A) 46
B) -46
C) 92
D) -92
Correct Answer: C
Solution :
[c] \[\frac{1}{2}{{N}_{2}}(g)+\frac{3}{2}{{H}_{2}}(g)\to N{{H}_{3}}(g)\] \[\Delta {{H}_{r}}=-46\,kJ/mol\] \[2N{{H}_{3}}(g)\to {{N}_{2}}(g)+3{{H}_{2}}(g)\] \[\Delta {{H}_{r}}=-2\Delta {{H}_{r}}\] \[=-2(-46)=92kJ\]You need to login to perform this action.
You will be redirected in
3 sec