26)Based on the following equations: 2 CO2(g) → O2(g) 2 CO(g) ΔH° = 566.0 kJ 2 C(graphite) O2(g) → 2 CO(g) ΔH° = - 221.0 kJ Determine the enthalpy change (ΔH°) for the following reaction: C(graphite) O2(g) → CO2(g) A)393.5 kJ/mol B)- 393.5 kJ/mol C)345.0 kJ/mol D)- 787.0 kJ/mol E)172.5 kJ/mol

To determine the enthalpy change (ΔH°) for the reaction, we will use Hess's Law, which states that the total enthalpy change for a reaction is the same whether it occurs in one step or in a series of steps.

So, let's write down the given equations:

[tex]2CO_{2}(g)[/tex] ⇒ [tex]2CO_{2}(g) + O_{2} (g)[/tex]
[tex]2C(graphite) + O_{2}[/tex] ⇒ [tex]2CO(g)[/tex]

Now, the desired reaction:

[tex]C(graphite) + O_{2}[/tex] ⇒ [tex]CO_{2}(g)[/tex]

Now that we have everything, we can start solving.

Flip the second given equation and add the first and second equations to cancel out [tex]CO_{2}(g)[/tex] and [tex]O_{2}(g)[/tex]:

[tex]2CO_{2}(g)[/tex] ⇒ [tex]2CO(g) + O_{2}(g)[/tex]
[tex]-2C(graphite) - O_{2}(g)[/tex] ⇒ [tex]-2CO_{2}(g)[/tex]

Solve:

3. [tex]2CO_{2}(g) - 2C(graphite) = 0[/tex]

Sum the enthalpy changes:

ΔH[tex]^[/tex]°3 = ΔH[tex]^[/tex]°1 + ΔH[tex]^[/tex]°2

ΔH[tex]^[/tex]°3 = 566 kJ - 221 kJ

ΔH[tex]^[/tex]°3 = 345 kJ

So, the enthalpy change is C) 345.0 kJ/mol.