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tatuchka [14]
2 years ago
7

The change in internal energy for the combustion of 1.0 molmol of octane at a pressure of 1.0 atmatm is -5084.1 kJkJ . Part A If

the change in enthalpy is -5074.4 kJkJ , how much work is done during the combustion
Chemistry
1 answer:
Dovator [93]2 years ago
3 0

Answer:

W = -10.3 kJ

Explanation:

During combustion, the system performs work and releases heat. Therefore, the change in internal energy is negative, and the change in enthalpy, which is equal to heat at constant pressure, is also negative. Work is then calculated by rearranging the equation for the change in internal energy:

w=ΔE−qp=−5084.3 kJ−(−5074.0 kJ)

The release of heat is much greater than the work performed by the system on its surroundings. The potential energy stored in the bonds of octane explains why considerably large amounts of energy can be lost by the system during combustion.

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rank the four gases (air, exhaled air, gas produced from the decomposition of H2O2, gas from decomposition of NaHCO3, in order o
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Answer: H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)


Initial important note:


Although NaHCO₃ contents oxygen atoms, and you can calculate its compositoin, the resulting gas does not containg pure oxygen gas (O₂). For the comparisson it is not useful to calculate the content of oxygent atoms, but the concentration of O₂ gas. As such, the gas from NaHCO₃ contains 0% of pure O₂, that is why it is ranked last.


1) Air:


Source: internet


Approximate 23%. It is variable, because air is not a pure substance but a mixture of gases, whose compositon is not unique.


2) Exhaled air:


Source: internet.


Approximate 13%. The compositon of the air changes in our lungs, due to the respiration process: we inhale fresh air with around 23% of oxygen, part of this oxygen pass to the cells (lungs - blood - heart - cells) and then it is exhaled with a lower content of air and a greater content of CO₂


3) Air from the decomposition of H₂O₂.


In this case we can do a chemical calculation, since we can state the chemical equation of the reaction:


i) Chemical Equation:


H₂O₂ (g) → H₂ (g) + O₂ (g)


ii) mole ratio of the products 1 mol H₂ : 1 mol O₂


iii) convert moles into mass (grams)


1 mol H₂ × 2 × 1.008 g/mol = 2.016 g


1 mol O₂ × 2 × 15.999 g/mol = 31.998 g


Composition, % = [31.998 g / (2.016 g + 31.998 g) ] × 100 ≈ 94%



4) Air from the decomposition of NaHCO₃:


i) chemical equation:


2 NaHCO₃(s) → Na₂CO₃(s) + CO₂(g) + H₂O(g)


ii) mole ratio: take into account only the gases in the products:


1 mol CO₂ (g) : 1 mol H₂O


iii) mass in grams


CO₂: molar mass ia approximately 44.01 g/mol


H₂O: molar mass is approximately 18.02 g/mol


iii) Those gases although have oxygen atoms, do not hae free oxygen gas, which is what we are compariing. That means, that from the decomposition of NaHCO₃ you get 0% oxygen gas.


5) The result is:


H₂O₂ (94%) > Air (23%) > Exhaled air (13%) > NaHCO₃ (0%)

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