For the reaction I2 + Br2 <-> 2IBr Kc=280 at 150C . Suppose that 0.500 mol IBr in a 1.00-L flask is allowed to reach equilibrium at 150C.
What is the equilibrium concentration of IBr
What is the equilibrium concentration of I2
What is the equilibrium concentration of Br2
(IBr) = 0.5 mol/K = 0.5M
...........I2 + Br2 ==> 2IBr
I..........0.....0.......0.5
C..........x.....x........-2x
E..........x.....x......0.5-2x
Write the Kc expression, substitute the E line, solve for x and evaluate 0.5-2x and x. Post your work if you get stuck.
To find the equilibrium concentrations of IBr, I2, and Br2, we will need to use the information given about the equilibrium constant (Kc) and the initial concentration of IBr. Here's how we can approach this problem:
Step 1: Write the balanced equation for the reaction:
I2 + Br2 ↔ 2IBr
Step 2: Define the initial concentrations:
The problem states that 0.500 mol of IBr is present in a 1.00-L flask. This means the initial concentration of IBr is 0.500 M (0.500 mol / 1.00 L = 0.500 M). Since there is no initial concentration provided for I2 and Br2, we will assume they are both zero.
Step 3: Set up the equilibrium expression:
The equilibrium expression for this reaction is:
Kc = [IBr]^2 / ([I2] * [Br2])
Step 4: Determine the equilibrium concentrations:
Let's assume that 'x' mol/L of IBr reacts to form I2 and Br2. As a result, the concentrations at equilibrium would be:
[IBr] = 0.500 M - 2x (since 2 mol of IBr is consumed to produce 1 mol of I2 and 1 mol of Br2)
[I2] = x M
[Br2] = x M
Step 5: Plug the equilibrium concentrations into the equilibrium expression:
Kc = (0.500 M - 2x)^2 / (x * x)
Step 6: Solve for 'x' using the quadratic equation or approximation:
Substitute the given value of Kc (280) and solve the equation for 'x'. Once you find the value of 'x', you can substitute it back into the expressions for [IBr], [I2], and [Br2] to obtain the equilibrium concentrations.
Note: The equilibrium constant, Kc, is given as 280 at 150°C, which indicates a significant amount of reactivity in both forward and reverse directions.
To answer these questions, we can use the concept of stoichiometry and the given equilibrium constant (Kc) value.
1. Equilibrium concentration of IBr:
Since the balanced equation for the reaction is I2 + Br2 ⇌ 2IBr, we know that for every 2 moles of IBr formed, 1 mole of I2 and 1 mole of Br2 react.
Let's assume that the equilibrium concentration of IBr is x M. Since the stoichiometry shows that 1 mole of I2 reacts to form 2 moles of IBr, the equilibrium concentration of I2 will be (0.500 - 2x) M.
2. Equilibrium concentration of I2:
Using the information from the previous step, we know that the equilibrium concentration of I2 is (0.500 - 2x) M.
3. Equilibrium concentration of Br2:
Again, referring to the stoichiometry, the equilibrium concentration of Br2 will be (0.500 - x) M.
Note: The values of x can be calculated using the equilibrium expression and the given equilibrium constant (Kc). However, the given question does not provide any additional information on the initial concentrations or the reaction conditions, such as pressure or volume. Therefore, we cannot determine the exact values of these concentrations without more information.
Please provide any additional information if available, and I will be happy to help you calculate the equilibrium concentrations of IBr, I2, and Br2.