Question

Using the reaction below at 448 degrees Celsius, calculate the keq if a flask contains 0.1207 M hydrogen, 0.0402 M of iodine, and 0.00272 M of hydroiodic acid

the reaction is H2+I2 ----> 2HI​

Answer 1

The expression for the equilibrium constant, K, can be written as:

K = [HI]^2 / ([H2][I2])

where the square brackets denote the molar concentrations of the species at equilibrium.

At the start of the reaction, the concentrations of H2, I2, and HI are:

[H2] = 0.1207 M
[I2] = 0.0402 M
[HI] = 0.00272 M

At equilibrium, the concentrations will change by some amount, let's call them x and y for I2 and H2 respectively. The concentration of HI will increase by 2x, since 2 moles of HI are formed for every mole of I2 that reacts and every mole of H2 that reacts. The new concentrations at equilibrium will be:

[H2] = 0.1207 - y
[I2] = 0.0402 - x
[HI] = 0.00272 + 2x

The balanced equation tells us that 1 mole of I2 reacts for every 1 mole of H2 that reacts, so we can set up an equation for x in terms of y:

x = y

Now we can substitute these expressions into the equilibrium constant expression:

K = ([HI]^2) / ([H2][I2])
K = (0.00272 + 2x)^2 / ((0.1207 - y)(0.0402 - x))

Substituting x = y, we get:

K = (0.00272 + 4y)^2 / ((0.1207 - y)(0.0402 - y))

At equilibrium, the reaction quotient Q will be equal to K. We can use this fact to solve for y:

K = Q
K = ([HI]^2) / ([H2][I2])
K = (0.00272 + 2x)^2 / ((0.1207 - y)(0.0402 - x))
K = (0.00272 + 4y)^2 / ((0.1207 - y)(0.0402 - y))

(0.00272 + 4y)^2 = K(0.1207 - y)(0.0402 - y)

We can solve this quadratic equation for y using the quadratic formula:

y = [-(0.00272)^2 ± sqrt((0.00272)^4 - 4K(0.1207)(