Question

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 rate at which dihydrogen is being produced. Give your answer in kilograms per second. Round your answer to significant digits..

Answer 1

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.

Answer: The rate at which dihydrogen is being produced is 0.12 kg/sec

Explanation:

The balanced chemical equation is ;

$CH_4+H_2O\rightarrow 3H_2+CO$

According to ideal gas equation:

$PV=nRT$

P = pressure of gas = 0.96 atm

V = Volume of gas = 924 L

n = number of moles

R = gas constant =$0.0821Latm/Kmol$

T =temperature =$261^0C=(261+273)K=534K$

$n=\frac{PV}{RT}$

$n=\frac{0.96atm\times 924L}{0.0820 L atm/K mol\times 534K}=20.2moles$

According to stoichiometry:

1 mole of methane produces = 3 moles of hydrogen

Thus 20.2 moles of methane produces = $\frac{3}{1}\times 20.2=60.6$ moles of hydrogen

Mass of hydrogen =$moles\times {\text {Molar mass}}=60.6mol\times 2g/mol=121.2g=0.12kg$

Thus the rate at which dihydrogen is being produced is 0.12 kg/sec