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Chapter 3: Problem 53

(a) If a rocket were fueled with kerosine and liquid oxygen, what weight of oxygen would be required for every liter of kerosine? (Assume kerosine to have the average composition of \(\mathrm{n}-\mathrm{C}_{14} \mathrm{H}_{30}\) and a density of \(0.764 \mathrm{~g} / \mathrm{m} 1\) ). (b) How much energy would be required to fragment the molecules in one liter of kerosine? (Assume \(157 \mathrm{kcal}\) mole for each \(-\mathrm{CH}_{2}\) - group and \(186 \mathrm{kcal}\) mole for each \(-\mathrm{CH}_{3}-\) group.)

Short Answer

Expert verified
(a) The weight of oxygen required for every 1 liter of kerosine is approximately 2628 g. (b) The energy required to fragment the molecules in 1 liter of kerosine is approximately 9,300 kcal.

Step by step solution

01

Write down the balanced chemical equation for the combustion of kerosine

The average composition of kerosine is given as \(\mathrm{n}-\mathrm{C}_{14} \mathrm{H}_{30}\). Its combustion with oxygen, O₂, would produce carbon dioxide (CO₂) and water (H₂O). The balanced equation for this reaction is: $$\mathrm{C}_{14} \mathrm{H}_{30} + 21.5 \, \mathrm{O}_{2} \rightarrow 14\, \mathrm{CO}_{2} + 15\, \mathrm{H}_{2}\mathrm{O}$$
02

Calculate the moles of oxygen required for complete combustion of kerosine

We know that 1 mole of \(\mathrm{n}-\mathrm{C}_{14} \mathrm{H}_{30}\) requires 21.5 moles of oxygen for complete combustion. Now, let's find the mass of kerosine in 1 mole. Molar mass of \(\mathrm{C}_{14} \mathrm{H}_{30} = 14 \times 12.01 (\mathrm{C}) + 30 \times 1.008 (\mathrm{H}) = 198.42 \, \mathrm{g/mol}\) For 1 mole (\(198.42 \mathrm{g}\)) of kerosine, we need \(21.5 \times 32 (\mathrm{O}_{2}) = 688 \, \mathrm{g}\) of oxygen.
03

Determine the mass of oxygen required for 1 liter of kerosine

Given the density of kerosine as \(0.764 \, \mathrm{g/mL}\), we can find the mass of kerosine in 1 liter: Mass of 1 L of kerosine = \(1000 \, \mathrm{mL} \times 0.764 \, \mathrm{g/mL} = 764\, \mathrm{g}\) Now we find the mass of oxygen required for this mass of kerosine: For \(198.42\, \mathrm{g}\) of kerosine, we need \(688\, \mathrm{g}\) of oxygen. So, for \(764 \, \mathrm{g}\) of kerosine, we will need \(\frac{764 \times 688}{198.42} \approx 2628 \, \mathrm{g}\) of oxygen.
04

Answer (a)

The weight of oxygen required for every 1 liter of kerosine is approximately 2628 g.
05

Determine the total number of each type of bond in kerosine

In \(\mathrm{n}-\mathrm{C}_{14} \mathrm{H}_{30}\), there are 13 \(-\mathrm{CH}_{2}-\) groups and 2 \(-\mathrm{CH}_{3}-\) groups.
06

Calculate the energy required to fragment the bonds in one mole of kerosine

Now we will find the total energy required to fragment these bonds using the given energy values: Energy required for \(-\mathrm{CH}_{2}-\) groups: \(13 \times 157 \, \mathrm{kcal/mol} = 2041 \, \mathrm{kcal/mol}\) Energy required for \(-\mathrm{CH}_{3}-\) groups: \(2 \times 186 \, \mathrm{kcal/mol} = 372 \, \mathrm{kcal/mol}\) So the total energy required to fragment 1 mole of kerosine is: \(2041 + 372 = 2413 \, \mathrm{kcal/mol}\)
07

Calculate the energy required to fragment 1 liter of kerosine

Finally, let's find the energy required to fragment 1 liter of kerosine: Moles of kerosine in 1 liter = \(\frac{764}{198.42} \approx 3.85\, \mathrm{moles}\) Energy required to fragment 1 liter of kerosine: \(3.85 \times 2413 \approx 9,300 \, \mathrm{kcal}\)
08

Answer (b)

The energy required to fragment the molecules in 1 liter of kerosine is approximately 9,300 kcal.

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