Question

. Given the following data:

Answer 1

Answer:

$\boxed{\text{-220 kJ}}$

Explanation:

We have two equations:

(I)   C(s) + O₂(g) ⟶ CO₂(g);     ΔH = -393 kJ

(II) 2CO(g) + O₂(g) → 2C0₂(g); ΔH = -566 kJ  

From these, we must devise the target equation:

(III) 2C(s) + O₂(g) → 2CO(g); ΔH = ?

The target equation has 2C on the left, so you double Equation (I).

When you double an equation, you double its ΔH.

(IV) 2C(s) + 2O₂(g) ⟶ 2CO₂(g); ΔH = -786 kJ

Equation (IV) has 2CO₂ on the right, and that is not in the target equation.

You need an equation with 2CO₂ on the left, so you reverse Equation (II).  

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

(V)  2CO₂(g) → 2CO(g) + O₂(g); ΔH = 566 kJ

Now, you add equations (IV) and (V), 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 (III):

(IV) 2C(s) + 2O₂(g) ⟶ 2CO₂(g);      ΔH = -786 kJ

(V) 2CO₂(g) → 2CO(g) + O₂(g);       ΔH =   566 kJ

(III) 2C(s) +O₂ ⟶ 2CO;                    ΔH = -220 kJ

$\Delta H \text{ for the reaction is } \boxed{\textbf{-220 kJ}}$