Question bank — Enthalpy H = U + PV; ΔH for reactions at constant P

False — the first clause holds when $U>0$ (then $P,V>0\Rightarrow PV>0\Rightarrow H=U+PV>U$), but the conclusion is a non-sequitur: what compares $\Delta H$ and $\Delta U$ is their difference $\Delta H-\Delta U=\Delta (PV)$, whose sign is set by whether gas moles rise or fall, so $\Delta H$ can be greater than, less than, or equal to $\Delta U$.

False — it holds only at constant pressure (that is what the $P$ subscript on $q$ demands); at constant volume the heat is $q_V=\Delta U$ instead, because no $PV$ expansion work is done.

True — "exothermic" is defined as heat leaving the system, and $\Delta H$ is measured from the system's viewpoint, so it must be negative.

False — a hot beaker means the surroundings gained heat, so the system lost it; that is exothermic, $\Delta H<0$.

True — condensed phases barely change volume, so $\Delta (PV)\approx 0$ and the two energy changes nearly coincide.

False — being a state function is exactly the reason $\Delta H$ depends only on initial and final states, which is what makes Hess's Law work.

False — $\Delta H^{\circ }$ is pinned to standard conditions (1 bar, 298 K, 1 M); $\Delta H$ at other conditions can differ.

False — a bomb is rigid (constant volume), so it measures $q_V=\Delta U$; you must add $\Delta (n_{\text{gas}})RT$ (assuming ideal gas, constant $T$) to reach $\Delta H$.

Error: with $\Delta V=V_{\text{final}}-V_{\text{initial}}$, expansion means volume increases, so $\Delta V>0$; then $w=-P\Delta V<0$, meaning the system loses energy doing positive work on the surroundings.

Error: only gaseous moles count in $\Delta n_{\text{gas}}$ (the formula rests on $PV=nRT$, which applies to the gas phase), so here $\Delta n_{\text{gas}}=1-0=+1$.

Error: the sign of $\Delta U$ depends on $\Delta (PV)$; a strongly negative $\Delta (PV)$ could in principle leave $\Delta U$ with a different magnitude or sign, so you cannot assert its sign without computing.

Error: conservation makes them equal and opposite, $\Delta H_{\text{system}}=-\Delta H_{\text{surroundings}}$; heat leaving one enters the other.

Error: moles go from $4\to 2$, a decrease, so $\Delta V<0$; the surroundings compress the system, and work is done on it ($w>0$), not by it.

Error: spontaneity is decided by $\Delta G=\Delta H-T\Delta S$ (see Entropy and Gibs Free Energy), not by $\Delta H$ alone; entropy and temperature also matter.

Error: the standard formation enthalpy of an element in its reference form is zero by definition (see Standard Enthalpy of Formation).

Because most reactions run in open vessels at constant atmospheric pressure, where the measured heat $q_P$ is $\Delta H$ directly, sparing you from tracking expansion work separately.

Track the differentials (recall $d$ = exact, $\delta$ = path-dependent, from the intuition box above). The First Law reads $dU=\delta q+w$, and for $PV$ work $w=-PdV$, so $dU=\delta q-PdV$. Differentiating the definition $H=U+PV$ gives $dH=dU+PdV+VdP=\delta q+VdP$, so at constant $P$ ($dP=0$) we get $dH=\delta q$, i.e. $\Delta H=q_P$. The $+PV$ was chosen precisely so the $-PdV$ term cancels; $U-PV$ would give $dH=\delta q-2PdV$ at constant $P$, doubling the work instead of cancelling it.

When gas moles fall, $\Delta V<0$ and the surroundings do work on the system ($w>0$); part of $\Delta H$'s drop is that inward push, so $\Delta U=\Delta H-\Delta (PV)$ comes out closer to zero.

Bond enthalpies are averages over many molecules and ignore phase changes and the $PV$ contribution, so they approximate rather than reproduce the exact $\Delta H$.

Because it comes from $PV=nRT$, which is a gas law. Quantitatively, one mole of ideal gas at 298 K, 1 bar occupies about 24.8 L, whereas one mole of liquid water is only ~0.018 L — over a thousand times smaller — so the condensed-phase $PV$ change is negligible next to the gas term (see PV Work and Expansion and the size comparison in Figure 1).

Because $\Delta H^{\circ }$ scales with how the equation is balanced; doubling the coefficients doubles $\Delta H^{\circ }$, so the value is meaningless without its stoichiometry.

Then $\Delta (PV)=0$, so $\Delta H=\Delta U$ exactly — the enthalpy and internal-energy changes are identical.

It breaks; with $\Delta V=0$ no $PV$ work is done, so the measured heat is $q_V=\Delta U$, and you recover $\Delta H$ afterward via the calorimetry correction $\Delta H=\Delta U+\Delta n_{\text{gas}}RT$ (ideal gas, constant $T$).

Yes — at high enough $T$ the $-T\Delta S$ term outweighs the positive $\Delta H$, making $\Delta G<0$; this is why ${\text{CaCO}}_3$ decomposes only when strongly heated.

It vanishes, since $RT=0$, so $\Delta H=\Delta U$ regardless of how many gas moles change — the temperature factor switches off the work term.

No — that shortcut assumes a single constant $T$; when $T$ changes appreciably the $PV$ product shifts for all gas moles (not just $\Delta n_{\text{gas}}$), so you must evaluate $PV$ at the actual initial and final states rather than plug in one $RT$.

It equals the heat absorbed at constant $P$, i.e. $q_P$ — the latent heat of fusion — because melting is a constant-pressure process where $q_P=\Delta H$ even though no chemical bonds break.

In principle yes, if $\Delta (PV)$ is large enough to flip the sum's sign; in practice for most reactions $\Delta (PV)$ is small (a few kJ), so they usually share a sign.