How can production of hydrogen from water gas be increased by using water gas shift reaction?
Water gas is produced when superheated steam is passed over red hot coke or coal at 1270 K in presence of nickel as catalyst.
$$\begin{aligned} & \mathrm{C}(\mathrm{s}) \\ & \text { Coke } \end{aligned}+\underset{\text { Steam }}{\mathrm{H}_2 \mathrm{O}(\mathrm{g})}+121.3 \mathrm{~kJ} \xrightarrow[\text { Nickel }]{1270 \mathrm{~K}} \underbrace{\mathrm{CO}(\mathrm{g})+\mathrm{H}_2(\mathrm{~g})}_{\text {Water gas }}$$
It is inconvinient to obtain pure $\mathrm{H}_2$ from water gas as CO is difficult to remove. Hence, to increase the production of $\mathrm{H}_2$ from water gas, CO is oxidised to $\mathrm{CO}_2$ by mixing it with more steam and passing the mixture over $\mathrm{FeCrO}_4$ catalyst at 673 K .
$\underbrace{\mathrm{CO}(g)+\mathrm{H}_2(g)}_{\text {Water gas }}+\mathrm{H}_{\text {Steam }}^{\mathrm{H}_2 \mathrm{O}}(\mathrm{g}) \xrightarrow[\mathrm{FeCrO}_4]{673 \mathrm{~K}} \mathrm{CO}_2(g)+2 \mathrm{H}_2(g)$
This is called water-gas shiff reaction. Carbon dioxide is removed by scrubbing with mixture of sodium arsenite solution or by passing the mixture through water under 30 atm pressure when $\mathrm{CO}_2$ dissolves leaving behind $\mathrm{H}_2$ which is collected.
What are metallic/interstitial hydrides? How do they differ from molecular hydrides?
Metallic/interstitial hydrides are formed by many $d$-block and $f$-block elements. These hydrides conduct heat and electricity.
Unlike saline hydride, they are almost always non-stoichiometric, being deficient in hydrogen. e.g., $\mathrm{LaH}_{2.87}, \mathrm{YbH}_{2.55}, \mathrm{TiH}_{1.5-1.8}, \mathrm{ZrH}_{1.3-1.75}, \mathrm{VH}_{0.56}, \mathrm{NiH}_{0.6-0.7}, \mathrm{PdH}_{0.6-0.8}$ etc. In such hydrides, the law of constant composition does not hold good.
Comparision between molecular and metallic hydrides
| Molecular hydrides | Metallic hydrides |
|---|---|
| These are mainly formed by $p$-block elements and some $s$-block elements (Be and Mg ). | These are formed by group 3, 4, 5 (Sc, Ti, V, Y, Zr, $\mathrm{Nb}, \mathrm{La}, \mathrm{Hf}, \mathrm{Ta}, \mathrm{Ac}$ etc..) 10, 11, 12 (Pd, Cu, Zn etc..) and f-block elements (Ce, Eu, Yb, Th, U etc.) |
| Those are usually volatile compounds having low melting and boiling point. | These are hard, have a metallic lustre. |
| It conduct electricity. | These do not conduct electricity. |
Name the classes of hydrides to which $\mathrm{H}_2 \mathrm{O}, \mathrm{B}_2 \mathrm{H}_6$ and NaH belong.
$\mathrm{H}_2 \mathrm{O}-$ Covalent or molecular hydride (electron rich hydride).
$\mathrm{B}_2 \mathrm{H}_6-$ Covalent or molecular hydride (electron deficient hydride).
NaH - lonic or saline hydride.
If same mass of liquid water and a piece of ice is taken, then why is the density of ice less than that of liquid water?
In ice, molecules of $\mathrm{H}_2 \mathrm{O}$ are not packed so closely as in liquid water. There exists vacant spaces in the crystal lattice. This results in larger volume and lower density (density = mass/volume).
In other words, density of ice is lower than liquid water and hence ice floats on water.
Complete the following equations
(i) $\mathrm{PbS}(s)+\mathrm{H}_2 \mathrm{O}_2(a q) \longrightarrow$
(ii) $\mathrm{CO}(\mathrm{g})+2 \mathrm{H}_2(\mathrm{g}) \xrightarrow[\text { catalyst }]{\text { Cobalt }}$
(i) When PbS react with hydrogen peroxide, then $\mathrm{PbSO}_4$ and water are formed.
$$\mathrm{PbS}(\mathrm{s})+4 \mathrm{H}_2 \mathrm{O}_2(a q) \longrightarrow \mathrm{PbSO}_4+4 \mathrm{H}_2 \mathrm{O}$$
(ii) When carbon mono-oxide reacts with hydrogen in the presence of cobalt catalyst, then methanol is formed.
$$\mathrm{CO}(g)+2 \mathrm{H}_2(g) \xrightarrow[\text { catalyst }]{\text { Cobalt }} \mathrm{CH}_3 \mathrm{OH}(l)$$