A) \[{{H}_{2}}(g)\,+B{{r}_{2}}(g)\xrightarrow{\,}2HBr(g)\]
B) \[C(s)+2{{H}_{2}}O(g)\xrightarrow{\,}\,2{{H}_{2}}(g)\,+C{{O}_{2}}(g)\]
C) \[PC{{l}_{5}}(g)\,\xrightarrow{\,}\,PC{{l}_{3}}(g)\,+C{{l}_{2}}(g)\]
D) \[2CO(g)+{{O}_{2}}(g)\,\xrightarrow{\,}\,2C{{O}_{2}}(g)\]
Correct Answer: A
Solution :
As we know that \[\Delta H=\Delta E+P\Delta V\] \[or\Delta E=\Delta E+\Delta nRT\] where \[\Delta \,\,H\to \] change in enthalpy of system (standard heat at constant pressure) \[\Delta \,E\to \] Change in internal energy of system (Standard heat at constant volume) \[\Delta \,n\to \] no. of gaseous moles of product - no. of gaseous moles of reactant \[R\to \] gas constant \[T\to \]absolute temperature If \[\Delta \]n = 0 for reactions which is carried out in an open container, therefore \[\Delta H=\Delta E\] So for reaction (1) \[\Delta n=2-2=0\] Hence, for reaction (1), \[\Delta H=\Delta E\]
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