Given the following thermodynamic reactions: CaO (s) + Cl2 (g) ---> CaOCl2 delta H° = -110.9...

Given the following thermodynamic reactions: CaO (s) + Cl2 (g) ---> CaOCl2 delta H° = -110.9 KJ/ H20 (l) + CaOCl2 (s) + NaBr (s) ---> 2NaCl (s) + Ca(OH)2 (s) + Br2 (l) delta H° = -60.2 KJ/ Ca(OH)2 (s) ---> CaO (s) + H2O (l) delta H° = +65.1 KJ. Calculate the value of delta(Δ)H° (in kilojoules) for the reaction 1/2 Cl2 (g) + NaBr (s) ---> NaCl (s) +1/2 Br2 (l)

Expert Solution

Given CaO (s) + Cl₂ (g) -----> CaOCl₂ ; ΔH°₁ = -110.9 KJ/mol -------------------(1)

H₂O(l) + CaOCl₂ (s) + 2NaBr (s) ---> 2NaCl (s) + Ca(OH)₂ (s) + Br₂ (l) ; ΔH°₂ = -60.2 KJ/mol -----(2)

Ca(OH)₂ (s) ---> CaO (s) + H₂O (l) ; ΔH°₃ = +65.1 KJ/mol ------------------------(3)

1/2 Cl₂ (g) + NaBr (s) ---> NaCl (s) +1/2 Br₂ (l) ;ΔH = ? -------------------------(4)

Equation (4) can be obtained from the given three equations as follows :

Eqn(4) = [(1/2) xEqn(1)] + [(1/2) xEqn(2)]+[(1/2)xEqn(3)]

= (1/2) x [ Eqn(1) + Eqn(2) + Eqn(3) ]

So ΔH = (1/2) x[ΔH°₁ + ΔH°₂ +ΔH°₃]

= (1/2) x [(-110.9 KJ/mol) +( -60.2 KJ/mol)+ ( +65.1 KJ/mol )]

= -53.0 kJ

Therefore the ΔH of the reaction is -53.0 kJ

queen_honey_blossom answered 6 months ago

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