# Question f4c22

##### 1 Answer
Aug 20, 2016

Given

$C \left(g r\right) + {O}_{2} \left(g\right) \to C {O}_{2} \left(g\right) , \Delta {H}_{1}^{\circ} = - 393.5 {\text{ kJmol}}^{-} 1. \ldots \left(1\right)$

$C O + \frac{1}{2} {O}_{2} \left(g\right) \to C {O}_{2} \left(g\right) , \Delta {H}_{2}^{\circ} = - 283.0 {\text{ kJmol}}^{-} 1. . . \left(2\right)$

$C O + C {l}_{2} \left(g\right) \to C O C {l}_{2} \left(g\right) , \Delta {H}_{3}^{\circ} = - 108.0 {\text{ kJmol}}^{-} 1. . . \left(3\right)$

$\text{We are to determine "DeltaH_f^@" of " COCl_2(g)" for the reaction:}$

C(gr)+1/2O_2(g)+Cl_2(g)->COCl_2(g),DeltaH_f^@=?#

It is obvious that adding(1) and (3) ,then subtracting (2) we can easily get $\Delta {H}_{f}^{\circ}$ of the reaction.

So

$\Delta {H}_{f}^{\circ} = \Delta {H}_{1}^{\circ} + \Delta {H}_{3}^{\circ} - \Delta {H}_{2}^{\circ}$

$\Delta {H}_{f}^{\circ} = \left(- 393.5 - 108.0 - \left(- 283.0\right)\right) {\text{ kJmol}}^{-} 1$

$= - 218.5 k J {\text{mol}}^{-} 1$