A) \[\text{C}{{\text{H}}_{\text{3}}}\text{CON}{{\text{H}}_{2}}\xrightarrow{NaN{{O}_{2}}/HCl}X\]
B) \[\text{C}{{\text{H}}_{\text{3}}}\text{COOH}\]
C) \[\overset{+}{\mathop{\text{C}{{\text{H}}_{\text{3}}}\text{CON}{{\text{H}}_{\text{3}}}{{\text{C}}^{-}}}}\,\]
D) \[\text{C}{{\text{H}}_{\text{3}}}\text{N}{{\text{H}}_{\text{2}}}\]
Correct Answer: D
Solution :
We know that,\[\Delta H=\Delta E+\Delta {{n}_{g}}RT\]For reaction, \[{{N}_{2}}{{O}_{4}}\xrightarrow{{}}2N{{O}_{2}}\] \[\Delta {{n}_{g}}=2-1=1\] \[\therefore \] \[\Delta H=\Delta E+RT\] Or \[\Delta H>\Delta E\]
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