Question #4b204

1 Answer
Mar 3, 2017

Answer:

You get the answer by applying Hess's Law.

Explanation:

You are given two equations:

#bb"(1)" "P(yellow)" + "½O"_2 → "½P"_2"O"_5; Δ_text(c)H = "-9.91 kJ/mol"#
#bb"(2)" "P(red)" color(white)(ml)+ "½O"_2 → "½P"_2"O"_5; Δ_text(c)H = "-8.78 kJ/mol"#

From these, you must devise the target equation

#"P(yellow) → P(red)"; color(white)(mmmml)Δ_text(r)H = ?#

The target equation has #"P(yellow)"# on the left, so you start by writing equation (1) normally.

#bb"(1)"color(white)(l) "P(yellow)" + "½O"_2 → "½P"_2"O"_5; Δ_text(c)H = color(white)(ll)"-9.91 kJ/mol"#

The target equation has #"P(red)"# on the right, so you write equation (2) in reverse.

#bb"(3)"color(white)(l) "½P"_2"O"_5 → "P(red)" + "½O"_2 ;color(white)(mll) Δ_text(c)H = "+8.78 kJ/mol"#

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

Then you add equations (1) and (3), canceling species that appear on opposite sides of the reaction arrows.

When you add two equations, you add their #ΔH# values.

This gives us the target equation (4):

#bb"(1)"color(white)(l) "P(yellow)" + color(red)(cancel(color(black)("½O"_2))) → color(red)(cancel(color(black)("½P"_2"O"_5))); Δ_text(c)H = color(white)(ll)"-9.91 kJ/mol"#
#bb"(3)"color(white)(l) color(red)(cancel(color(black)("½P"_2"O"_5))) → "P(red)" + color(red)(cancel(color(black)("½O"_2))) ;color(white)(mll) Δ_text(c)H = "+8.78 kJ/mol"#
#bb"(4)" stackrel(——————————)(color(white)(l)"P(yellow) → P(red)"); color(white)(mmmmml)stackrel(———————————)(Δ_text(r)H = color(white)(ll)"-1.13 kJ/mol")#