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Your trouble comes from the two opposed concepts used in the world to define the work $\pu{p\Delta V}$. For theoretical scientists, the work is positive when work is done on the system, when the gas is compressed by an external force. And in such a compression, the volume decreases, so $\pu{\Delta V}$ is negative. So in order for the work to be positive, ...


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Reason lies in definition of enthalpy of reaction. Enthalpy of reaction is heat exchanged between our system in which reaction happens and surroundings when reaction is carried at constant temperature and pressure. If reaction is exothermic, it releases heat and increases temperature of our system and so to keep it at the same temperature you need to give ...


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Good question. Enthalpy of reaction dependence on temperature is given by Kirchoff's equation (you can check it) and it is true that enthalpy of reaction changes with temperature. For most reactions if temperature change isn't too big, enthalpy of reaction won't change much so we can regard it as constant in that temperature span on which we are plotting lnK ...


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Out of the four equations that you have, you need to find $\Delta H^\circ_\mathrm f$ of $\ce{MgO}$, or rather, the $\Delta H$ value of the reaction $$\ce{Mg(s) + 1/2O2(g) -> MgO(s)} \tag{1} \label{1}$$ Why is this? This is because of the definition of heat of formation is as follows: Standard heat of formation of a compound is the change of enthalpy ...


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why is dH=dE at constant volume? It isn't, and nowhere in the problem or answer is this implied. First of all some definitions. For a combustion reaction, the enthalpy change can be equated with the heat of combustion at constant pressure, whereas the internal energy is the heat of combustion at constant volume: $$\Delta U = q_V ~~~~~\text{constant volume} \...


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