At 800 K the equilibrium constant for I2(g) 2 I(g) is Kc = 3.1 10-5. If an equilibrium mixture in a 10.0-L vessel contains 3.56 10-2 g of I(g), how many grams of I2 are in the mixture?
...........I2 ==> 2I
equil mols I = 0.0356g/about 127 = about 2.8E-4. M = mols/L = about 2.8E-5. You need to confirm these numbers.
Kc = (I)^2/(I2)
You know equil (I) and Kc, calculate (I2).
AMDG i need some help bkim and zurak
To find the number of grams of I2 in the mixture, we need to use the equilibrium constant expression and the given information.
The equilibrium constant expression for the reaction I2(g) ⇌ 2 I(g) is given by:
Kc = [I(g)]^2 / [I2(g)]
Now let's substitute the given values into the equation.
Kc = 3.1 × 10^(-5)
[I(g)] = 3.56 × 10^(-2) g
[I2(g)] = ? (the unknown mass of I2)
We can rearrange the equilibrium constant expression to solve for [I2(g)]:
[I2(g)] = [I(g)]^2 / Kc
Plugging in the known values:
[I2(g)] = (3.56 × 10^(-2) g)^2 / (3.1 × 10^(-5))
Simplifying the equation:
[I2(g)] = 12.903 g / (3.1 × 10^(-5))
Calculating the result:
[I2(g)] ≈ 416,548.39 g
Therefore, the number of grams of I2 in the mixture is approximately 416,548.39 g.