Answer:
[tex]\large \boxed{ \text{B. 2 kJ}}[/tex]
Explanation:
We have two equations:
1. C(s, graphite) + O₂(g) ⟶ CO₂(g); ∆H = -394 kJ
2. C(s, diamond) + O₂(g) ⟶CO₂(g); ∆H = -396 kJ
From these, we must devise the target equation:
3. C(s, graphite) ⟶ C(s,diamond); ΔH = ?
The target equation has C(s, graphite) on the left, so you rewrite Equation 1.
4. C(s, graphite) + O₂(g) ⟶ CO₂(g); ∆H = -394 kJ
Equation 4 has CO₂ on the right, and that is not in the target equation.
You need an equation with CO₂ on the left, so you reverse Equation 2.
When you reverse an equation, you reverse the sign of its ΔH.
5. CO₂(g)⟶ C(s, diamond) + O₂(g); ∆H = +396 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. C(s, graphite) + O₂(g) ⟶ CO₂(g); ∆H = -394 kJ
5. CO₂(g) ⟶ C(s, diamond) + O₂(g); ∆H = +396 kJ
3. C(s, graphite) ⟶ C(s, diamond); ΔH = 2 kJ
[tex]\Delta H \text{ for the reaction is $\large \boxed{\textbf{2 kJ/mol}}$}[/tex]
