Answered

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]



Answer :

GinnyAnswer
To determine the heat of combustion, per mole, of glucose, we need to analyze the given chemical reaction and the provided enthalpy change.

The balanced chemical equation for the combustion of glucose is:
[tex]\[ C_6H_{12}O_6(s) + 6 O_2(g) \rightarrow 6 CO_2(g) + 6 H_2O(l) \][/tex]

It is given that the enthalpy change for this reaction is [tex]\( -2,840 \, \text{kJ} \)[/tex]. This means that when one mole of glucose ( [tex]\( C_6H_{12}O_6 \)[/tex] ) is completely combusted, the reaction releases [tex]\( -2,840 \, \text{kJ} \)[/tex] of energy.

The problem asks for the heat of combustion per mole of glucose. Since the enthalpy change given [tex]\( -2,840 \, \text{kJ} \)[/tex] is already for the combustion of one mole of glucose, this is the heat of combustion per mole.

Hence, the heat of combustion, per mole, of glucose is:
[tex]\[ -2,840 \, \text{kJ/mol} \][/tex]

Therefore, the correct answer is:
[tex]\[ -2,840 \, \text{kJ/mol} \][/tex]