A) \[-23.0\text{ }kJ\text{ }mo{{l}^{-1}}\]
B) \[-46\text{ }kJ\text{ }mo{{l}^{-1}}\]
C) \[-92.0\text{ }kJ\text{ }mo{{l}^{-1}}\]
D) \[46\text{ }kJ\text{ }mo{{l}^{-1}}\]
Correct Answer: B
Solution :
\[2N{{H}_{3}}(g)\xrightarrow{{}}{{N}_{2}}(g)+3{{H}_{2}}(g);\] \[\Delta H=+92\,kJ\] \[{{N}_{2}}(g)+3{{H}_{2}}(g)\xrightarrow{{}}2N{{H}_{3}}(g);\] \[\Delta H=-92\,kJ\] \[\frac{1}{2}{{N}_{2}}(g)+\frac{3}{2}{{H}_{2}}(g)\xrightarrow{{}}N{{H}_{3}}(g);\] \[\Delta H=-\,46\,kJ\] Hence, enthalpy of formation of\[N{{H}_{3}}\] is \[-\text{ }46\text{ }kJ\] \[mo{{l}^{-1}}\].
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