Should temperature and pressure conditions for two reactions be same for them to combine for enthalpy calculations? I was solving a question in which I had to calculate Enthalpy of formation of $\ce{CS_2(l)}$ using the given Enthalpy of combustion (not in standard state) of $\ce{C(s)}$, $\ce{S(s)}$ and $\ce{CS_2(l)}$.
To calculate the $\Delta H_f$ of $\ce{CS_2(l)}$, we combine the corresponding equation of combustion of these chemicals  in appropriate order to form the equation :
$$\ce{C(s) + 2S(g) -&gt; CS_2(l) }$$
Should the temperature and pressure conditions of the combustion equations be same for me to combine them mathematically?
yes they should be for a simple reason: when you do make C and S react together  you are going to assume that the temperature and the pressure of the box you're using for the reaction is the same. In particular you can even think of putting for example the hot solid in the box with the cold gas, when you're talking of entalphy of formation you're basically speaking of the equilibrium situation. You will reach equilibrium only once al the macroscopical quantities are perfectly equal in all the box you're working with. If this was not the case, the equations you would have to solve are much more complicated than those you are usually seeing .