There are only three ways we can increase its internal energy (which is the sum of its internal potential energy and internal kinetic energy): flow heat into it; do work on it; or add matter to it (and the converse for decreasing its internal energy). If the system is closed (no matter can flow in or out), we are left with only the first two, which naturally leads to the following equation:

ΔU=q+w

Now, suppose, It turns out that we can construct such a function by adding pV to U. We call that function the enthalpy, H:

H=U+pV=>ΔH=ΔU+Δ(pV)=ΔU+pΔV+VΔp

At constant pressure, Δp=0, so:

ΔHp=ΔU+pΔV=qp+wp+pΔV

But at constant p, psys=pext. Thus, if we only have pV work:

wp=−pextΔV=−pΔV

Hence:

ΔHp=qp−pΔV+pΔV=qp

Let's compare a reaction at constant pressure with one at constant volume. At constant V, with pV-work only, qV=ΔU (because, with pV-work only, wV=0). Suppose the reaction is exothermic, and the volume of the products is greater than that of the reactants, i.e., ΔV>0. You can see from the equations that ΔH will be less negative than ΔU, i.e., less heat is evolved at constant pressure.