can calculate the initial moles of HI.
n(HI) = [HI] × V = 3.03 M × 1.00 L = 3.03 mol
Step 3: Calculate the moles of H2 and I2 produced from the decomposition of HI:
Since the stoichiometry of the reaction is 2:1:1 (2 moles of HI produces 1 mole of H2 and 1 mole of I2), we can determine the moles of H2 and I2 produced using the initial moles of HI.
n(H2) = 0.359 mol
n(I2) = 0.359 mol
Step 4: Calculate the moles of HI remaining after the decomposition:
Since each mole of HI decomposes to produce 1 mole of H2 and 1 mole of I2, the number of moles of HI remaining is the initial moles minus the moles of H2 and I2 produced.
n(HI) = 3.03 mol - (0.359 mol + 0.359 mol) = 2.312 mol
Step 5: Calculate the total moles of HI after the introduction of 1.00 mol:
Since an additional 1.00 mol of HI is introduced, the total moles of HI becomes the previous moles of HI remaining plus the additional moles introduced.
Total moles of HI = 2.312 mol + 1.00 mol = 3.312 mol
Step 6: Calculate the new equilibrium concentrations using the total moles of HI in the same volume:
Now that we know the total moles of HI, H2, and I2 at equilibrium, we can calculate the equilibrium concentrations using the volume.
[HI] = n(HI) / V = 3.312 mol / 1.00 L = 3.312 M
[H2] = n(H2) / V = 0.359 mol / 1.00 L = 0.359 M
[I2] = n(I2) / V = 0.359 mol / 1.00 L = 0.359 M
Therefore, the equilibrium concentrations after the new equilibrium has been reached are:
[HI] = 3.312 M
[H2] = 0.359 M
[I2] = 0.359 M
To find the equilibrium concentrations after the introduction of an additional 1.00 mol of HI, we need to first calculate the change in the number of moles for each species in the balanced equation.
From the balanced equation: 2HI(g) ⇌ H2(g) + I2(g), we can see that 2 moles of HI produce 1 mole of H2 and 1 mole of I2.
Given that an additional 1.00 mol of HI is introduced, we can expect the following changes in the number of moles for each species:
[H2]: Initially, there is 0.359 M of H2. With the introduction of 1.00 mol of HI, the number of moles of H2 will increase by half of that amount since the stoichiometry ratio is 1:2. So, the change in the number of moles of H2 will be (1.00 mol HI) / 2 = 0.50 mol. Thus, the new number of moles of H2 is 0.359 M + 0.50 mol / 1.00 L = 0.859 M.
[I2]: Initially, there is 0.359 M of I2. Since the stoichiometry ratio is 1:1, the change in the number of moles of I2 will be equal to the change in the number of moles of H2, which is 0.50 mol. Thus, the new number of moles of I2 is 0.359 M + 0.50 mol / 1.00 L = 0.859 M.
[HI]: Initially, there is 3.03 M of HI. With the introduction of 1.00 mol of HI, the number of moles of HI will increase by that amount. So, the new number of moles of HI is 3.03 M + 1.00 mol / 1.00 L = 4.03 M.
Therefore, the equilibrium concentrations after the new equilibrium has been reached are:
[HI] = 4.03 M
[H2] = 0.859 M
[I2] = 0.859 M