A) \[{{H}_{2}}(g)\xrightarrow{{}}2{{H}_{\text{atom}}}\] \[\Delta {{H}^{o}}=436\,kJ,\Delta {{S}^{o}}=90.7\,\text{eu}\]
B) \[\frac{1}{2}{{N}_{2}}(g)+\frac{1}{2}{{O}_{2}}(g)\xrightarrow{{}}NO(g);\] \[\Delta {{H}^{o}}=90.3\,kJ,\Delta {{S}^{o}}=3.0\,\text{eu}\]
C) \[2N{{O}_{2}}(g)\xrightarrow{{}}{{N}_{2}}{{O}_{4}}(g)\] \[\Delta {{H}^{o}}=-56.0\,kJ\,\Delta {{S}^{o}}=-17.7\text{eu}\]
D) \[{{H}_{2}}{{O}_{2}}(g)\xrightarrow{{}}{{H}_{2}}O(l)+\frac{1}{2}{{O}_{2}}(g)\] \[\Delta {{H}^{o}}=-98.3\,kJ\Delta {{S}^{o}}=80.0\,\text{eu}\]
Correct Answer: D
Solution :
For a spontaneous process, \[\Delta G\]must be negative. \[\Rightarrow \] \[\Delta G=\Delta H-T\Delta S\] \[\therefore \]For negative value of \[\Delta G\]at all temperatures, \[\Delta H=-ve\] \[\Delta S=+\,ve\]
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