The Keq for the reaction
2 HI(g) H2(g) + I2(g) has a value of 1.85 x 10-2 at 425°C. If 0.18 mol of HI is placed in a 2.0 L flask and allowed to come to equilibrium at this temperature, what will be the equilibrium [I2] ?
To solve this problem, we can use the equation for the equilibrium constant (Keq), which is:
Keq = [H2][I2] / [HI]^2
Given that Keq = 1.85 x 10^-2 and the initial concentration of HI is 0.18 mol in a 2.0 L flask, we can calculate the initial concentration of HI:
[HI] = 0.18 mol / 2.0 L = 0.09 M
Now, let's assign the equilibrium concentration of I2 as x (in moles per liter). Since 2 moles of HI form 1 mole of I2 in the reaction, the change in concentration of HI is -2x and the change in concentration of I2 is x. Thus, the equilibrium concentration of HI will be:
[HI] = 0.09 M - 2x
Using this information, we can substitute the values into the equilibrium constant equation:
1.85 x 10^-2 = x / (0.09 M - 2x)^2
Simplifying the equation:
1.85 x 10^-2 * (0.09 M - 2x)^2 = x
Expanding and rearranging the equation:
1.85 x 10^-2 * (0.0081 M^2 - 0.36 Mx + 4x^2) = x
0.015 Mx^2 - 0.333 Mx + 0.01467 M = x
Rearranging the equation to set it equal to zero:
0.015 Mx^2 - 1.333 Mx + 0.01467 M = 0
Now we can solve the quadratic equation using the quadratic formula:
x = (-(-1.333 M) ± sqrt((-1.333 M)^2 - 4 * 0.015 M * 0.01467 M)) / (2 * 0.015 M)
Calculating the values under the square root:
x = (-(-1.333 M) ± sqrt(1.777889421 M^2 - 8.7805e-5 M^2)) / (0.03 M)
x = (-(-1.333 M) ± sqrt(1.777800541 M^2)) / (0.03 M)
x = (-(-1.333 M) ± 1.334 M) / (0.03 M)
x1 = (2.667 M) / (0.03 M) = 88.9 M
x2 = (-0.001 M) / (0.03 M) = -0.033 M
Since the concentration of a substance cannot be negative, the equilibrium concentration of I2 is 0.033 M.