Questions: 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 663 liters per second of methane are consumed when the reaction is run at 216 °C and the methane is supplied at 0.21 atm. Calculate the rate at which dihydrogen is being produced. Give your answer in kilograms per second. Round your answer to 2 significant digits. kg/s

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 663 liters per second of methane are consumed when the reaction is run at 216 °C and the methane is supplied at 0.21 atm. Calculate the rate at which dihydrogen is being produced. Give your answer in kilograms per second. Round your answer to 2 significant digits.
kg/s

Transcript text: The reform reaction between steam and gaseous methane $\left(\mathrm{CH}_{4}\right)$ 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 663. liters per second of methane are consumed when the reaction is run at $216 .{ }^{\circ} \mathrm{C}$ and the methane is supplied at 0.21 atm . Calculate the rate at which dihydrogen is being produced. Give your answer in kilograms per second. Round your answer to 2 significant digits. $\square \frac{\mathrm{kg}}{\mathrm{s}}$

Solution

Solution Steps

Step 1: Write the Balanced Chemical Equation

The balanced chemical equation for the reform reaction between steam and methane is: \[ \text{CH}_4(g) + \text{H}_2\text{O}(g) \rightarrow \text{CO}(g) + 3\text{H}_2(g) \] This equation shows that 1 mole of methane produces 3 moles of dihydrogen.

Step 2: Calculate the Moles of Methane Consumed

First, we need to convert the volume of methane consumed to moles using the ideal gas law: \[ PV = nRT \] Given:

$ P = 0.21 \, \text{atm} $
$ V = 663 \, \text{L/s} $
$ R = 0.0821 \, \text{L atm/mol K} $
$ T = 216^\circ \text{C} = 489 \, \text{K} $

Rearranging the ideal gas law to solve for $ n $ (moles per second): \[ n = \frac{PV}{RT} = \frac{0.21 \times 663}{0.0821 \times 489} \]

Calculating: \[ n \approx \frac{139.23}{40.1469} \approx 3.467 \, \text{mol/s} \]

Step 3: Calculate the Moles of Dihydrogen Produced

From the balanced equation, 1 mole of methane produces 3 moles of dihydrogen. Therefore, the rate of dihydrogen production is: \[ 3 \times 3.467 \approx 10.401 \, \text{mol/s} \]

Step 4: Convert Moles of Dihydrogen to Kilograms

The molar mass of dihydrogen ($\text{H}_2$) is approximately 2.016 g/mol. To convert moles per second to kilograms per second: \[ \text{mass rate} = 10.401 \, \text{mol/s} \times 2.016 \, \text{g/mol} \times \frac{1 \, \text{kg}}{1000 \, \text{g}} \]

Calculating: \[ \text{mass rate} \approx 20.976 \, \text{g/s} \approx 0.0210 \, \text{kg/s} \]