Consider the reaction
`SO_(2)Cl_(2) hArr SO_(2)(g)+Cl_(2)(g)`
at `375^(@)C`, the value of equilibrium constant for the reaction is `0.0032`. It was observed that the concentration of the three species is `0.050 mol L^(-1)` each at a certain instant. Discuss what will happen in the reaction vessel?

Explanation

In this question, concentration of three species, i.e., `SO_(2)Cl_(2)(g), SO_(2)(g) and Cl_(2)(g)` each is given, but it is not mentioned that whether the system is at equilibrium or not. So first check it.
Find reaction coefficient for given equation.
`Q=([SO_(2)][Cl_(2)])/([SO_(2)Cl_(2)])=((0.05)(0.05))/((0.05))=0.05`
`rArr Q ne K`, so system is not at equilibrium state.
As `Q gt K_(eq)`, the concentrations must adjust till `Q=K_(eq)` for equilibrium. This can happen only if reaction shifts backwards, and products recombine to give back reactants. Hence in the reaction vessel, the system will move backward so that it can achieve equilibrium state.