Final answer:
The pressure of CO₂ produced from the glucose fermentation reaction at 200°C in a sealed container can be calculated using the ideal gas law, which results in a pressure of approximately 2 atm.
Step-by-step explanation:
The question is about calculating the pressure of CO₂ produced in the glucose fermentation reaction using the ideal gas law. From the information given, 2.13 grams of glucose (C₆H₁₂O₆) produced 2 moles of ethanol (C₂H₆O) and 2 moles of CO₂. To calculate the pressure of CO₂, we can use the ideal gas law, which is PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant (0.0821 atm·L/mol·K for the units atm, L, and K), and T is the temperature in Kelvin.
To solve for P we rearrange the equation: P = nRT/V. Since the reaction is happening at 200°C, we first need to convert the temperature to Kelvin by adding 273.15 to it, resulting in T = 473.15 K. We are given that 2 moles of CO₂ are produced, so we will calculate the pressure of just those 2 moles.
As the specific volume wasn't given, we should assume standard conditions, which means using the molar volume of a gas at STP (22.414 L), but we must remember that STP is defined at 0°C (273.15 K), so to eliminate the need for volume, we can cancel it out by using the same conditions for R, and understanding that at a given temperature and constant R the pressure is directly proportional to moles. Therefore, if 1 mole of gas at STP has a pressure of 1 atm, then 2 moles will have a pressure of 2 atm.