Question

Use the thermochemical equations shown below to determine the enthalpy for the reaction: COCl2(g) + H2O(l) --> CH2Cl2(l) + O2(g) ½H2(g) + ½Cl2(g) --> HCl(g) ΔH=-46kJ H2O(g) + Cl2(g) --> 2HCl(g) + ½O2(g) ΔH=-21KJ CH2Cl2(l) + H2(g) + O2(g) --> COCl2(g) + 2H2O(l) ΔH = 80.5 kJ

Answer 1

Answer:

ΔH° (Enthalpy change) for the reaction is equal to (-151.5 KJ)

Explanation:

Determine the enthalpy for the reaction

           COCl₂ (g) + H₂O(l) --> CH₂Cl₂(l) + O₂(g)

It becomes easier to find the unknown value of enthalpy of a particular reaction by applying the Hess's law on the given reaction with the known value of standard enthalpy change.

$\frac{1}{2}$ H₂(g) + $\frac{1}{2}$Cl₂(g) --> HCl(g) ΔH=-46 kJ -----------------------------(1)

H₂O(g) + Cl₂(g) --> 2 HCl(g) + $\frac{1}{2}$O₂ (g) ΔH=-21 KJ-------------------(2)

CH₂Cl₂(l) + H₂(g) + $\frac{3}{2}$O₂(g) --> COCl₂(g) + 2 H₂O(l) ΔH = 80.5 kJ ----------(3)

Equation (1) Multiply by factor 2 we get

H₂(g) + Cl₂(g)  ---> 2 HCl (g)   ΔH = - 46 x 2 = - 92 KJ ------------(4)

Revers the equation (2) and (3) we get

2 HCl(g) + $\frac{1}{2}$O₂ (g) --> H₂O(g) + Cl₂(g)  ΔH=+21 KJ ----------------(5)

COCl₂(g) + 2 H₂O(l) --> CH₂Cl₂(l) + H₂(g) + $\frac{3}{2}$O₂(g) ΔH = - 80.5 kJ  -----------(6)

Now add these three equations i.e., (4) + (5) + (6)

we can get these equation

                COCl₂ (g) + H₂O(l) --> CH₂Cl₂(l) + O₂(g)

The enthalpy change during the reaction = -92 + 21 -80.5 = - 151.5 KJ