Thermodynamics Question 103
Question: For the reaction $ H _{2}(g)+\frac{1}{2}O _{2}(g)\to H _{2}O(l),\Delta H=-285.8kJmo{{l}^{-1}} $ $ \Delta S=-0.163kJmo{{l}^{-1}}{{K}^{-1}}. $ What is the value of free energy change at $ 27^{o}C $ for the reaction? [KCET 1999]
Options:
A) $ -236.9kJmo{{l}^{-1}} $
B) $ -281.4kJmo{{l}^{-1}} $
C) $ -334.7kJmo{{l}^{-1}} $
D) $ +334.7kJmo{{l}^{-1}} $
Show Answer
Answer:
Correct Answer: A
Solution:
$ \Delta G=\Delta H-T\Delta S,T=27+273=300K $
$ \Delta G=(-285.8)-(300)(-0.163)=-236.9kJmo{{l}^{-1}} $
