Question

1. consider the following reaction: (ℓ) + 4.5 o₂ (g) → + 2 co₂ (g) + 2 h₂o (g)

a) calculate the overall atom efficiency (oae)
b) calculate the e - factor
c) in the lab you make 7.5 g, but the theoretical yield was 8 g. what is the % yield for this reaction?
d) comment on the green nature of this reaction for the following reaction:

Explanation:

Step1: Calculate molar - masses

Let's assume the reactant (benzene) has a molar - mass \(M_{benzene}=78\ g/mol\), the product (maleic anhydride) has a molar - mass \(M_{maleic\ anhydride}=98\ g/mol\), \(M_{CO_2}=44\ g/mol\), and \(M_{H_2O}=18\ g/mol\).

Step2: Calculate the overall atom efficiency (OAE)

The formula for OAE is \(OAE=\frac{\text{Mass of atoms in desired product}}{\text{Mass of atoms in all reactants}}\times100\%\).
The mass of atoms in the desired product (maleic anhydride) is \(m_{product}=98\ g/mol\).
The mass of atoms in all reactants: Benzene (\(C_6H_6\)) has a molar - mass of \(78\ g/mol\) and oxygen (\(4.5\ mol\) of \(O_2\)) has a mass of \(4.5\times32 = 144\ g/mol\). The total mass of reactants \(m_{reactants}=78 + 144=222\ g/mol\).
\(OAE=\frac{98}{222}\times100\%\approx44.1\%\)

Step3: Calculate the E - factor

The E - factor is defined as \(\text{E - factor}=\frac{\text{Mass of waste}}{\text{Mass of product}}\).
The mass of waste: \(2\ mol\) of \(CO_2\) has a mass of \(2\times44 = 88\ g/mol\) and \(2\ mol\) of \(H_2O\) has a mass of \(2\times18 = 36\ g/mol\). The total mass of waste \(m_{waste}=88 + 36=124\ g/mol\). The mass of the product (maleic anhydride) is \(m_{product}=98\ g/mol\).
\(\text{E - factor}=\frac{124}{98}\approx1.27\)

Step4: Calculate the percentage yield

The formula for percentage yield is \(\%\text{Yield}=\frac{\text{Actual yield}}{\text{Theoretical yield}}\times100\%\).
Given actual yield \(=7.5\ g\) and theoretical yield \( = 8\ g\).
\(\%\text{Yield}=\frac{7.5}{8}\times100\% = 93.75\%\)

Step5: Comment on the green nature

The OAE is relatively low (\(44.1\%\)), indicating that a significant portion of the atoms in the reactants do not end up in the desired product. The E - factor of \(1.27\) means that for every gram of product, there is about \(1.27\) grams of waste. The reaction is not very green as it has low atom - efficiency and a non - negligible amount of waste.

Answer:

a) \(44.1\%\)
b) \(1.27\)
c) \(93.75\%\)
d) The reaction is not very green due to low atom - efficiency (\(44.1\%\)) and a non - negligible E - factor (\(1.27\)).