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Fynjy0 [20]
3 years ago
9

Hydrocarbons, compounds containing only carbon and hydrogen, are important in fuels.The heat of combustion of cyclopentane, C5H1

0, is 786.6 kcal/mol.Write a balanced equation for the complete combustion of cyclopentane.
Chemistry
1 answer:
shtirl [24]3 years ago
7 0

Answer : The balanced equation for the complete combustion of cyclopentane will be :

2C_5H_{10}+15O_2\rightarrow 10CO_2+10H_2O

Explanation :

As we know that the hydrocarbons can undergo complete or incomplete combustion that depends on the amount of oxygen available.

Complete combustion : It takes place when there is a good supply of oxygen. In this, hydrocarbon react with the oxygen to give carbon dioxide and water as a product.

The general reaction will be :

\text{Hydrocarbon}+O_2\rightarrow CO_2+H_2O

Incomplete combustion : It takes place when there is a poor supply of oxygen. In this, hydrocarbon react with the oxygen to give carbon monoxide, carbon dioxide and water as a product.

The general reaction will be :

\text{Hydrocarbon}+O_2\text{(limited)}\rightarrow CO+CO_2+H_2O

As per question, the complete combustion of cyclopentane react with oxygen to gives carbon dioxide and water as a product.

Therefore, the balanced chemical reaction will be :

2C_5H_{10}+15O_2\rightarrow 10CO_2+10H_2O

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Chloroacetic acid, ClCH2COOH, has a pKa of 2.87. What are [H3O+], pH, [ClCH2COO−], and [ClCH2COOH] in 1.55 M ClCH2COOH?
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first to all, this is an acid base reaction where the chloroacetic acid is being dissociated in water. Therefore, is an equilibrium reaction.

ClCH₂COOH has a pKa of 2.87 so, the Ka would be:

pKa = -logKa ---> Ka = antlog(-pka)

Ka = antlog(-2.87)

Ka = 1.35x10⁻³

Now that we know the Ka, we need to write the chemical reaction and then, an ICE chart:

      ClCH₂COOH + H₂O <----------> ClCH₂COO⁻ + H₃O⁺    Ka = 1.35x10⁻³

i)            1.55                                             0                0

c)             -x                                             +x               +x

e)         1.55-x                                            x                 x

Writting now the equilibrium reaction:

Ka = [H₃O⁺] [ClCH₂COO⁻] / [ClCH₂COOH]

Replacing the values of the chart:

1.35x10⁻³ = x² / 1.55-x

1.35x10⁻³(1.55-x) = x²

2.0925x10⁻³ - 1.35x10⁻³x = x²

x² + 1.35x10⁻³x - 2.0925x10⁻³ = 0  --> a = 1; b = 1.35x10⁻³x; c = 2.0925x10⁻³

From here we use the general equation for solve x in a quadratic equation which is:

x = -b±√(b² - 4ac) / 2a

Replacing the values we have:

x = -1.35x10⁻³ ±√(1.35x10⁻³)² - 4*1*(-2.0925x10⁻³) / 2

x = -1.35x10⁻³ ±√(8.37x10⁻³) / 2

x = -1.35x10⁻³ ± 0.091 / 2

x1 = -1.35x10⁻³ + 0.091 / 2 = 0.045 M

x2 = -1.35x10⁻³ - 0.091 / 2 = -0.046 M

In this case, we will take the positive value of x, in this case, x1.

With this value, the equilibrium concentrations are the following:

[H₃O⁺] = [ClCH₂COO⁻] = 0.045 M

[ClCH₂COOH] = 1.55 - 0.045 = 1.505 M

Finally the pH:

pH = -log[H₃O⁺]

pH = -log(0.045)

pH = 1.35

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