MOLES of sucrose \[=\frac{34.2}{342}=0.1\] \[-{{(\DELTA G)}_{T.P}}=\]USEFUL WORK done by the system \[-\Delta G=-\Delta H+T\Delta S\] \[=+(6000\times 0.1)+\frac{180\times 0.1\times 300}{1000}\] \[=605.4\text{ }kJ\]

"> MOLES of sucrose \[=\frac{34.2}{342}=0.1\] \[-{{(\DELTA G)}_{T.P}}=\]USEFUL WORK done by the system \[-\Delta G=-\Delta H+T\Delta S\] \[=+(6000\times 0.1)+\frac{180\times 0.1\times 300}{1000}\] \[=605.4\text{ }kJ\]

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Combustion of sucrose is used by aerobic organisms for providing energy for the life sustaining processes. If all the capturing of energy from the reaction is done through electrical process (non P-V work) then calculate maximum available energy which can be captured by combustion of 34.2 g of sucrose Given: \[\Delta {{H}_{combustion}}\left( sucrose \right)=-\text{ }6000\text{ }kJ\text{ }mo{{l}^{-1}}\] \[\Delta {{S}_{combustion}}=180J/Kmol\] and body temperature is 300 K

NEET Chemistry in NEET 1 year ago

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[d] No. of MOLES of sucrose \[=\frac{34.2}{342}=0.1\] \[-{{(\DELTA G)}_{T.P}}=\]USEFUL WORK done by the system \[-\Delta G=-\Delta H+T\Delta S\] \[=+(6000\times 0.1)+\frac{180\times 0.1\times 300}{1000}\] \[=605.4\text{ }kJ\]

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Posted on 03 Aug 2024, this text provides information on NEET related to Chemistry in NEET. Please note that while accuracy is prioritized, the data presented might not be entirely correct or up-to-date. This information is offered for general knowledge and informational purposes only, and should not be considered as a substitute for professional advice.

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