# What is the standard enthalpy change for the reaction?

## Given the standard enthalpy changes for the following two reactions: (1) Zn(s) + Cl2(g)ZnCl2(s)...... ΔH° = -415.0 kJ (2) Pb(s) + Cl2(g)PbCl2(s)......ΔH° = -359.4 kJ what is the standard enthalpy change for the reaction: (3) Zn(s) + PbCl2(s)ZnCl2(s) + Pb(s)......ΔH° = ?

Jun 12, 2018

ΔH^@ = "-55.6 kJ"

#### Explanation:

What you have

You have two equations:

$\boldsymbol{\text{(1)"color(white)(m) "Zn(s)" + "Cl"_2"(g)" → "ZnCl"_2"(s)"; DeltaH^@ = "-415.0 kJ}}$

$\boldsymbol{\text{(2)"color(white)(m) "Pb(s)" + "Cl"_2"(g)" → "PbCl"_2"(s)"; DeltaH^@ = "-359.4 kJ}}$

From these, you must devise the target equation:

bb"(3)"color(white)(m)color(blue)("Zn(s)" + "PbCl"_2"(s)" rarr "ZnCl"_2"(s) + Pb(s)"; ΔH^@ = ?)

What you must do

The target equation has $\text{Zn(s)}$ on the left, so you rewrite equation (1).

$\boldsymbol{\text{(4)"color(white)(m) "Zn(s)" + "Cl"_2"(g)" → "ZnCl"_2"(s)"; DeltaH^@ = "-415.0 kJ}}$

Equation (4) has $\text{Cl"_2"(g)}$ on the left, and that is not in the target equation.

You need an equation with $\text{Cl"_2"(g)}$ on the right.

Reverse Equation (2).

When you reverse an equation, you reverse the sign of its  ΔH.

$\boldsymbol{\text{(5)"color(white)(m)"PbCl"_2"(s)" → "Pb(s)" + "Cl"_2"(g)"; DeltaH^@ = "+359.4 kJ}}$

Now, you add equations (4) and (5), cancelling species that appear on opposite sides of the reaction arrows.

When you add equations, you add their ΔH values.

This gives the target equation (6):

color(white)(mmmmmmmmmmmmmmmmmmmmmmll)ul(ΔH^@"/kJ")
$\boldsymbol{\text{(4)"color(white)(m) "Zn(s)" + color(red)(cancel(color(black)("Cl"_2"(g)"))) → "ZnCl"_2"(s)";color(white)(mmmmmmll) "-415.0}}$
bb"(5)"color(white)(m)ul("PbCl"_2"(s)" → "Pb(s)" + color(red)(cancel(color(black)("Cl"_2"(g)")));color(white)(mmmmm))color(white)(m)ul("+359.4")
bb"(6)"color(blue)(color(white)(m) "Zn(s)" + "PbCl"_2"(g)" → "ZnCl"_2"(s)" + "Pb(s)";color(white)(mm) "-55.6")

ΔH^@ = "-55.6 kJ"

Jun 13, 2018

$\Delta \text{H"^@=-55.6color(white)(x)"kJ}$
Delta"H"^@=Sigma[Delta"H"_f^@("products")]-Sigma[Delta"H"_f^@("reactants")]
$\Delta \text{H"^@=-415-(-359.4)=-55.6color(white)(x)"kJ}$