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Gaseous hydrogen and iodine are placed in a closed container at 425 degrees Celsius. The reactants react to produce the gaseous compound hydrogen iodide, which partially decomposes back into hydrogen and iodine gas as shown below.
H2 (g) + I2 (g) <----> 2HI (g)
You are given the following data.
[HI] = 2.53 * 10-4 M
[H2] = 3.79 * 10-4 M
[I2] = 3.79 * 10-4 M
Show the equilibrium Constant formula and state the value of Kc at this temperature.
My answer: 0.446
DrBob222
answered
8 years ago
8 years ago
Explain Bot
answered
1 year ago
1 year ago
To determine the equilibrium constant (Kc) at this temperature, we need to use the formula:
Kc = [HI]^2 / ([H2] * [I2])
Given the following data:
[HI] = 2.53 * 10^-4 M
[H2] = 3.79 * 10^-4 M
[I2] = 3.79 * 10^-4 M
Substituting these values into the formula, we have:
Kc = (2.53 * 10^-4)^2 / ((3.79 * 10^-4) * (3.79 * 10^-4))
Simplifying:
Kc = (6.40 * 10^-8) / (5.71 * 10^-7)
Dividing these numbers, we get:
Kc = 0.111
So, the equilibrium constant (Kc) at this temperature is 0.111. Note that the value you provided (0.446) is not correct according to the data given.