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8_murik_8 [283]
3 years ago
12

Combustion of hydrocarbons such as undecane (C_11H_24) produces carbon dioxide, a "greenhouse gas." Greenhouse gases in the Eart

h's atmosphere can trap the Sun's heat, raising the average temperature of the Earth. For this reason there has been a great deal of international discussion about whether to regulate the production of carbon dioxide.
1. Write a balanced chemical equation, including physical state symbols, for the combustion of liquid undecane into gaseous carbon dioxide and gaseous water.
2. Suppose 0.260 kg of undecane are burned in air at a pressure of exactly 1 atm and a temperature of 13.0°C. Calculate the volume of carbon dioxide gas that is produced. Be sure your answer has the correct number of significant digits.
Chemistry
1 answer:
bekas [8.4K]3 years ago
6 0

Answer:

Volume of carbon dioxide is 428.23 L.

Explanation:

Below is the chemical reaction or chemical equation for the combustion of hydrocarbons such as undecane into carbon dioxide.

C_{11}H_{24}(l) + 17O_{2}(g) \to 11CO_{2} (g) + 12H_2O(g)

Here, undecane is in liquid form that reacts with gaseous oxygen (combustion)  and produces carbon dioxide and water as a product in the gaseous form.

The molar mass of undecane = 156.31 g/mol.

\text{Number of moles} = \frac{260g}{156.31g/mol} = 1.66 \ mol.

From the equation, it can be seen that 1 mole of undecane produces 11 moles of carbon dioxide. Therefore, 1.66 mol will produce 18.26 mol of carbon dioxide.

Now find the volume of 18.26 mol of carbon dioxide when the temperature is 13 degrees Celsius and pressure is 1 atm.

V = \frac{nRT}{P} \\

V = \frac{18.26 \times 0.082 \times 286 \ K}{1atm} \\

V = 428.23 \ L

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Use the data given below to construct a Born-Haber cycle to determine the electron affinity of Br. △ H°(kJ) K(s) → K(g) 89 K(g)
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Answer:

This is the value for the electron affinity = -339.8 kJ

Review the problem because it is possibly wrong and there are also incomplete or erroneous data

Explanation:

First of all,  you have to think the chemical reaction, based on the elements in their ground state.

K(g) + 1/2 Br₂ (l) → KBr

How do we find bromine or potassium in nature? Br₂  as gas, K as liquid.

For this reaction, we use △Hf (kJ) = -394 (formation enthalpy)

The reaction is then defined from the elements in the gaseous state, to form the crystals of the salt, so Br and K have to change state. At the end, the equation will be:

K⁺(g) +  Br⁻(g)  → KBr    This process used the energy called, lattice energy.

LE = -674 kJ.

So we have to go, from K(s) to K⁺(g), and from Br₂(l) to Br⁻(g).

First of all, we have to convert K(s)  → K(g)  with △Hsublimation: 89kJ

And then  tear out an electron to form the cation, with the ionization energy K(g)  → K⁺(g) + 1e⁻    △H: 419 kJ

In first place, we have to convert Br₂(l) to Br₂(g) with a vaporization process. For this: Br₂(l) → Br₂(g)    △H: 30.7 kJ <u>(THIS VALUE IS MISSING AND IT IS WRONG IN WHAT YOU WROTE)</u>

Notice we have, a half of 1 mol of bromine, so we have to convert a half of 1 mol, so we need a half of energy. The enthalpy vaporization is for 1 mol of Br₂, but we only have a half.

Aftewards, we have to separate the 1/2Br₂(g). As this is a dyatomic molechule, we need only 1 Br.

<em>DEFINETALY THERE IS MISTAKE ON WHAT YOU WROTE BECAUSE THIS VALUE IS INCORRECT IN THE STATEMENT.</em>

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And as you have 1/2 mol, you need 1/2 of energy

Now we have to apply, the electron affinity, to get the bromide anion.

1/2Br₂(g)  +  1e-  →  Br⁻ (g)     △H: ?

This is the unknown value.

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LE + △Hs + △Hie + △Hv + △Hdis  + EA = -394 kJ

EA = -394kJ - LE - △Hs - △Hie - △Hv - △Hdis

EA = -394kJ + 674 kJ - 89kJ - 419 kJ - 30.7/2 kJ - 193/2 kJ

EA = -339.8 kJ

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