For the reaction `A(g)+B(g) to C(s)+D(l),DeltaH=-233.5kJ and DeltaS=-466.1J*K^(-1)`. At what temperature, equilibrium will be established? In whichi directions the reaction will proceed above and below the temperature?

Explanation

At constant temperature and pressure, the equilibrium temperature of a reaction, `T=(DeltaH)/(DeltaS)`.
given: `DeltaH=-233xx10^(3)J and DeltaS=-466.1J*K^(-1)`
Therefore, `T=(233.5xx10^(3))/(466.1)K=500.9K=227.9^(@)C`
`therefore`At `227.9^(@)C`, the reaction will attain equilibrium.
(1) When `T gt 500.9K`, the magnitude of `T Delta S` is greater than that of `DeltaH`. then accoding to the equation the reaction will be non-spontaneous above 500.9 K.
(2) When `T lt 500.9K`, magnitude of `TDeltaS` is less than that of `DeltaH`. accoding to the equation `DeltaG=DeltaH-TDeltaS,DeltaG` will be negative. therefore, the reaction will be spontaneous below 500.9 K.