Chapter 17, Problem 66A

Interpretation Introduction

Interpretation: The enthalpy change of the chemical reaction 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 66A

Thus, the enthalpy changes for the reaction ${\text{PbCl}}_{\text{2}}{\text{(s)+Cl}}_{\text{2}}\text{(g)}\to {\text{PbCl}}_{\text{4}}\text{(l)}$ is $\text{+30}\text{.2kJ}$

Explanation of Solution

The chemical reaction for which enthalpy change is to be calculated

  ${\text{PbCl}}_{\text{2}}{\text{(s)+Cl}}_{\text{2}}\text{(g)}\to {\text{PbCl}}_{\text{4}}\text{(l)}$   $(1)$

The thermochemical equations given are

  ${\text{Pb(s)+2Cl}}_{\text{2}}\text{(g)}\to {\text{PbCl}}_{\text{4}}\text{(l) }\Delta \text{H= -329}\text{.2 kJ}$   $(2)$

  ${\text{Pb(s)+Cl}}_{\text{2}}\text{(g)}\to {\text{PbCl}}_2\text{(s) }\Delta \text{H= -359}\text{.4 kJ}$   $(3)$

To get the given chemical reaction, we must reverse the reaction $(3)$ which will change the sign of enthalpy.

  ${\text{PbCl}}_2\text{(s)}\to {\text{Pb(s)+Cl}}_{\text{2}}\text{(g) }\Delta \text{H= +359}\text{.4 kJ}$   $(4)$

Now add the two reactions $(2)$ and $(4)$ , we get $(1)$ and the enthalpies also get added

  ${\text{PbCl}}_2{\text{(s)+Cl}}_2(g)\to {\text{PbCl}}_4\text{(l) }\Delta \text{H= +30}\text{.2kJ}$