A) 600 kJ
B) 594.6 kJ
C) 5.4 kJ
D) 605.4 kJ
Correct Answer: D
Solution :
[d] No. of moles of sucrose \[=\frac{34.2}{342}=0.1\] \[-{{(\Delta G)}_{T.P}}=\]useful work done by the system \[-\Delta G=-\Delta H+T\Delta S\] \[=+(6000\times 0.1)+\frac{180\times 0.1\times 300}{1000}\] \[=605.4\text{ }kJ\]You need to login to perform this action.
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