If $Delta H = Q + W$ (assuming continuous conditions), climate there room two terms affiliated in calculating $Delta H$, only among which measures warmth gained/released. It is possible then because that $Delta H$ come be an adverse (if $W$ were very negative), even with $Q$ gift positive? (And vice versa too.)

So why carry out we say that the authorize of $Delta H$ shows whether a reaction is exo- or endo- thermic?


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Your initial equation is incorrect. For a process occurring in a closed system, the equation should read$$Delta U=Q+W$$where U is the internal energy of the system, Q is the heat included to the system, and W is the work done by the next site on the system, $W=-intPdV$. If the process takes location at a consistent pressure, then $W=-PDelta V$, and also the adjust in internal energy becomes:$$Delta U=Q-PDelta V$$. But, native the meaning of enthalpy, we have actually $Delta H=Delta U+Delta (PV)$. So, finally, $$Delta H=Q$$So, for a procedure carried the end at consistent pressure, if the heat added to the system is optimistic (endothermic), $Delta H$ is positive and also if the heat added to the device is an adverse (exothermic, warm removed indigenous system), $Delta H$ is negative.

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The prize $Delta$ (here) simply way difference, e.g.:$$Delta H=H_2-H_1$$

So, enthalpy of the end state minus enthalpy that the initial state.

In the case of an exothermic reaction the system has actually lost enthalpy, so:

$$H_2

It is feasible then for $ΔH$ to be negative (if $W$ were an extremely negative), also with $Q$ gift positive? (And evil versa too.)

Yes.


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