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I have been studying some basic chapter of thermodynamics in chemistry and physics when I came across these two statements: $$\Delta Q=\Delta U+\Delta W$$ $$\Delta H=\Delta U+P\Delta V$$ So is it true $\Delta Q=\Delta H$ (since $\Delta W=P\Delta V$)

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  • $\begingroup$ Yes, chemists often refer to the 'heat of reaction' while physicists tend to call it the 'enthalpy of reaction'. But it's the same thing. $\endgroup$ – lemon Mar 4 '18 at 18:50
  • $\begingroup$ @lemon thanks sir actually the boss i am study are not clarifying things well $\endgroup$ – user181463 Mar 4 '18 at 18:51
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    $\begingroup$ Actually they are not the same thing. Enthalpy is a state variable. Heat is not. The only time they are the same idea is when the system is under constant pressure. Then the heat will be equal to the change in enthalpy. So for these purposes you can think of them as the same thing, but it's not always the case. $\endgroup$ – Aaron Stevens Mar 4 '18 at 19:21
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To supplement Aaron's comment, we can always write Q=ΔU+W, where Q is heat transfer to the system, U is the internal energy of the system, and W is the work done by the system. (It doesn't make much sense to add Δ to Q or W because they're not state functions whose differences are relevant.)

We can also always write H=U+PV for all states by definition, where H is the enthalpy. If the pressure is constant, then we can integrate the differential form dH=dU+PdV+VdP=dU+PdV between two states to obtain ΔH=ΔU+PΔV. In such a case, the heat transfer Q into the system indeed corresponds to ΔH.

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