The rate of the dihydrogen production is 0.0175 kg/sec for the given reaction.
The given reaction
[tex]\bold {CH_4 + H_2O \rightarrow CO + 3H_2}[/tex]
One mole of Methane is react with one mole of water to form 1 mole of CO and 3 moles of Hydrogen.
The rate of hydrogen production can be calculated from ideal gas law
[tex]\bold {pV = nRT}[/tex]
Where,
p- pressure = 0,86 atm · 101,325 kPa/atm = 87,14 kPa.
V- volume of methane = 159 L
T - temperature= 294°C = 567 K.
R - gas constant = 8,314 J/K·mol.
n - number of moles = ?
Put the values in the formula, sole it for n,
[tex]\bold{n(CH₄) = \dfrac {pV} {RT}}\\\\\bold {n(CH₄) = \dfrac {87,14 kPa \times 159 L} {8,314 J/K.mol \times 567 K}.}\\\\\bold {n(CH₄) = 2,93 mol}[/tex]
Since, the molar ratio of methane to hydrogen is 3:1.
SO,
[tex]\bold {n(H_2) = 2,93 mol \times 3 }\\\\\bold {n (H_2) = 8,79 mol.}\\[/tex]
Hence, the mass of the dihydrogen produced,
[tex]\bold{ m(H_2) = 8.79 mol \times 2 g/mol.}\\\\\bold {m(H_2) = 17.58 g \times 0.001 kg/g}\\\\\bold {m(H_2) = 0.01758 kg}[/tex]
Therefore, the rate of the dihydrogen production is 0.0175 kg/sec for the given reaction.
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