Question

given the following data:

$p_{4}(s)+6cl_{2}(g)\to4pcl_{3}(g)$ $\delta_{r}h=-2439\\ kj$

$4pcl_{5}(g)\to p_{4}(s)+10cl_{2}(g)$ $\delta_{r}h=+3438\\ kj$

using hesss law, the enthalpy of reaction for the following equation: $pcl_{5}(g)\to pcl_{3}(g)+cl_{2}(g)$ is calculated to be kj.

Explanation:

Step1: Manipulate the given equations

Let's call the first equation \(E_1\): \(P_4(s)+6Cl_2(g)\to4PCl_3(g)\), \(\Delta_rH_1 = - 2439\space kJ\)
The second equation \(E_2\): \(4PCl_5(g)\to P_4(s)+10Cl_2(g)\), \(\Delta_rH_2=+3438\space kJ\)

We want to find \(\Delta H\) for \(PCl_5(g)\to PCl_3(g)+Cl_2(g)\).

If we add \(E_1\) and \(E_2\):
\((P_4(s)+6Cl_2(g))+(4PCl_5(g))\to(4PCl_3(g))+(P_4(s)+10Cl_2(g))\)

Simplifying the left - hand side and right - hand side (canceling out \(P_4(s)\)) gives \(4PCl_5(g)\to4PCl_3(g)+4Cl_2(g)\)

By Hess's Law, \(\Delta H\) for \(4PCl_5(g)\to4PCl_3(g)+4Cl_2(g)\) is \(\Delta H=\Delta_rH_1+\Delta_rH_2\)

\(\Delta H=-2439 + 3438=999\space kJ\)

Step2: Scale the equation

The equation \(4PCl_5(g)\to4PCl_3(g)+4Cl_2(g)\) has a \(\Delta H = 999\space kJ\).

We want the equation \(PCl_5(g)\to PCl_3(g)+Cl_2(g)\). If we divide the equation \(4PCl_5(g)\to4PCl_3(g)+4Cl_2(g)\) by \(4\) throughout, the enthalpy change is also divided by \(4\)

\(\Delta H=\frac{999}{4}=249.75\space kJ\)

Answer:

\(249.75\)