Question

The equilibrium constant $\left(\mathrm{K}\right)$ for the reaction $2\mathrm{HI}\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{I}}_2\left(\mathrm{g}\right)$ at room temperature is $5.7$ and that of $698\mathrm{K}$ is $2.8\times {10}^{-2}$. this implies that the forward reaction is:

Answer 1

Step 1: Equilibrium constant :

• “It expresses the relationship between products and reactants at equilibrium.”
• It is denoted by $\mathrm{K}$.

Step 2: Given information:

• The given reaction is, $2\mathrm{HI}\left(\mathrm{g}\right)\rightleftharpoons {\mathrm{H}}_2\left(\mathrm{g}\right)+{\mathrm{I}}_2\left(\mathrm{g}\right)$
• At room temperature, the equilibrium constant, $\mathrm{K}=5.7$
• At $698\mathrm{K}$, the equilibrium constant, $\mathrm{K}=2.8\times {10}^{-2}$

Step 3: Observations from the given information:

• On increasing the temperature, the value of $\mathrm{K}$ decreases.
• This implies that at high temperature, the reaction proceeds in a backward direction or in forward at room temperature.
• At room temperature, $\mathrm{HI}$ is less stable than ${\mathrm{H}}_2$​ and ${\mathrm{I}}_2$.