For the reaction H2+I2, Kc=55.3 at 700K. In a 2. L flask containing an equilibrium mixture of three gases, there are .056g H2 and 4.36 g I2. What is the mass of the I2 in the flask?
To find the mass of I2 in the flask, we need to use the equilibrium constant (Kc) and the initial amounts of H2 and I2 in the mixture.
First, let's find the number of moles of H2 and I2 in the mixture. We can use the molar mass of each compound to convert the given masses to moles.
Molar mass of H2 = 2.02 g/mol
Molar mass of I2 = 253.8 g/mol
Number of moles of H2 = Mass of H2 / Molar mass of H2
= 0.056 g / 2.02 g/mol
= 0.0277 mol
Number of moles of I2 = Mass of I2 / Molar mass of I2
= 4.36 g / 253.8 g/mol
= 0.0172 mol
Now, using the equation for the reaction, H2 + I2 ⇌ 2HI, we can write the expression for the equilibrium constant:
Kc = [HI]^2 / ([H2] * [I2])
Since there is no given concentration of HI, we need to calculate it using the given moles of H2 and I2 and the volume of the flask.
Given:
Equilibrium constant (Kc) = 55.3
Volume of the flask = 2 L
[H2] = moles of H2 / Volume of the flask
= 0.0277 mol / 2 L
= 0.01385 mol/L
[I2] = moles of I2 / Volume of the flask
= 0.0172 mol / 2 L
= 0.0086 mol/L
Now we can substitute the calculated values into the equilibrium constant expression:
55.3 = ([HI]^2) / (0.01385 * 0.0086)
[HI]^2 = 55.3 * (0.01385 * 0.0086)
[HI]^2 = 0.06395
[HI] ≈ √(0.06395)
[HI] ≈ 0.2528 mol/L
Now, to find the mass of I2 in the flask, we can use the new concentration of HI and the equilibrium expression:
[HI] = moles of HI / Volume of the flask
= mass of HI / molar mass of HI / Volume of the flask
= (mass of I2 / Molar mass of I2) / Volume of the flask
= (mass of I2 / 253.8 g/mol) / 2 L
Rearranging the equation, we can solve for the mass of I2:
mass of I2 = [HI] * Molar mass of I2 * Volume of the flask
= 0.2528 mol/L * 253.8 g/mol * 2 L
= 128.33 g
Thus, the mass of I2 in the flask is approximately 128.33 grams.