Consider a bipropellant rocket using a fuel-oxidizer combination for which the stoichiometric reaction is written 2H2 + O2 2H2O. The actual reaction occurring in the chamber is a fuel-rich one, written 5H2 + O2 ? 2H2O + 3H2. Determine the corresponding equivalence ratio f for this case and show that the use of either moles or masses for this calculation yields the same value of f.
The correct answer and explanation is :
To determine the equivalence ratio (denoted ( f )) for this fuel-rich reaction, we need to understand the difference between the stoichiometric and actual reaction and use it to find the ratio of fuel-to-oxidizer.
Step 1: Understand the Stoichiometric Reaction
The stoichiometric reaction for hydrogen and oxygen is:
[
2H_2 + O_2 \rightarrow 2H_2O
]
In this case, 2 moles of hydrogen (H₂) react with 1 mole of oxygen (O₂) to form 2 moles of water (H₂O).
Step 2: The Actual Fuel-Rich Reaction
The actual reaction provided in the problem is fuel-rich:
[
5H_2 + O_2 \rightarrow 2H_2O + 3H_2
]
Here, 5 moles of hydrogen react with 1 mole of oxygen, but unlike the stoichiometric reaction, additional hydrogen (3 moles) is produced.
Step 3: Equivalence Ratio
The equivalence ratio ( f ) compares the actual fuel-to-oxidizer ratio to the stoichiometric fuel-to-oxidizer ratio.
For moles:
- In the stoichiometric case, 2 moles of H₂ react with 1 mole of O₂, so the stoichiometric fuel-to-oxidizer ratio is ( \frac{2}{1} ) or 2.
- In the actual fuel-rich reaction, 5 moles of H₂ react with 1 mole of O₂, so the actual fuel-to-oxidizer ratio is ( \frac{5}{1} ) or 5.
The equivalence ratio ( f ) is defined as:
[
f = \frac{\text{Actual fuel-to-oxidizer ratio}}{\text{Stoichiometric fuel-to-oxidizer ratio}}
]
So,
[
f = \frac{5}{2} = 2.5
]
For masses:
To calculate the equivalence ratio using masses, we use the molar masses of hydrogen and oxygen. The molar mass of hydrogen (H₂) is 2 g/mol, and the molar mass of oxygen (O₂) is 32 g/mol.
- Stoichiometric fuel-to-oxidizer ratio by mass:
[
\text{Fuel mass} = 2 \text{ mol H}_2 \times 2 \text{ g/mol} = 4 \text{ g of H}_2
]
[
\text{Oxidizer mass} = 1 \text{ mol O}_2 \times 32 \text{ g/mol} = 32 \text{ g of O}_2
]
Thus, the stoichiometric fuel-to-oxidizer ratio by mass is ( \frac{4}{32} = \frac{1}{8} ). - Actual fuel-to-oxidizer ratio by mass:
[
\text{Fuel mass} = 5 \text{ mol H}_2 \times 2 \text{ g/mol} = 10 \text{ g of H}_2
]
[
\text{Oxidizer mass} = 1 \text{ mol O}_2 \times 32 \text{ g/mol} = 32 \text{ g of O}_2
]
Thus, the actual fuel-to-oxidizer ratio by mass is ( \frac{10}{32} = \frac{5}{16} ).
The equivalence ratio ( f ) by mass is:
[
f = \frac{\frac{5}{16}}{\frac{1}{8}} = 2.5
]
Conclusion:
In both cases, whether we use moles or masses, the equivalence ratio ( f ) is 2.5. This confirms that the use of moles or masses in the calculation leads to the same result.