Suppose a 500.mL flask is filled with 1.2mol of H2 and 2.0mol of HI. The following reaction becomes possible:+H2gI2g 2HIgThe equilibrium constant K for this reaction is 2.01 at the temperature of the flask.Calculate the equilibrium molarity of H2. Round your answer to two decimal places.
Question
Suppose a 500.mL flask is filled with 1.2mol of H2 and 2.0mol of HI. The following reaction becomes possible:+H2gI2g 2HIgThe equilibrium constant K for this reaction is 2.01 at the temperature of the flask.Calculate the equilibrium molarity of H2. Round your answer to two decimal places.
Solution
To solve this problem, we first need to understand the reaction and what the equilibrium constant (K) tells us. The reaction is:
H2 + I2 ⇌ 2HI
The equilibrium constant expression for this reaction is:
K = [HI]^2 / ([H2][I2])
We are told that K = 2.01. We are also told that initially, we have 1.2 mol of H2 and 2.0 mol of HI in a 500 mL flask.
First, we need to convert these amounts to molarity (M), which is moles/volume in liters.
[H2] initial = 1.2 mol / 0.5 L = 2.4 M [HI] initial = 2.0 mol / 0.5 L = 4.0 M
Since I2 is not mentioned, we can assume that initially, [I2] = 0.
At equilibrium, the concentrations will change. We can express these changes as:
[H2] = 2.4 - x [HI] = 4.0 + 2x [I2] = x
We can now substitute these into the equilibrium expression and solve for x:
2.01 = (4.0 + 2x)^2 / ((2.4 - x) * x)
This is a quadratic equation in the form of ax^2 + bx + c = 0. Solving this equation will give us the value of x, which represents the change in molarity of the substances as the reaction reaches equilibrium.
Once we have the value of x, we can substitute it back into the expression [H2] = 2.4 - x to find the equilibrium molarity of H2.
Please note that solving the quadratic equation might require the use of the quadratic formula or a calculator with the ability to solve quadratic equations.
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