Exercises — Preparation, properties, uses of dihydrogen

(a) False. Concentrated ${\text{H}}_2{\text{SO}}_4$ is an oxidising acid — it oxidises any ${\text{H}}_2$ formed back to water and gives off ${\text{SO}}_2$ instead. We always use dilute acid. (b) True. In a metal hydride the metal is the electron donor, so hydrogen accepts an electron and becomes ${\text{H}}^{-}$. That is why ${\text{CaH}}_2+{\text{H}}_2\text{O}$ works as an acid–base reaction: ${\text{H}}^{-}+{\text{H}}^{+}\to {\text{H}}_2$. (c) False. Step 1 (reforming) is endothermic ($+206\text{kJ/mol}$), step 2 (shift) is exothermic ($-41\text{kJ/mol}$); the sum is $+165\text{kJ/mol}$ — still endothermic. External heat is required.

At the cathode (reduction): $2{\text{H}}_2\text{O}+2e^{-}\to {\text{H}}_2+2{\text{OH}}^{-}$. Two electrons produce one ${\text{H}}_2$ molecule. This number — call it $z=2$, the electrons per molecule in Faraday's law (Tool 3) — is the one we plug in later; losing it is the single most common electrolysis error.

Step 1 — Equation (WHAT/WHY): we need the mole ratio, because balanced equations count in moles not grams. $\text{Zn}+2\text{HCl}\to {\text{ZnCl}}_2+{\text{H}}_2$ One Zn gives one ${\text{H}}_2$ (ratio $1:1$). Step 2 — Moles of Zn (use Tool 1, $n=m/M$): we divide the mass we have by the mass of one mole so that grams cancel and a count of moles remains. $n_{\text{Zn}}=\frac{6.5\text{g}}{65\text{g/mol}}=0.1\text{mol}$ Step 3 — Moles of ${\text{H}}_2$: the balanced $1:1$ ratio means each mole of Zn hands us one mole of ${\text{H}}_2$, so $n_{{\text{H}}_2}=0.1\text{mol}$. Step 4 — Volume at STP (use Tool 2, $V=n\times 22.4$): at STP one mole of any gas fills $22.4\text{L}$, so multiplying moles by that molar volume converts a mole count into litres. $V=0.1\times 22.4=\boxed{2.24\text{L}}$

Step 1 — Charge (use $Q=It$): current is coulombs per second, so we first turn minutes into seconds, then multiply rate by time to get total charge. $Q=It=2\times (30\times 60)=2\times 1800=3600\text{C}$ Step 2 — Moles of ${\text{H}}_2$ (use Tool 3, Faraday's law): dividing charge $Q$ by $F$ gives moles of electrons; dividing again by $z=2$ (electrons per ${\text{H}}_2$ molecule) gives moles of ${\text{H}}_2$. $n=\frac{Q}{zF}=\frac{3600}{2\times 96485}=0.018656\text{mol}$ Step 3 — Mass (Tool 1 in reverse, $m=n\times M$): multiplying a mole count by grams-per-mole gives back grams. $m=n\times M=0.018656\times 2=\boxed{0.0373\text{g}}$

Step 1 — Equation: note each ${\text{CaH}}_2$ yields two ${\text{H}}_2$ (one from the hydride ions, one from water's hydrogens). ${\text{CaH}}_2+2{\text{H}}_2\text{O}\to {\text{Ca(OH)}}_2+2{\text{H}}_2$ Step 2 — Moles of ${\text{CaH}}_2$ (Tool 1, $n=m/M$): divide grams by grams-per-mole to count moles. $n=\frac{0.84\text{g}}{42\text{g/mol}}=0.02\text{mol}$ Step 3 — Moles of ${\text{H}}_2$: the balanced $1:2$ ratio doubles the count, so $n_{{\text{H}}_2}=0.04\text{mol}$. Step 4 — Volume (Tool 2, $V=n\times 22.4$ at STP): convert the mole count to litres using the molar volume. $V=0.04\times 22.4=\boxed{0.896\text{L}}$

