You wish to know the enthalpy change for the formation of liquid $PC{l}_3$ from the elements.

$P_4(s)+6C{l}_2(g)4PC{l}_3(l)\frac{,}{{?}_{\frac{?}{?}}{?}_r}H=$

The enthalpy change for the formation of $PC{l}_5$ from the elements can be determined experimentally, as can the enthalpy change for the reaction of $PC{l}_3(l)$ with more chlorine to give $PC{l}_5(s)$ :

$P_4(s)+10C{l}_2(g)4PC{l}_5(s)\frac{?}{{?}_{\frac{?}{?}}{?}_r}$

$H=-1774\mathrm{.}0k\frac{J}{m}ol-rxn$

$PC{l}_3(l)+C{l}_2(g)PC{l}_5(s)\frac{?}{{?}_{\frac{?}{?}}{?}_r}$

$H=-123\mathrm{.}8k\frac{J}{m}ol-rxn$

Use these data to calculate the enthalpy change for the formation of $2\mathrm{.}90$mol of $PC{l}_3(l)$ from phosphorus and chlorine.

Enthalpy change $=$

kJ