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Jobisdone [24]
3 years ago
10

During a reaction, the enthalpy of formation of an intermediate is 90.3 kJ/mol. During the reaction, 2 moles of the intermediate

are formed as a reactant. What is the enthalpy value for this step of the reaction?
A. -180.6 kJ

B. 90.3 kJ

C. -90.3 kJ

D. 180.6 kJ​
Chemistry
1 answer:
Soloha48 [4]3 years ago
7 0

Answer:

180.6 kJ​

Explanation:

The enthalpy of reaction refers to the energy absorbed or released during a reaction. If heat is absorbed in a reaction, the enthalpy of reaction is positive. If heat is released in a reaction, the enthalpy of reaction his negative.

Since the energy absorbed when one mole of the intermediate is formed is 90.3 kJ/mol, then, when two moles of intermediate is formed, 2 × 90.3 kJ/mol = 180.6 kJ​ of energy is absorbed.

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Ethyl butyrate, CH3CH2CH2CO2CH2CH3, is an artificial fruit flavor commonly used in the food industry for such flavors as orange
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Answer:

A. 10.0 grams of ethyl butyrate would be synthesized.

B. 57.5% was the percent yield.

C. 7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

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CH_3CH_2CH_2CO_2H(l)+CH_2CH_3OH(l)+H^+\rightarrow CH_3CH_2CH_2CO_2CH_2CH_3(l)+H_2O(l)

A

Moles of butanoic acid = \frac{7.60 g}{88 g/mol}=0.08636 mol

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

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Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 100%

Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

100\%=\frac{\text{Experimental yield}}{10.0 g}\times 100

Experimental yield = 10.0 g

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B

Theoretical yield of ethyl butyrate  = 10.0 g

Experimental yield ethyl butyrate = 5.75 g

Percentage yield of the reaction = ?

Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

=\frac{5.75 g}{10.0 g}\times 100=57.5\%

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C

Moles of butanoic acid = \frac{7.60 g}{88 g/mol}=0.08636 mol

According to reaction ,1 mole of butanoic acid gives 1 mol of ethyl butyrate,then 0.08636 mol of butanoic acid will give :

\frac{1}{1}\times 0.08636 mol=0.08636 mol of ethyl butyrate

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Theoretical yield = 10.0 g

Experimental yield = ?

Percentage yield of the reaction = 78.0%

Yield\%=\frac{\text{Experimental yield}}{\text{Theoretical yield}}\times 100

78.0\%=\frac{\text{Experimental yield}}{10.0 g}\times 100

Experimental yield = 7.80 g

7.80 grams of ethyl butyrate would be produced from 7.60 g of butanoic acid.

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