for the reaction H2+I2--2HI at a given temperature it was found that an equilibrium mixture in a 10mL container consisted of 8.0 mol of HI, 1.0 mol of H2 nd 1.0 mole of I2. what is the equilibrium constant at that temperature?
M HI = mols/L = ?
M H2 = mols/L = ?
M I2 = mols/L = ?
Then substitute the M of each into the Keq expression and solve for K.
To determine the equilibrium constant (K) for the given reaction, you can use the molar concentrations of the reactants and products at equilibrium. The balanced equation for the reaction H2 + I2 ⇌ 2HI gives us the stoichiometric coefficients, which tell us the ratio of moles between the reactants and products.
In this case, the initial concentrations are not provided, but we are given the mole amounts in the equilibrium mixture:
[HI] = 8.0 mol
[H2] = 1.0 mol
[I2] = 1.0 mol
Since the reaction is in a 10 mL container, we need to convert these mole amounts to molar concentrations. Molar concentration (C) is expressed as moles per liter (mol/L).
To calculate the molar concentration of a substance in solution, you need to divide the number of moles by the volume of the solution in liters.
First, convert the volume of the container from mL to L:
10 mL = 10/1000 L = 0.01 L
Now, calculate the molar concentrations:
[HI] = 8.0 mol / 0.01 L = 800 mol/L
[H2] = 1.0 mol / 0.01 L = 100 mol/L
[I2] = 1.0 mol / 0.01 L = 100 mol/L
Now, substitute the molar concentrations into the equilibrium constant expression:
K = ([HI]^2) / ([H2] * [I2])
= (800 mol/L)^2 / (100 mol/L * 100 mol/L)
= 640,000 mol^2/L^2 / 10,000 mol^2/L^2
= 64
Therefore, the equilibrium constant (K) at the given temperature is 64.