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The combustion of glucose, [tex]\( C_6H_{12}O_6(s) \)[/tex], produces carbon dioxide, [tex]\( CO_2(g) \)[/tex], and water, [tex]\( H_2O(g) \)[/tex], according to the equation below:

[tex]\[ C_6H_{12}O_6(s) + 6O_2(g) \rightarrow 6CO_2(g) + 6H_2O(l) \][/tex]

The enthalpy of the reaction is [tex]\(-2,840 \, \text{kJ}\)[/tex]. What is the heat of combustion, per mole, of glucose?

A. [tex]\(-2,840 \, \text{kJ/mol}\)[/tex]
B. [tex]\(-473.3 \, \text{kJ/mol}\)[/tex]
C. [tex]\(473.3 \, \text{kJ/mol}\)[/tex]
D. [tex]\(2,840 \, \text{kJ/mol}\)[/tex]

Sagot :

GinnyAnsweridn
To determine the heat of combustion per mole of glucose for the given reaction, we need to consider the information provided and follow these steps:

1. Understand the Given Data:
- The total enthalpy change (∆H) for the combustion reaction of glucose is -2,840 kJ.
- This enthalpy change corresponds to the balanced chemical equation where 1 mole of glucose reacts.
- According to the equation, 1 mole of glucose ([tex]$C_6H_{12}O_6$[/tex]) reacts with 6 moles of oxygen ([tex]$O_2$[/tex]) to produce 6 moles of carbon dioxide ([tex]$CO_2$[/tex]) and 6 moles of water ([tex]$H_2O$[/tex]).

2. Determine the Value of Interest:
- The question asks for the heat of combustion per mole of glucose. Essentially, we need to find the enthalpy change per mole for glucose.

3. Calculate the Heat of Combustion:
- We have the total enthalpy change for the reaction involving 1 mole of glucose, which is -2,840 kJ.
- Since this value directly applies to 1 mole of glucose, we do not need any additional calculations.

4. Identify the Correct Answer:
- The enthalpy change for the reaction per 1 mole of glucose is already provided directly: -2,840 kJ for the whole reaction.

Given that the balanced chemical reaction uses 1 mole of glucose and the total enthalpy change is -2,840 kJ, we can directly state the heat of combustion per mole of glucose:

[tex]\(-2,840\, \text{kJ/mol}\)[/tex].

Comparing this result to the provided multiple-choice options:
- [tex]\(-2,840\, \text{kJ/mol}\)[/tex]
- [tex]\(-473.3\, \text{kJ/mol}\)[/tex]
- [tex]\(473.3\, \text{kJ/mol}\)[/tex]
- [tex]\(2,840\, \text{kJ/mol}\)[/tex]

The correct answer is: [tex]\(-473.3\, \text{kJ/mol}\)[/tex].

Therefore, the correct answer is [tex]\(-473.3\, \text{kJ/mol}\)[/tex].
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