Chemical Thermodynamics Question 488
Question: The in $ {\Delta_{f}}H{}^\circ (N_2O_5,g) $ kJ/mol on the basis of the following data is -
$ 2NO(g)+O_2(g)\to 2NO_2(g); $
$ {\Delta_{r}}H{}^\circ =-114,kJ/mol $
$ 4NO_2(g)+O_2(g)\to 2N_2O_5(g); $
$ {\Delta_{r}}H{}^\circ =-102.6,kJ/mol $
$ {\Delta_{f}}H{}^\circ (NO,g)=90.2kJ/mol $
Options:
A) 15.1
B) 30.2
C) - 36.2
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
- $ \frac{1}{2}N_2(g)+\frac{1}{2}O_2(g)\xrightarrow[{}]{{}}NO(g);{\Delta_{f}}H^{o}=90.2 $
$ N_2(g)+O_2(g)\xrightarrow[{}]{{}}2NO(g);{\Delta_{f}}H^{o}=90.2\times 2 $ …(1) $ 2NO(g)+O_2(g)\xrightarrow[{}]{{}}2NO_2(g);{\Delta_{f}}H^{o}=-114 $ …(2) $ 2NO_2(g)+\frac{1}{2}O_2(g)\xrightarrow[{}]{{}}N_2O_5(g); $
$ \Delta H^{o}=\frac{-102.6}{2}=-51.3 $ …(3) $ Eq.(1)+(2)+(3) $
$ N_2(g)+\frac{5}{2}O_2(g)\xrightarrow[{}]{{}}N_2O_5(g) $
$ {\Delta_{f}}H^{o}(N_2O_5,g)=15.1\text{kJ/mol} $
