Question

If a 0.50 mol of hydrogen and 0.70 mol of iodine are initially introduced into a 500 mL reaction chamber, calculate the concentrations of all entities at equilibrium. K =46.0 at 490°C H2 + I2 <------> 2HI

Answer 1

The equilibrium concentrations are [H2]= 0.126 M, [I2]= 0.526 M and [HI]= 1.748 M.

- First, to calculate the concentrations of all entities at equilibrium, we need to analyze the initial state, the reaction and the equilibrium state:

Remember that thw units of concentrations must be in molarity (M), so we need to convert the 0.50 mol and 0.70 mol.

`H_2+I_2\rightleftarrows2HI^{}_{}`

Initial: 1 1.4 0

Reaction: -x -x 2x

Equilibrium: 1-x 1.4-x 2x

- Second, we write the formula for the equilibrium constant (K):

`\begin{gathered} K=(\lbrack HI\rbrack^2)/(\lbrack H_2\rbrack.\lbrack I_2\rbrack) \\ 46.0=\text{ }((2x)^2)/((1-x).(1.4-x)) \\ 46.0=\frac{4x^2}{1.4-x+x^2\text{ -1.4x}} \\ 64.4-46x+46x^2-64.4x=4x^2 \\ 46x^2-110.4x+64.4=4x^2 \\ 46.0x^2-110.4x+64.4-4x^2=0 \\ 42x^2\text{ - 110.4x + 64.4= 0} \end{gathered}`

- Now, we solve the quadratic equation with the Bhaskara formula:

In this case, a=42, b=-110.4 and c=64.4.

`\begin{gathered} x=\text{ }\frac{-b\pm\sqrt[]{b^2-4.a.c}}{2.a} \\ x=\text{ }\frac{-(-110.4)\pm\sqrt[]{(-110.4)^2-4.42.(64.4)}}{2.42} \end{gathered}`

We find x1=0.874 and x2=1.755. The correct value for x is x=0.874. We can't use x=1.755 because it is a quantity greater that 1.4.

- Finally, we replace the value of x and solve:

H2 concentration:

[H2]= 1-x

[H2]= 1-0.874

[H2]= 0.126 M

I2 concentration:

[I2]= 1.4-x

[I2]= 1.4-0.874

[I2]= 0.526 M

HI concentration:

[HI]= 2x

[HI]= 2 . 0.874

[HI]= 1.748 M

So, the equilibrium concentrations are [H2]= 0.126 M, [I2]= 0.526 M and [HI]= 1.748 M.