If $Delta H = Q + W$ (assuming continuous conditions), climate there room two terms connected in calculating $Delta H$, only among which measures warmth gained/released. The is feasible then for $Delta H$ to be an unfavorable (if $W$ were really negative), also with $Q$ gift positive? (And angry versa too.)

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


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Your initial equation is incorrect. Because that 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 also W is the occupational done by the surroundings on the system, $W=-intPdV$. If the procedure takes place at a constant pressure, then $W=-PDelta V$, and the adjust in internal energy becomes:$$Delta U=Q-PDelta V$$. But, indigenous the meaning of enthalpy, we have $Delta H=Delta U+Delta (PV)$. So, finally, $$Delta H=Q$$So, because that a procedure carried the end at constant pressure, if the heat included to the mechanism is positive (endothermic), $Delta H$ is positive and also if the heat added to the mechanism is negative (exothermic, heat removed indigenous system), $Delta H$ is negative.

You are watching: If delta h is negative exothermic


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

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

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

$$H_2

It is feasible then for $ΔH$ come be negative (if $W$ were very negative), even with $Q$ gift positive? (And angry versa too.)

Yes.


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