Given C_6H_{12}O_6 + 6O_2 rightarrow 6CO_2 + | vedclass

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(B) The molar mass of glucose $(C_6H_{12}O_6)$ is $180 \ 	ext{g/mol}$.For $1 \ 	ext{mol}$ $(180 \ 	ext{g})$ of glucose,the enthalpy of combustion i

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$0.64$

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(B) The molar mass of glucose $(C_6H_{12}O_6)$ is $180 \ 	ext{g/mol}$.For $1 \ 	ext{mol}$ $(180 \ 	ext{g})$ of glucose,the enthalpy of combustion is $\Delta H = -72 \ 	ext{kcal}$.This means the energy released during the combustion of $180 \ 	ext{g}$ of glucose is $72 \ 	ext{kcal}$.Therefore,the energy released for $1.6 \ 	ext{g}$ of glucose is $\frac{72 	imes 1.6}{180} = 0.64 \ 	ext{kcal}$.Since the formation reaction is the reverse of the combustion reaction,the energy required for the formation of $1.6 \ 	ext{g}$ of glucose is $+0.64 \ 	ext{kcal}$.

Given $C_6H_{12}O_6 + 6O_2 \rightarrow 6CO_2 + 6H_2O : \Delta H = -72 \ \text{kcal}$,calculate the energy required for the formation of $1.6 \ \text{g}$ of glucose.

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