Chapter 17, Problem 78A

Interpretation Introduction

Interpretation: The enthalpy change for the formation of phosphorus pentachloride is to be calculated.

Concept Introduction: Hess’s Law states that the overall change in enthalpy for the solution is the sum of all changes independent of the different phases or steps of a reaction.

Answer to Problem 78A

The enthalpy changes for the reaction ${\text{2P(s)+5Cl}}_{\text{2}}\text{(g)}\to 2{\text{PCl}}_5\text{(s)}$ is $\text{-887}\text{.0 kJ}$

Explanation of Solution

The chemical reaction for which enthalpy change is to be calculated

  ${\text{2P(s)+5Cl}}_{\text{2}}\text{(g)}\to 2{\text{PCl}}_5\text{(s)}$   $(1)$

The thermochemical equations given are

  ${\text{PCl}}_5\text{(s)}\to {\text{PCl}}_3{\text{(g)+Cl}}_{\text{2}}\text{(g) }\Delta \text{H= +156}\text{.5 kJ}$   $(2)$

  ${\text{2P(s)+3Cl}}_2(g)\to 2{\text{PCl}}_3\text{(g) }\Delta \text{H= -574}\text{.0kJ}$   $(3)$

To get the required chemical reaction, first we must reverse the reaction $(2)$ which will change the sign of enthalpy.

  ${\text{PCl}}_3{\text{(g)+Cl}}_{\text{2}}\text{(g)}\to {\text{PCl}}_5\text{(s) }\Delta \text{H= -156}\text{.5 kJ}$   $(4)$

Now we have to multiply the reaction $(4)$ by $2$ which will also change enthalpy by a factor of $2$

  ${\text{2PCl}}_3{\text{(g)+2Cl}}_{\text{2}}\text{(g)}\to 2{\text{PCl}}_5\text{(s) }\Delta \text{H= -2}\times \text{156}\text{.5 kJ=-313}\text{.0 kJ}$

  $\text{(5)}$

Now adding the two reactions $(3)$ and $\text{(5)}$ , we get $(1)$ and the enthalpies also get added

  $\begin{array}{l}{\text{2PCl}}_3{\text{(g)+2Cl}}_{\text{2}}\text{(g)}\to 2{\text{PCl}}_5\text{(s) }\Delta \text{H= -313}\text{.0 kJ}\\ {\text{2P(s)+3Cl}}_2(g)\to 2{\text{PCl}}_3\text{(g) }\Delta \text{H= -574}\text{.0 kJ}\\ \overline{\underset{¯}{2P\left(s\right)+5C{l}_2\left(g\right)\to {\text{2PCl}}_{\text{5}}\left(\text{s}\right)\Delta \text{H= -887}\text{.0 kJ}}}\end{array}$

The enthalpy change for the reaction ${\text{2P(s)+5Cl}}_{\text{2}}\text{(g)}\to 2{\text{PCl}}_5\text{(s)}$ is $\text{-887}\text{.0 kJ}$