Chemistry, 04.06.2020 13:30 zaylencollins55

The reform reaction between steam and gaseous methane () produces "synthesis gas," a mixture of carbon monoxide gas and dihydrogen gas. Synthesis gas is one of the most widely used industrial chemicals, and is the major industrial source of hydrogen. Suppose a chemical engineer studying a new catalyst for the reform reaction finds that liters per second of methane are consumed when the reaction is run at and . Calculate the r

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Answer from: keshan3000

The rate at which dihydrogen gas is being produced = 0.018 kg/s

Explanation:

Firstly, we write the balanced equation for the production of the synthesis gas

CH₄ + H₂O → CO + 3H₂

The rate of consumption of CH₄ is 159 litres per second. With the reaction ran at T = 294°C and a pressure of 0.86 atm

Using the ideal gas equation, we can convert the volumetric rate of consumption of methane to molar rate of consumption

PV = nRT

PV' = n'RT

P = pressure = 0.86 atm = 87,139.5 Pa

V' = 159 L/s = 0.159 m³/s

n' = ?

R = molar gas constant = 8.314 J/mol.K

T = absolute temperature in Kelvin = 294°C = 567.15 K

87,139.5 × 0.159 = n' × 8.314 × 567.15

n' = (87,139.5 × 0.159) ÷ (8.314 × 567.15)

n' = 2.9383547773 mol/s = 2.938 mol/s

From the stoichiometry of this reaction,

1 mole of methane gives 3 moles of dihydrogen gas

2.938 mol/s of methane will correspond to (3 × 2.938) mol/s of dihydrogen gas, that is, 8.815 mol/s.

Mass flowrate = (molar flowrate) × (molar mass)

Molar flowrate = 8.815 mol/s

Molar mass of dihydrogen gas = 2 g/mol = 0.002 kg/mol

Mass flowrate = 8.815 × 0.002 = 0.0176301287 kg/s = 0.018 kg/s to 2 s.f.

Hope this Helps!!!

Answer from: jaidxxxz

Complete Question:

The reform reaction between steam and gaseous methane (CH4) produces "synthesis gas," a mixture of carbon monoxide gas and dihydrogen gas. Synthesis gas is one of the most widely used industrial chemicals, and is the major industrial source of hydrogen.

Suppose a chemical engineer studying a new catalyst for the reform reaction finds that 924. liters per second of methane are consumed when the reaction is run at 261.°C and 0.96atm. Calculate the rate at which dihydrogen is being produced. Give your answer in kilograms per second. Round your answer to 2 significant digits.

Rate at which H₂ is produced = 0.12 kg/s

Explanation:

Volume of methane produced, V = 924 litres

Temperature, T = 261.°C = 261 + 273

T = 534 K

Pressure, P = 0.96 atm

Gas constant, R = 0.0821 L-atm/ K-mol

We will calculate the number of moles of methane used in the reaction.

$n_{methane} = \frac{PV}{RT} \\n_{methane} = \frac{0.96 * 924}{0.0821 * 534} \\n_{methane} = \frac{887.04}{43.8414}\\n_{methane} = 20.23 moles$

Equation of reaction:

$CH_{4} + H_{2} O \rightarrow CO + 3H_{2}$

From the reaction above :

1 mole of methane produced 3 moles of H₂

20.23 moles of methane will produce (20.23 * 3 ) moles of H₂

Number of moles of H₂, $n_{H_{2} } = 60.69 moles$

That is 60.69 moles of hydrogen is produced per second.

Number of moles = Mass/ Molar mass

$n_{H_{2} } = \frac{Mass_{H_{2} }}{Molar mass_{H_{2} }} \\Mass_{H_{2}} = n_{H_{2} } * Molar mass_{H_{2} }\\Mass_{H_{2}} = 60.69 * 2.016\\Mass_{H_{2}} = 122.35 g$

Rate at which H₂ is produced = 122.35 g/s = 0.12 kg/s

Answer from: Quest

D-no effect on the rate

Answer from: Quest

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