Step 1 — Overall reaction (WHY: chains both stages): the reforming step and the Water Gas Shift Reaction add together; cancelling the intermediate $\text{CO}$ gives the net equation below, in which one ${\text{CH}}_4$ yields four ${\text{H}}_2$. ${\text{CH}}_4+2{\text{H}}_2\text{O}\to {\text{CO}}_2+4{\text{H}}_2$ Step 2 — Moles of ${\text{CH}}_4$ (Tool 1, $n=m/M$): convert the fed mass to a mole count so the ratio applies. $n_{{\text{CH}}_4}=\frac{80000\text{g}}{16\text{g/mol}}=5000\text{mol}$ Step 3 — Theoretical moles of ${\text{H}}_2$: the $1:4$ ratio multiplies the count by four. $n_{{\text{H}}_2}=4\times 5000=20000\text{mol}$ Step 4 — Theoretical mass ($m=n\times M$): turn the mole count back into mass. $m=20000\times 2=40000\text{g}=40\text{kg}$ Step 5 — Apply efficiency: only $75\%$ of the feed actually converts, so scale the ideal answer down. $m_{\text{actual}}=40\times 0.75=\boxed{30\text{kg/hour}}$

Step 1 — Moles of ${\text{H}}_2$ wanted (Tool 2 run backwards, $n=V/22.4$): at STP, dividing a gas volume by the molar volume $22.4\text{L/mol}$ counts how many moles that volume holds. $n_{{\text{H}}_2}=\frac{67.2}{22.4}=3\text{mol}$ Step 2 — Mole ratio (from balanced equation): the equation shows $2\text{Al}:3{\text{H}}_2$, so $n_{\text{Al}}=\frac{2}{3}\times 3=2\text{mol}$ Step 3 — Mass (Tool 1 in reverse, $m=n\times M$): multiply the mole count by grams-per-mole. $m=2\times 27=\boxed{54\text{g}}$

(a) Moles of ${\text{H}}_2$ (Tool 3): first the charge, then Faraday's law with $z=2$. $Q=10\times 3600=36000\text{C},n_{{\text{H}}_2}=\frac{36000}{2\times 96485}=0.18656\text{mol}$ (b) Volume of ${\text{H}}_2$ (Tool 2, at STP): $V=0.18656\times 22.4=4.179\text{L}$ (c) Volume of ${\text{O}}_2$: at the anode, oxygen is oxidised out of water. Written in the reduction direction so we can compare electron counts cleanly: ${\text{O}}_2+4{\text{H}}^{+}+4e^{-}\to 2{\text{H}}_2\text{O}$ Reading it as the reverse (the direction that actually happens at the anode, $2{\text{H}}_2\text{O}\to {\text{O}}_2+4{\text{H}}^{+}+4e^{-}$) tells us that building one ${\text{O}}_2$ releases 4 electrons, i.e. $z=4$ for oxygen. Same charge $36000\text{C}$ flows, but now we divide by $z=4$: $n_{{\text{O}}_2}=\frac{36000}{4\times 96485}=0.09328\text{mol},V_{{\text{O}}_2}=0.09328\times 22.4=2.089\text{L}$ Ratio explained: $V_{{\text{H}}_2}:V_{{\text{O}}_2}=4.179:2.089=2:1$. This matches $2{\text{H}}_2\text{O}\to 2{\text{H}}_2+{\text{O}}_2$ — twice as much hydrogen gas by volume, because ${\text{H}}_2$ needs only half the electrons per molecule ($z=2$) that ${\text{O}}_2$ does ($z=4$), so the same electron flow builds twice as many ${\text{H}}_2$ molecules. The figure below shows the two collected volumes side by side.

First, the one idea we borrow from thermodynamics: $\Delta H$ is the heat content change per the way the equation is written. A positive $\Delta H$ means the reaction absorbs heat (endothermic), so a plant running it must supply that heat from outside. Here the $+165\text{kJ}$ is quoted per mole of ${\text{CH}}_4$, so we must count moles of ${\text{CH}}_4$, not ${\text{H}}_2$. Step 1 — Moles of ${\text{H}}_2$ per hour (Tool 1 reverse, $n=m/M$): $n_{{\text{H}}_2}=\frac{8000\text{g}}{2\text{g/mol}}=4000\text{mol}$ Step 2 — Moles of ${\text{CH}}_4$: the ratio $1{\text{CH}}_4:4{\text{H}}_2$ means dividing by four. $n_{{\text{CH}}_4}=\frac{4000}{4}=1000\text{mol}$ Step 3 — Heat (multiply moles of ${\text{CH}}_4$ by $\Delta H$): endothermic, so this heat must be added. $Q_{\text{heat}}=1000\times 165=\boxed{165000\text{kJ/hour}=1.65\times 10^5\text{kJ/hour}}$

