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Given

$$\ce{2 SO2(g) + O2(g) <=> 2 SO3(g)} \quad \Delta_\mathrm{r}H = \pu{-196.0 kJ mol^-1}$$

and $\Delta_\mathrm{f}H(\ce{SO2}) = \pu{-296.0 kJ mol^-1},$ find $\Delta_\mathrm{f}H(\ce{SO3}).$

I have chosen to solve this by constructing a Hess cycle and have come up with the following:

Hess cycle

This method has lead me to the correct procedure of solving the equation: $$\Delta_\mathrm{r}H = -2\Delta_\mathrm{f}H(\ce{SO2}) + 2\Delta_\mathrm{f}H(\ce{SO3})$$

However, it is not entirely complete in my opinion and I am unsure on how to proceed in completing it, specifically on what I should place below the two arrows.

I am aware that it should be some combination of $\ce{S + O2}$ but am unaware on how exactly the coefficients would work.

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    $\begingroup$ Sounds like you are overthinking the problem: there is no need to go back to the elements. Also, please note that enthalpy is not a dimensionless quantity (see the edit). $\endgroup$ – andselisk Oct 1 '20 at 20:51
  • $\begingroup$ @andselisk but if I am showing the $\ce{\Delta H_f}$ shouldn't I also show the reaction for it aswell? $\endgroup$ – Filthyscrub Oct 2 '20 at 19:51
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You did the right thing. However the method you used sounds confusing sometimes, it gave me a hard time too. Hess's law says that the resultant enthalpy change in a reaction is same whether it occurs in one or several steps .

We can use that to our advantage by assuming a hypothetical situation where the given reaction proceed in a number of steps for our convinience.

In your example, we disassociated $\ce{SO_2}$ molecule to corresponding standard state constituents and the enthalpy change associated with it will be $-\Delta H_f(\ce{SO_2})$. $$\ce{2SO_2(g) + O_2(g) ->[-2\Delta H_f(SO_2)] 2S(s) + 2O_2(g) + O_2(g)->[2\Delta H_f(SO_3)] 2SO_3(g)}$$ I took the following reaction through above mentioned 2 steps $$\ce{2SO_2(g) + O_2(g)->[\Delta_r H] 2SO_3(g)}$$ According to Hess law, the Enthalpy of reaction $\Delta_r H$ will be the summation of the all enthalpy changes occuring in the hypothetical intermediate steps. $$\Delta_r H = -2\Delta H_f(\ce{SO_2}) + 2\Delta H_f(\ce{SO_3})$$

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  • $\begingroup$ In the first reaction where the $SO_2$ is converted into free state elements, you didn't consider the chemical changes of the molecular oxygen, why? $\endgroup$ – Buraian Oct 7 '20 at 17:36
  • $\begingroup$ Heat of formation is the amount of heat absorbed or evolved when one mole of a compound is formed from its constituent elements, each substance being in its standard state. $\ce O_2$ is already in it's standard state and by convention, the enthalpy of substances in their standard state is taken as zero. $\endgroup$ – Eyy boss Oct 7 '20 at 17:57
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    $\begingroup$ Oh nice, that makes sense. Thank you! $\endgroup$ – Buraian Oct 7 '20 at 17:58

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