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!

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

What reaction?

∆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`

I sincerely apologize. Here's the reaction:

2C(s) + O2(g)→2CO(g)

Oh, chemistry! That's where all the reactions happen. Let's get those numbers dancing, shall we?

First, we need to balance the equation. Let's assume we have one mole of carbon and one mole of carbon monoxide:

C + CO → CO2

Since the enthalpy of combustion of solid carbon is -393.7 kJ/mol and the enthalpy of combustion of carbon monoxide is -283.3 kJ/mol, we can plug in these values:

∆H = Σ(∆H(products)) - Σ(∆H(reactants))
∆H = (∆H(CO2)) - (∆H(C) + ∆H(CO))

∆H = (-393.7 kJ/mol) - (-393.7 kJ/mol + -283.3 kJ/mol)

∆H = -393.7 kJ/mol + 393.7 kJ/mol + 283.3 kJ/mol

Cancelling out the kJ/mol terms, we get:

∆H = -283.3 kJ/mol

Voila! The enthalpy change for the reaction is -283.3 kJ/mol. That's quite the fiery reaction, isn't it? Keep those chemical equations burning!

To calculate ∆H for the reaction, we need to use the concept of Hess's Law, which states that the enthalpy change of a reaction depends only on the initial and final states and is independent of the pathway taken.

The balanced equation for the reaction is:

C (s) + CO (g) → 2 CO2 (g)

According to Hess's Law, we can manipulate the given enthalpy values to obtain the enthalpy change (∆H) for the reaction.

Step 1: Reverse the equation

Since we have the enthalpy change for the combustion of carbon monoxide, we need to reverse the equation to match the reaction we want:

CO2 (g) → CO (g)

Step 2: Calculate the enthalpy change for the reversed reaction

The enthalpy change for the reversed reaction is the negative of the enthalpy change for the combustion of carbon monoxide:

∆H1 = -(-283.3 kJ/mol) = 283.3 kJ/mol CO

Step 3: Multiply the equation by the appropriate factor

In the reversed reaction, one mole of carbon dioxide is formed, but in the desired reaction, two moles of carbon dioxide are formed. Therefore, we need to multiply the reaction by a factor of 2:

2 CO2 (g) → 2 CO (g)

Step 4: Calculate the enthalpy change for the desired reaction

The enthalpy change for the desired reaction is:

∆H2 = 2 * ∆H1 = 2 * 283.3 kJ/mol CO = 566.6 kJ/mol CO2

Therefore, the ∆H for the reaction C (s) + CO (g) → 2 CO2 (g) is 566.6 kJ/mol CO2.