Consider the reaction:
2HI(g) ⇌ H2(g) + I2(g)
at 620 °C where K = 1.63 x 10¯3. Given that [H2]o = 0.300 M, [I2]o = 0.150 M and [HI]o = 0.400 M, calculate the equilibrium concentrations of HI, H2, and I2.
Let's denote the equilibrium concentrations of HI, H2, and I2 as [HI], [H2], and [I2] respectively.
Based on the reaction equation, we can set up the equilibrium expression:
K = ([H2][I2]) / [HI]^2
Given:
[K] = 1.63 x 10^-3
[H2]o = 0.300 M
[I2]o = 0.150 M
[HI]o = 0.400 M
At equilibrium, let's denote the change in concentration of HI, H2, and I2 as x.
Hence, at equilibrium:
[HI] = [HI]o - x
[H2] = [H2]o + x
[I2] = [I2]o + x
Substituting these values into the equilibrium expression, we have:
K = (([H2]o + x)([I2]o + x)) / ([HI]o - x)^2
Now let's substitute the given values into the equation:
1.63 x 10^-3 = ((0.300 + x)(0.150 + x)) / (0.400 - x)^2
Expanding and rearranging the equation, we have:
1.63 x 10^-3 * (0.400 - x)^2 = (0.300 + x)(0.150 + x)
Now let's solve for x:
1.63 x 10^-3 * (0.1600 - 0.800x + x^2) = 0.045 + 0.150x + 0.300x + x^2
0.002608 - 0.0104x + 0.00163x^2 = 0.045 + 0.150x + 0.300x + x^2
0.00163x^2 - 0.1604x - 0.052392 = 0
Now we can solve this quadratic equation using the quadratic formula:
x = (-(-0.1604) ± sqrt((-0.1604)^2 - 4(0.00163)(-0.052392))) / (2(0.00163))
x = (0.1604 ± sqrt(0.025792 + 0.043012543)) / 0.00326
x = (0.1604 ± sqrt(0.068804543)) / 0.00326
x ≈ 0.2264 / 0.00326 or x ≈ -0.3336 / 0.00326
x ≈ 69.46 or x ≈ -102.43
Since x cannot be negative (concentration can't be negative), we disregard the negative value.
Therefore, x ≈ 69.46
Now we can calculate the equilibrium concentrations:
[HI] = [HI]o - x
[HI] = 0.400 - 69.46
[HI] ≈ -69.06 M (disregard the negative value)
[H2] = [H2]o + x
[H2] = 0.300 + 69.46
[H2] ≈ 69.76 M
[I2] = [I2]o + x
[I2] = 0.150 + 69.46
[I2] ≈ 69.61 M
So, the equilibrium concentrations are approximately:
[HI] ≈ -69.06 M
[H2] ≈ 69.76 M
[I2 ] ≈ 69.61 M