Step 1 — Moles of each reactant: mass → moles for Zn (Tool 1); for the acid, molarity × volume-in-litres gives moles directly. $n_{\text{Zn}}=\frac{13}{65}=0.2\text{mol},n_{\text{HCl}}=0.100\text{L}\times 1\text{mol/L}=0.1\text{mol}$ Step 2 — Reaction demand: $\text{Zn}+2\text{HCl}\to {\text{ZnCl}}_2+{\text{H}}_2$. To consume all $0.2\text{mol}$ Zn you'd need $0.4\text{mol}$ HCl, but only $0.1\text{mol}$ HCl is present. Step 3 — Identify the limiter: HCl runs out first → HCl is limiting, Zn is in excess. Step 4 — ${\text{H}}_2$ from the limiter: $2\text{HCl}:1{\text{H}}_2$, so $n_{{\text{H}}_2}=\frac{0.1}{2}=0.05\text{mol}$ Step 5 — Volume (Tool 2, at STP): $V=0.05\times 22.4=\boxed{1.12\text{L}}$ Why this is L5: naively plugging the $13\text{g}$ of Zn into the $1:1$ shortcut (as in Ex 2.1) gives $4.48\text{L}$ — wrong, because there isn't enough acid. The formula must be told which reactant is scarce.

First recall the one law we borrow, Graham's Law of Diffusion: at the same temperature all gas molecules carry the same average kinetic energy, so the lighter ones must move faster. Quantitatively, effusion rate $\propto 1/\sqrt{M}$ — heavier mass in the denominator means slower escape. Step 1 — Apply the law: $\frac{r_{{\text{H}}_2}}{r_{{\text{O}}_2}}=\sqrt{\frac{M_{{\text{O}}_2}}{M_{{\text{H}}_2}}}=\sqrt{\frac{32}{2}}=\sqrt{16}=4$ Answer: ${\text{H}}_2$ effuses $\boxed{4}$ times faster than ${\text{O}}_2$. This is why stored ${\text{H}}_2$ leaks out of containers so readily — its tiny mass makes it the fastest-moving common gas.

Context in one line (from the hydrogen-economy discussion): the theoretical minimum voltage to split water is $1.23\text{V}$; any real cell needs more to overcome overpotential and resistance, and that excess is lost as heat — this is the core reason "green" ${\text{H}}_2$ is costly. (a) Mass of ${\text{H}}_2$ (Tool 3, voltage irrelevant): only charge matters for amount. $Q=50\times (2\times 3600)=360000\text{C},n=\frac{360000}{2\times 96485}=1.8656\text{mol}$ $m=1.8656\times 2=\boxed{3.731\text{g}}$ (b) Energy analysis: energy $=$ charge $\times$ voltage. The extra voltage above $1.23\text{V}$ does no chemistry — it is dissipated as heat.

• Actual electrical energy $=Q\times V=360000\times 2.0=720000\text{J}$
• Thermodynamic minimum $=Q\times 1.23=360000\times 1.23=442800\text{J}$
• Wasted fraction: $\frac{720000-442800}{720000}=\frac{277200}{720000}=0.385=\boxed{38.5\%}$ Meaning: about $38.5\%$ of the electricity does not end up as chemical bond energy — it is dissipated overcoming overpotential and cell resistance.

Ex 2.1 volume of ${\text{H}}_2$ from $6.5\text{g}$ Zn ::: $2.24\text{L}$ Ex 2.3 volume from $0.84\text{g}$ ${\text{CaH}}_2$ ::: $0.896\text{L}$ Ex 3.1 ${\text{H}}_2$ from $80\text{kg}$ ${\text{CH}}_4$ at $75\%$ ::: $30\text{kg/h}$ Ex 3.2 Al needed for $67.2\text{L}$ ${\text{H}}_2$ ::: $54\text{g}$ Ex 4.1 ${\text{H}}_2:{\text{O}}_2$ volume ratio ::: $2:1$ Ex 5.1 real ${\text{H}}_2$ volume (limiting HCl) ::: $1.12\text{L}$ Ex 5.2 how much faster ${\text{H}}_2$ effuses than ${\text{O}}_2$ ::: $4\times$ Ex 5.3 wasted energy fraction ::: $38.5\%$