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I am asked to find the enthalpy change of formation of the following: $$\ce{N2 + 1/2O2 -> N2O}$$

I am given the following enthalpies of reaction: \begin{align} \ce{C + N2O &-> CO + N2} &\Delta H_f =-193 \tag{1} \\ \ce{C + 1/2O2 &-> CO} &\Delta H_f=-111 \tag{2} \end{align}

How do I calculate enthalpy change of formation of the nitrous oxide?

I thought it would just be adding the two enthalpies together because I substitute the second equation into the first one, but this is wrong.

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If you start by reversing equation 1, you get $$\ce{CO + N2 ->C + N2O~~ {\Delta}H_{f}=+193}$$

Now, add this result with equation 2 and you get $$\ce{CO~+N2~+C~+1/2O2->C~+N2O~+CO~~{\Delta}H_{f}=+193-111=82 }$$

Cancelling like terms that exist on both sides of this equation yields $$\ce{N2~+1/2O2->N2O~~{\Delta}H_{f}=82}$$

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I took the first equation away from the second equation:

\begin{array}{lll} &\ce{C + 1/2O2 &-> CO} &\Delta H_f &= -111 \\ - &\ce{C + N2O &-> CO + N2} &\Delta H_f &= -193 \\ \hline = &\ce{1/2O2 - N2O &-> - N2} &\Delta H_f &= 82 \\ = &\ce{N2 + 1/2O2 &-> N2O } \end{array}

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Try adding, subtracting, multiplying by real numbers and reversing the equations as necessary to get the desired equation.

For example, since your target equation has $\ce{N_2}$ on the reactant side, I would flip the first given equation so that diatomic nitrogen ends up on the reactant side.

Then note that if you have the same molecule on both sides of the equation in equal quantities, you may remove it from the equation since these species will not affect the enthalpy of the reaction.

(Side note) Catalysts are similarly not consumed by a chemical reaction, yet they definitely affect the rate (kinetics) of the reaction.

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  • $\begingroup$ What then after flipping the target equation? $\endgroup$
    – Cobbles
    May 31, 2014 at 17:33
  • $\begingroup$ As you stated in your question, you can add enthalpies. This correctly implies you can add equations. $\endgroup$
    – Dissenter
    May 31, 2014 at 17:38
  • $\begingroup$ You flipped the target equation, but what enthalpies do you add? So far you've only flipped one equation. $\endgroup$
    – Cobbles
    May 31, 2014 at 17:42
  • $\begingroup$ I didn't flip the target equation. You add the given enthalpies with signs taken into consideration. Is this a HW question? $\endgroup$
    – Dissenter
    May 31, 2014 at 17:51
  • $\begingroup$ its fine i found a way to do it $\endgroup$
    – Cobbles
    May 31, 2014 at 17:54

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