Answered

At Westonci.ca, we make it easy for you to get the answers you need from a community of knowledgeable individuals. Our platform offers a seamless experience for finding reliable answers from a network of experienced professionals. Join our platform to connect with experts ready to provide precise answers to your questions in different areas.

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 :

GinnyAnswer
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].
We hope this was helpful. Please come back whenever you need more information or answers to your queries. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.