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")
