# In the reaction between HNO_2 and HCN: HNO_2(aq) + CN^(-)(aq) rightleftharpoons HCN(aq) + NO_2^(-)(aq), K = 1 x 10^6. What are the predominate species after the reaction reaches equilibrium?

May 14, 2017

Which side do you think is favoured by the equilibrium?

#### Explanation:

For $A + B r i g h t \le f t h a r p \infty n s C + D$, we define ${K}_{\text{eq}} = \frac{\left[C\right] \left[D\right]}{\left[A\right] \left[B\right]}$

If ${K}_{\text{eq}}$ is LARGE, which it most certainly is, then it follows that the quotient $\frac{\left[C\right] \left[D\right]}{\left[A\right] \left[B\right]}$ is also large, and this means that the concentrations of $C$ and $D$, the PRODUCTS are favoured at equilibrium.

Given that the following reaction has reached equilibrium:

$H N {O}_{2} \left(a q\right) + N \equiv {C}^{-} r i g h t \le f t h a r p \infty n s N {O}_{2}^{-} + H C \equiv N \left(a q\right)$,

.............it follows that there are large concentrations of $H C \equiv N$ and $\text{nitrite anion}$.

Why would you NOT want to be downwind of this reaction?