Question

Can someone answer this question?

The enthalpy of combustion of solid carbon to form carbon dioxide is -393.7 kJ/mol carbon, and the enthalpy of combustion of carbon monoxide to form carbon dioxide is -283.3 kj/mol CO. Use these data to calculate ∆H for the reaction.

Thank you! I appreciate it!

Answer 1

What reaction?

Answer 2

∆H for what reaction...?

Write the balanced reactions:
C + O2 ---> CO2 ; ∆H = -393.7 kJ/mol C

CO + 1/2 O2 ---> CO2 ; ∆H = -283.3 kJ/mol CO

If the problem asks for ∆H of formation for CO, we can make the second reaction become:
CO2 ---> CO + 1/2 O2 ; ∆H = +283.3 kJ/mol CO

And add this to the first equation:
CO2 ---> CO + 1/2 O2 ; ∆H = +283.3 kJ/mol
C + O2 ---> CO2 ; ∆H = -393.7 kJ/mol
--------------------------------------------------
C + 1/2 O2 ---> CO ; ∆H = -110.4 kJ/mol

hope this helps? `u`

Answer 3

I sincerely apologize. Here's the reaction:

2C(s) + O2(g)→2CO(g)

Answer 4

If the stoichiometric coefficients are multiplied by some number, then the value for ∆H (in kJ/mol) is also multiplied by that number.

We got earlier the ∆H formation for CO with fraction stoichiometric coefficients:
C + 1/2 O2 ---> CO ; ∆H = -110.4 kJ/mol

And we need ∆H for 2C(s) + O2(g) → 2CO(g). We just multiply everything by 2:

2 * [ C + 1/2 O2 ---> CO ; ∆H = -110.4 kJ/mol ]
= 2C(s) + O2(g) → 2CO(g) ; ∆H = -220.8 kJ