Given below are two reactions, involved in the commercial production of dihydrogen $(\mathrm{H}_{2})$. The two reactions are carried out at temperature "$\mathrm{T}_{1}$" and "$\mathrm{T}_{2}$" respectively $\mathrm{C}(\mathrm{s})+\mathrm{H}_{2}\mathrm{O}(\mathrm{g}) \xrightarrow{\mathrm{T}_{1}} \mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g})$ $\mathrm{CO}(\mathrm{g})+\mathrm{H}_{2}\mathrm{O}(\mathrm{g}) \xrightarrow[\text{Catalyst}]{\mathrm{T}_{2}} \mathrm{CO}_{2}(\mathrm{g})+\mathrm{H}_{2}(\mathrm{~g})$ The temperature $\mathrm{T}_{1}$ and $\mathrm{T}_{2}$ are correctly related as

(A) $\mathrm{T}_{1}>\mathrm{T}_{2}$

(B) $\mathrm{T}_{1}=\mathrm{T}_{2}$

(C) $\mathrm{T}_{1}=100 \mathrm{~K}, \mathrm{T}_{2}=1270 \mathrm{~K}$

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Hydrogen

Given below are two reactions, involved in the commercial production of dihydrogen

(H_{2})

.
The two reactions are carried out at temperature "

T_{1}

" and "

T_{2}

" respectively

C (s) + H_{2} O (g) T_{1} CO (g) + H_{2} (g)

CO (g) + H_{2} O (g) T_{2} Catalyst CO_{2} (g) + H_{2} (g)

The temperature

T_{1}

and

T_{2}

are correctly related as

(A) $T_{1} > T_{2}$ ✓ Correct
(B) $T_{1} = T_{2}$
(C) $T_{1} = 100 K, T_{2} = 1270 K$
(D) $T_{1} < T_{2}$

Answer: (A)

Solution

The first step is the water-gas reaction, in which white-hot coke is treated with steam. It is strongly endothermic and needs a very high temperature, around 1270 K, to run. The carbon monoxide produced is then converted to further hydrogen by the water-gas shift reaction, which is carried out over an iron-chromate catalyst at a distinctly lower temperature, around 673 K, where the equilibrium favours carbon dioxide and hydrogen. Since the first reaction demands a much higher temperature than the catalysed second one,

T_{1}

is greater than

T_{2}

JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Single correct option

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JEE Main · 2023 · 10 Apr 2023 · Shift 1 · Hydrogen

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