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Kazeer [188]
4 years ago
11

Complete the acid-base reaction between butyric acid HC4H7O2 and potassium hydroxide KOH.

Chemistry
1 answer:
Lilit [14]4 years ago
6 0

An acid-base reaction or a neutralization reaction is a <u>chemical reaction that occurs between an acid and a base producing a salt and water</u>. The acids and bases can be strong or weak depending on their degree of ionization in water.

Butyric acid is a weak acid and in water it is ionized in the following way, loosing a proton (H+):

HC4H7O2 (aq) ⇆ H+ (aq) + C4H7O2- (aq)

On the other hand, potassium hydroxide is a strong base, so it will be completely ionized in water:

KOH(aq) → K+(aq) + OH-(aq)

Then the <u>net acid-base reaction</u> between butyric acid and KOH is:

HC4H7O2 (aq) + OH- (aq) ⇆ H2O + C4H7O2- (aq)

It is valid to consider only the OH- produced from the ionization of KOH in water since, as mentioned, this molecule is completely ionized. Also, we do not include the K + in the net equation since it is a spectator ion, it does not undergo chemical changes.

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Answer: D

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The higher the energy density of a fuel, the greater the amount of energy it has stored.

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95. Using the standard enthalpy of formation data in Appendix G, calculate the bond energy of the carbon-sulfur double bond in C
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Answer:

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Explanation:

The following data will be used to calculate the average C-S bond energy in CS2(l).

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Enthalpy of formation of CS2(l)

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CS2(l) ---> CS2(g)

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So we must construct it stepwise.

1: C(s) ---> C(g) ΔH = 715 kJ

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adding 1 + 2 = 3

ΔH = 715 + 446

= 1161 kJ

3: C(s) + 2S(s) --> C(g) + 2S(g) ΔH = 1161 kJ

4: C(s) + 2S(s) --> CS2(l) ΔH = 88 kJ

adding (reversed 3) from 4 = 5

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5: C(g) + 2S(g) --> CS2(l) ΔH = -1073 kJ

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adding 5 + 6 = 7

ΔH = -1073 + 27

= -1046 kJ

7. C(g) + 2S(g) --> CS2(g) ΔH = -1046 kJ

Reverse and divide by 2 for C-S bond enthalpy

= -(-1046)/2

= +523 kJ.

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