| The in \[{{\Delta }_{f}}H{}^\circ ({{N}_{2}}{{O}_{5}},g)\] kJ/mol on the basis of the following data is - |
| \[2NO(g)+{{O}_{2}}(g)\to 2N{{O}_{2}}(g);\]\[{{\Delta }_{r}}H{}^\circ =-114\,kJ/mol\] |
| \[4N{{O}_{2}}(g)+{{O}_{2}}(g)\to 2{{N}_{2}}{{O}_{5}}(g);\]\[{{\Delta }_{r}}H{}^\circ =-102.6\,kJ/mol\] |
| \[{{\Delta }_{f}}H{}^\circ (NO,g)=90.2kJ/mol\] |
A) 15.1
B) 30.2
C) - 36.2
D) None of these
Correct Answer: A
Solution :
| [a] \[\frac{1}{2}{{N}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow[{}]{{}}NO(g);{{\Delta }_{f}}{{H}^{o}}=90.2\] | |
| \[{{N}_{2}}(g)+{{O}_{2}}(g)\xrightarrow[{}]{{}}2NO(g);{{\Delta }_{f}}{{H}^{o}}=90.2\times 2\] | ...(1) |
| \[2NO(g)+{{O}_{2}}(g)\xrightarrow[{}]{{}}2N{{O}_{2}}(g);{{\Delta }_{f}}{{H}^{o}}=-114\] | ...(2) |
| \[2N{{O}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow[{}]{{}}{{N}_{2}}{{O}_{5}}(g);\]\[\Delta {{H}^{o}}=\frac{-102.6}{2}=-51.3\] | ...(3) |
| \[Eq.(1)+(2)+(3)\] |
| \[{{N}_{2}}(g)+\frac{5}{2}{{O}_{2}}(g)\xrightarrow[{}]{{}}{{N}_{2}}{{O}_{5}}(g)\] |
| \[{{\Delta }_{f}}{{H}^{o}}({{N}_{2}}{{O}_{5}},g)=15.1\text{kJ/mol}\] |
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