A) \[2HI(g)\to {{H}_{2}}(g)+{{I}_{2}}(g)\]
B) \[2S{{O}_{2}}(g)+{{O}_{2}}(g)\to 2S{{O}_{3}}(g)\]
C) \[{{N}_{2}}(g)+3{{H}_{2}}(g)\to 2N{{H}_{3}}(g)\]
D) \[2NH{{O}_{2}}(g)\to {{N}_{2}}{{O}_{4}}(g)\]
Correct Answer: A
Solution :
For the reaction, \[2HI(g)\to {{H}_{2}}(g),\Delta H\] is equal to \[\Delta U\] \[\Delta H=\Delta U=\Delta nRT\] For the reaction, \[2HI(g)\to {{H}_{2}}(g)+{{I}_{2}}(g),\Delta n=1+1-2=0\] Hence, \[\Delta H=\Delta U+(0)RT\] \[\Delta H=\Delta U\] Note: \[\Delta n\] is the difference between the number of moles of gaseous products and the number of moles of gaseous reactants.
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