Question

Calculate the enthalpy change,,ΔH=∑ Hp - ∑ HR for the following reaction using equations 1, 2 C graphite(s) --> C diamond(s) The following is known. 1. Cgraphite(s) + O2(g) --> CO2(g) ∆H = -394 kJ 2. Cdiamond(s) + O2(g) --> CO2(g) ∆H = -396 kJ A. -763kJ B. 2kJ C. 790kJ D. -2kJ

Answer 1

Answer:

$\large \boxed{ \text{B. 2 kJ}}$

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

$\Delta H \text{ for the reaction is \large \boxed{\textbf{2 kJ/mol}} }$