Question

Given the standard enthalpy changes for the following two reactions:-
(1) $Zn(s)+C{{l}_{2}}(g)\to ZnC{{l}_{2}}(s)\text{ }\Delta {{\text{H}}^{\circ }}\text{=-415}\text{.0kJ}$
(2) $Pb(s)+C{{l}_{2}}(g)\to PbC{{l}_{2}}(s)\text{ }\Delta {{\text{H}}^{\circ }}=-359.4kJ$
What is the standard enthalpy change for the reaction?
(3) $Zn(s)+PbC{{l}_{2}}(s)\to ZnC{{l}_{2}}(s)+Pb(s)\,\text{ }\Delta {{\text{H}}^{\circ }}=?$

Answer 1

We know the enthalpy change of the two reactions and from that we have to find the enthalpy change for the third reaction. So first we will reverse the given equation (2), as in the final equation it is on the reactant side and then add the equations (1) and (2), you will get equation(3) and the enthalpy change for that very reaction. Now using this answer the given statement accordingly.

Complete answer:
First of all let’s discuss what is standard enthalpy of formation. It is defined as the change in the enthalpy during the formation of one mole of the substance from its constituents elements involved in the reaction. It is denoted by ${{\Delta }_{f}}{{H}^{\circ }}$.
It tells us about the change involved in the enthalpy during the chemical reaction. It can be either exothermic or endothermic in nature.
Now considering the statement as;
We have to find the standard enthalpy change of the reaction;-
$Zn(s)+PbC{{l}_{2}}(s)\to ZnC{{l}_{2}}(s)+Pb(s)$ -----------(1)
Given;-
$Zn(s)+C{{l}_{2}}(g)\to ZnC{{l}_{2}}(s)\text{ }\Delta {{\text{H}}^{\circ }}\text{=-415}\text{.0kJ}$ ------------(2)
$Pb(s)+C{{l}_{2}}(g)\to PbC{{l}_{2}}(s)\text{ }\Delta {{\text{H}}^{\circ }}=-359.4kJ$------------(3)
From the equation(1), we can see that the lead chloride is on the reactant side, so we will reverse equation (3) as;
$PbC{{l}_{2}}(s)\to Pb(s)+C{{l}_{2}}(g)\text{ }\Delta {{\text{H}}^{\circ }}=359.4kJ$ -----------(4)
Now on adding equations (2) and (4), we will get equation (1), then the standard enthalpy of formation for the reaction will be;-
$Zn(s)+PbC{{l}_{2}}(s)\to ZnC{{l}_{2}}(s)+Pb(s)\,\text{ }\Delta {{\text{H}}^{\circ }}=-415.0+359.4=-55.6kJ$
Hence, the standard enthalpy change for the reaction is ;-
$Zn(s)+PbC{{l}_{2}}(s)\to ZnC{{l}_{2}}(s)+Pb(s)\,\text{ }\Delta {{\text{H}}^{\circ }}=-55.6kJ$

Note:
While calculating the standard enthalpy of formation ,each substance or compound must be in their standard states and standard enthalpy of formation can either be positive or negative in nature.