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° = ?

2 Answers
Jun 12, 2018

Answer:

#ΔH^@ = "-55.6 kJ"#

Explanation:

What you have

You have two equations:

#bb"(1)"color(white)(m) "Zn(s)" + "Cl"_2"(g)" → "ZnCl"_2"(s)"; DeltaH^@ = "-415.0 kJ"#

#bb"(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 #"Zn(s)"# on the left, so you rewrite equation (1).

#bb"(4)"color(white)(m) "Zn(s)" + "Cl"_2"(g)" → "ZnCl"_2"(s)"; DeltaH^@ = "-415.0 kJ"#

Equation (4) has #"Cl"_2"(g)"# on the left, and that is not in the target equation.

You need an equation with #"Cl"_2"(g)"# on the right.

Reverse Equation (2).

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

#bb"(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")#
#bb"(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

Answer:

#Delta"H"^@=-55.6color(white)(x)"kJ"#

Explanation:

When estimating the enthalpy of a reaction, given the standard enthalpies of formation, we can apply Hess' Law:

#Delta"H"^@=Sigma[Delta"H"_f^@("products")]-Sigma[Delta"H"_f^@("reactants")]#

#Delta"H"^@=-415-(-359.4)=-55.6color(white)(x)"kJ"#