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When a 3.80-g sample of liquid octane (c8h18) is burned in a bomb calorimeter, the temperature of the calorimeter rises by 27.3 °c. the heat capacity of the calorimeter, measured in a separate experiment, is 6.18 kj>°c. determine the δe for octane combustion in units of kj>mol octane?

Respuesta :

Dejanras

Answer is:  the δe for octane combustion is -5112.42 kJ/mol.

Chemical reaction: C₈H₁₈ + 25/2O₂ → 8CO₂ + 9H₂O.

m(C₈H₁₈) = 3.80 g; mass of octane.

n(C₈H₁₈) = m(C₈H₁₈) ÷ M(C₈H₁₈).

n(C₈H₁₈) = 3.8 g ÷ 114 g/mol.

n(C₈H₁₈) = 0.033 mol; amount of octane.

cp= 6.18 kJ/°C, specific heat capacity of the bomb calorimeter.  

Qcal = ΔT · cp.  

Qcal = 27.3°C · 6.18 kJ/°C.  

Qcal = 168.71 kJ; amount of heat absorbed.  

ΔE = Qcal ÷ n(C₈H₁₈).

ΔE = 168.71 kJ ÷ 0.033 mol.

ΔE = 5112.42 kJ/mol.

PKinakin

Answer:

[tex]-4.33 * 10^3 \frac{kJ}{mol}[/tex]

Explanation:

To determine the correct answer, use the equations  [tex]q_{cal} = C_{cal}[/tex] x ΔT  and  [tex]q_{cal} = - q_{rxn}[/tex].  ΔE  is equal to [tex]q_{rxn}[/tex] divided by the number of moles of octane that reacted.

[tex]q_{cal} = - q_{rxn}[/tex]

[tex]-C_{cal} \times \Delta T=-\frac{6.20 kJ}{\textdegree C} \times 26.9 \textdegree C=-167 kJ[/tex]

[tex]4.40 \text{g} \ \text{octane} \times \frac{1 \text{mol \ octane}}{114.23 \text{g \ octane}}=3.85 \times 10^{-2}\ \text {mol \ octane}[/tex]

[tex]\Delta E = \frac{-167kJ }{3.85\times10^{3}\text{mol \ octane}}=-4.33\times 10^3 \ \text{kJ/mol}[/tex]

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