An industrial fuel, ‘water gas’ which consists of a mixture of H 2 and CO can be made by…
Chemistry · JEE Advanced · NTA Exams — Equilibrium
An industrial fuel, ‘water gas’ which consists of a mixture of H2 and CO can be made by passing steam over red-hot carbon. The reaction is
\(\mathrm{C}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{~g}) \rightleftharpoons \mathrm{CO}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) ; \Delta \mathrm{H}=+131 \mathrm{~kJ}\)
The yield of CO and H2 at equilibrium would be shifted to the product side by
\(\mathrm{C}(\mathrm{~s})+\mathrm{H}_{2} \mathrm{O}(\mathrm{~g}) \rightleftharpoons \mathrm{CO}(\mathrm{~g})+\mathrm{H}_{2}(\mathrm{~g}) ; \Delta \mathrm{H}=+131 \mathrm{~kJ}\)
The yield of CO and H2 at equilibrium would be shifted to the product side by
- raising the relative pressure of steam
- adding hot carbon
- raising the temperature
- reducing the volume of the system
Answer
(A) raising the relative pressure of steam, (C) raising the temperature
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