Consider the chemical equations shown here.

3 equations. 1: upper N upper O gas plus upper O subscript 3 gas right arrow upper N upper O subscript 2 gas plus upper O subscript 2 gas. Delta H subscript 1 equals 198.9 kilojoules. 2: StartFraction 3 over 2 EndFraction upper O gas right arrow upper O subscript 3 gas. Delta H subscript 2 equals 142.3 kilojoules. 3: upper O gas right arrow one half upper O subscript 2 gas. Delta H subscript 3 equals negative 247.5 kilojoules.

What is DeltaHrxn for the reaction shown below?

NO(g)+O(g)->NO2(g)

-1300. kJ  

Explanation:

We have two equations:

1. P₄(s)       +3O₂(g) ⟶ P₄O₆(s);                 ΔH₁  = -1640.1 kJ

2. P₄O₁₀(s)               ⟶ P₄(s)      + 5O₂(g); ΔH₂ =  2940.1 kJ

From these, we must devise the target equation:

3. P₄O₆(s)  + 2O₂(g) ⟶ P₄O₁₀(s);               ΔH = ?

The target equation has P₄O₆(s) on the left, so you reverse Equation 1.

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

4. P₄O₆(s)                 ⟶ P₄(s)       +3O₂(g); ΔH₁  =  1640.1 kJ

Equation 4 has P₄ on the right. That is not in the target equation.

You need an equation with P₄ on the left, so you reverse Equation 2.  

5. P₄(s)      + 5O₂(g)  ⟶ P₄O₁₀(s);               ΔH₂ = -2940.1 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.

You get the target equation 3:

4. P₄O₆(s)                      ⟶ P₄(s)        + 3O₂(g);      ΔH₁ =   1640.1 kJ

5. P₄(s)       + 2(5)O₂(g) ⟶ P₄O₁₀(s);                       ΔH₂ = -2940.1 kJ

3. P₄O₆(s)   + 2O₂(g)     ⟶ P₄O₁₀(s);                       ΔH   = -1300.  kJ

ΔH for the reaction is -1300. kJ

dynamic equilibrium is the answer.

i think it is true since you can test the answer and can use observations to complete it.