# In which direction must the reaction proceed to reach equilibrium?

## K= $1.2 \cdot {10}^{3}$ at ${400}^{\circ} C$ for the reaction to produce the highly toxic nerve gas phosgene: CO(g) +$C {l}_{2}$(g) $\to$ $C O C {l}_{2}$(g) If the concentration of all 3 species is each 0.050 M, in which direction must the reaction proceed to reach equilibrium

May 7, 2018

The reaction will shift to the right in order to achieve equilibrium.

#### Explanation:

This is our equilibrium:

$C O \left(g\right) + C {l}_{2} \left(g\right) r i g h t \le f t h a r p \infty n s C O C {l}_{2} \left(g\right)$

From this, we can determine that the expression for ${K}_{c}$ is this:

${K}_{c} = \frac{\left[C O C {l}_{2}\right]}{\left[C O\right] \times \left[C {l}_{2}\right]} = 1.2 \times {10}^{3}$

Right now, all of our concentrations have a concentration of $\text{0.050 M}$. So, the value of our ${Q}_{c}$ expression would be:

$\frac{\left[C O C {l}_{2}\right]}{\left[C O\right] \times \left[C {l}_{2}\right]} = \left(\text{0.050 M")/("0.050 M" xx "0.050 M}\right) = 2.0 \times 10$

The value of ${Q}_{c}$ is much less than ${K}_{c}$, which means that the concentration of products will be much higher than reactants at equilibrium.

So, to produce more products and reduce the concentration of reactants, the forward/right reaction would be favoured.