Question

h2+cl2-> 2HCl if you start with .250 of H2 and Cl2 determine equilibrium concentrations if keq=2.51 x 1 x 10 -4 what is the answer

Answer 1

h2+cl2->2HCl if you start with .250 of H2 and Cl2 determine equilibrium concentrations if $\text{keq} = 2.51 \times 1 \times 10^{-4}$ what is the answer

Solution:

From the equation of reaction at starting point the concentrations of $\mathsf{H}_2$ and $\mathsf{Cl}_2$ were $0.250\mathrm{M}$ and concentration of HCl was 0. The $x$ moles of $\mathsf{H}_2$ and $\mathsf{Cl}_2$ have been reacted at equilibrium point and $2\times \mathsf{M}$ of HCl has been found. So, the equilibrium concentrations of $\mathsf{H}_2$ and $\mathsf{Cl}_2$ are $(0.25 - x)\mathsf{M}$ and HCl is $2\times \mathsf{M}$ .

The equilibrium constant of this reaction is:

$$K _ {c} = \frac {[ H C ] _ {\mathrm {e q}} ^ {2}}{[ H _ {2} ] _ {\mathrm {e q}} \cdot [ C l _ {2} ] _ {\mathrm {e q}}} = \frac {[ 2 x ] ^ {2}}{[ 0 . 2 5 - x ] \cdot [ 0 . 2 5 - x ]} = 2. 5 1 \cdot 1 0 ^ {- 4}$$ $$2. 5 1 \cdot 1 0 ^ {- 4} \cdot (0. 0 6 2 5 - 0. 5 x + x ^ {2}) = 4 x ^ {2}$$ $$3. 9 9 9 7 4 9 x ^ {2} + 1. 2 5 5 \cdot 1 0 ^ {- 4} x - 1. 5 6 \cdot 1 0 ^ {- 5} = 0$$

We calculate the x: $x_{1,2} = \frac{-1.255 \cdot 10^{-4} \pm \sqrt{1.255 \cdot 10^{-4} - 4 \cdot 3.999749 \cdot (-1.56 \cdot 10^{-5})}}{2 \cdot 3.999749}$

$x_{1} = 7.84 \cdot 10^{-6}, x_{2} = -3.92 \cdot 10^{-5}$ (the second gives negative concentrations so is wrong).

The equilibrium concentrations of $\mathsf{H}_2$ and $\mathsf{Cl}_2$ are: $0.25 - 7.84 \cdot 10^{-6} = 0.249992 \, \text{M}$ .

The equilibrium concentration of HCl is: $2 \cdot 7.84 \cdot 10^{-6} = 1.568 \cdot 10^{-5} \, \text{M}$ .

Answer:

The equilibrium concentrations of $\mathsf{H}_2$ and $\mathsf{Cl}_2$ are 0.249992 M and the equilibrium concentration of HCl is $1.568\cdot 10^{-5}$ .