The equilibrium constant, Kc, for the following reaction is 1.54E-2 at 643 K.
2HI(g) H2(g) + I2(g)
When a sufficiently large sample of HI(g) is introduced into an evacuated vessel at 643 K, the equilibrium concentration of I2(g) is found to be 0.351 M.
Calculate the concentration of HI in the equilibrium mixture. ___M
You really need to find the arrow key and use it.
.....2HI ==> H2 + I2
I....................
C....................
E......y......x....0.351
So if I2 is 0.351 that means H2 is 0.351 (and that does NOT means HI is 0.351).
Write the Kc expression, substitute for H2 and I2 and solve for HI.
To calculate the concentration of HI in the equilibrium mixture, we can use the equilibrium expression and the given equilibrium constant, Kc.
The equilibrium expression for the given reaction is:
Kc = [H2] * [I2] / [HI]^2
We are given the equilibrium concentration of I2(g) as 0.351 M. We need to find the concentration of HI.
Let's assume the equilibrium concentration of HI is x M. Since the stoichiometric coefficient for HI is 2 in the balanced equation, the concentration of H2 will also be x M.
Substituting these values into the equilibrium expression, we get:
1.54E-2 = (x) * (0.351) / (x)^2
Simplifying the equation, we get:
1.54E-2 * (x)^2 = (x) * (0.351)
Now, we can solve this quadratic equation to find the value of x (the concentration of HI).
1.54E-2 * (x)^2 - (x) * (0.351) = 0
Use the quadratic formula: x = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 1.54E-2, b = -0.351, and c = 0.
Solving the quadratic equation, we find two possible values for x: x = 0.0275 M and x = 0.
However, since we assumed the concentration of HI to be x M, the value x = 0 is not valid since it would mean no HI was present in the equilibrium mixture.
Therefore, the concentration of HI in the equilibrium mixture is 0.0275 M.