Purification And Characterisation Of Organic Compounds Question 140
Question: The value of $ {\Delta _{f}}G{}^\circ $ for formation of $ Cr_2O_3 $ is $ -540kJmo{{l}^{-1}} $ and that of $ Al_2O_3 $ is $ -827kJmo{{l}^{-1}} $ What is the value of $ {\Delta _{f}}G{}^\circ $ for the reaction- $ \frac{4}{3}Al(s)+\frac{2}{3}Cr_2O_3(s)\to \frac{2}{3}Al_2O_3(s)+\frac{4}{3}Cr(s). $
Options:
A) $ -574kJmo{{l}^{-1}} $
B) $ -287kJmo{{l}^{-1}} $
C) $ +574kJmo{{l}^{-1}} $
D) $ +287kJmo{{l}^{-1}} $
Show Answer
Answer:
Correct Answer: B
Solution:
[b] The two equation are: $ \frac{4}{3}Al(s)+O_2(g)\to \frac{2}{3}Al_2O_3(s), $
$ {\Delta _{f}}G{}^\circ =-827kJmo{{l}^{-1}} $ ….
(1)
$ \frac{4}{3}Cr(s)+O_2(g)\to \frac{2}{3}Cr_2O_3(s), $
$ {\Delta _{f}}G{}^\circ =-540kJmo{{l}^{-1}} $ ….
(2)
Subtracting equation (ii) from equation (i) we have,
$ \frac{4}{3}Al(s)+\frac{2}{3}Cr_2O_3(s),\to \frac{2}{3}Al_2O_3(s)+\frac{4}{3}Cr(s), $
$ {\Delta _{r}}{{G}^{{}^\circ }}=-287kJmo{{l}^{-1}} $
BETA
