A) +184 kJ
B) +23 kJ
C) +92 kJ
D) +46 kJ
Correct Answer: C
Solution :
\[2N{{H}_{3}}(g)\xrightarrow{{}}{{N}_{2}}(g)+3{{H}_{2}}(g)\] \[\Delta {{H}_{r}}=-\](\[2\times \] enthalpy of formation of \[N{{H}_{3}}\]) \[=-(2\times -46)=92\,kJ\]You need to login to perform this action.
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