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Consider the reaction SO(2)Cl(2) hArr ...

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?

Text Solution

Verified by Experts

In this equation, 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_(eq)`, so system is not at equilibrium state.
As `Q gt K_(eq)`, the concentration must adjust till `Q=K_(eq)` for equilibrium. This can happen only if reaction shifts backwards, and products recombine to give back reacyants. Hence, in the vessel, the system move backward so that it can achieve equilibrium state.
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