Question

Cesium is one of the most reactive alkali metals. When it comes into contact with water, it reacts to form cesium hydroxide and hydrogen gas. The gas forms a bubble around the solid, which brings the cesium to the surface, and exposes it to more water. This then causes more cesium to react, releasing more energy that ignites the hydrogen gas, causing explosions on the water surface. a) How much heat would be liberated if 0.150 mol of solid cesium was allowed to completely react with water under standard conditions? (Use textbook Resource section for data in this calculation.) b) Calculate DU° for this reaction under standard conditions at 298.15 K

Answer 1

(a) The balanced equation for the given reaction can be written as follows.

Cs(s) + H₂O(l) ----> CsOH(s) + 1/2 H₂(g)

$\Delta$H^o_reaction = {($\Delta$H^o)CsOH(s) + 1/2 * ($\Delta$H^o)_H2} - {($\Delta$H^o)_Cs(s) + ($\Delta$H^o)_H2O(l)}

= {-416.73 + 1/2 * 0} - {0 + (-285.83)}

= -130.9 kJ/mol

Therefore, the amount of heat released in the reaction = 130.9 kJ/mol * 0.15 mol = 19.635 kJ

(b) Formula: $\Delta$H^o = $\Delta$U^o + $\Delta$nRT

i.e. -130.9*1000 J/mol = $\Delta$U^o + (1/2 * 8.314 J/mol.K * 298.15 K)

i.e. -130900 J = $\Delta$U^o + 1234.41

i.e. $\Delta$U^o = -132.14 kJ/mol