Question bank — Hess's law — enthalpy is a state function; enthalpy cycles

False — heat $q$ is a path function in general; it only equals a state-function change under fixed conditions. At constant pressure $q_p=\Delta H$ and at constant volume $q_V=\Delta U$, and those are state functions. Change the path (e.g. do work along the way) and $q$ changes.

False — $\Delta H$ depends only on the initial and final states, not on rate or mechanism. Speed is kinetics; enthalpy change is thermodynamics. Same start and end ⟹ same $\Delta H$.

False — a catalyst lowers the activation barrier (an intermediate hump), but the reactant and product states are unchanged, so $\Delta H=H_{\text{final}}-H_{\text{initial}}$ is identical. It changes how fast, not how far down.

True — forming an element from itself has identical start and end states, so $\Delta H=0$ by the definition of formation enthalpy. See Standard states and conventions for what "standard state" pins down (e.g. graphite, not diamond, for carbon).

False — the standard state of carbon is graphite, so only graphite gets zero. Diamond is a different state of the same element, so its formation from graphite has a small nonzero $\Delta H$.

False — enthalpy is extensive, so doubling the amount doubles the change: $2A\to 2B$ has $\Delta H=-80$ kJ.

False — the steps can be hypothetical; Hess's law works because $H$ is a state function, so any algebraic path that sums to the target gives the right $\Delta H$, whether or not it's the real mechanism.

True — $\Delta H_{Z\to A}=H_A-H_Z=-(H_Z-H_A)$, so the magnitude is unchanged and only the sign flips.

False — spontaneity is decided by Gibbs free energy $\Delta G=\Delta H-T\Delta S$, not $\Delta H$ alone. An exothermic reaction can be non-spontaneous if entropy falls enough at high $T$.

You must first flip any reaction whose species sits on the wrong side (flip the sign too) and scale to match coefficients, then add. Raw addition only works if every species already cancels correctly.

When you scale the equation ×3 you must scale $\Delta H$ ×3, because $\Delta H$ is extensive — three times the stuff releases three times the heat. Balancing atoms and scaling $\Delta H$ travel together.

That $+498$ is the bond dissociation energy (atomising $O_2$ into $2O$), not the formation enthalpy. ${\Delta }_f{H}^{\circ }(O_2)=0$ because $O_2$ is the standard state of oxygen. See Bond enthalpies vs Enthalpy of formation.

Liquid and gaseous water are different states with different $H$; the gap is the enthalpy of vaporisation. Mixing phases silently adds/removes that vaporisation term and gives the wrong Enthalpy of combustion.

It's backwards: the correct form is products minus reactants, ${\Delta }_r{H}^{\circ }=\sum {\nu }_p{\Delta }_f{H}_{\text{prod}}^{\circ }-\sum {\nu }_r{\Delta }_f{H}_{\text{react}}^{\circ }$. Their version gives the exact negative of the true answer.

Multiply each formation enthalpy by its stoichiometric coefficient before summing: $2H_{2}O$ contributes $2\times {\Delta }_f{H}^{\circ }(H_{2}O)$. Skipping the 2 undercounts the enthalpy.

Because each intermediate appears once as a "final" of one step and once as an "initial" of the next; in the telescoping sum $\sum (H_{n+1}-H_n)$ those equal terms subtract to zero, leaving only the true endpoints.

CO over-oxidises to $C{O}_2$ before you can isolate the single step, so a clean measurement is impossible. Hess's law lets us subtract two measurable combustions to reach it without ever running the reaction alone.

$\Delta H$ tracks a state function ($H=U+PV$), so it depends only on endpoints; heat also carries the effects of work and process details along the way, which differ path to path unless a constraint (constant $P$) locks $q$ to $\Delta H$.

The first law makes internal energy $U$ a state function (energy is conserved and $\Delta U$ depends only on state); since $H=U+PV$ is built from state functions, $\Delta H$ is path-independent too — which is Hess's law.

Because the cycle closes: the sum of all steps around the loop must be zero (state function), so the one unknown (lattice enthalpy) is forced by all the other measurable steps.

Enthalpy is a state property, so $H_{2}O(l)$ and $H_{2}O(g)$ carry different values; without stating the phase the cancellation of "middlemen" species is not valid and the total is wrong.

Exactly zero, because initial and final states coincide; this is the degenerate case that makes ${\Delta }_f{H}^{\circ }$ of an element (formed from itself) equal zero.

Zero — $\Delta H+(-\Delta H)=0$, consistent with returning to the starting state; any closed loop of steps sums to zero.

No — cancellation requires the same species in the same state. Different phases don't cancel; you'd need an explicit vaporisation/condensation step to bridge them.

Yes — scaling by any factor (including $\frac{1}{2}$) is legal; $\Delta H$ scales by the same factor because it's extensive. This is exactly what the $\frac{1}{2}{O}_2$ in the CO example does.

No contradiction — reversing an exothermic step turns its contribution positive, so the aligned sum can be positive even when every printed step is negative.