1. Tardigrade
  2. Question
  3. Chemistry
  4. At 25° C and 1 atm pressure, the enthalpy of combustion of benzene (1) and acetylene ( g ) are -3268 kJ mol -1 and -1300 kJ mol -1, respectively. The change in enthalpy for the reaction 3 C 2 H 2( g ) arrow C 6 H 6( l ), is

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Q. At $25^{\circ} C$ and $1 $ atm pressure, the enthalpy of combustion of benzene (1) and acetylene ( $g )$ are $-3268 \, kJ \,mol ^{-1}$ and $-1300 \,kJ\, mol ^{-1}$, respectively. The change in enthalpy for the reaction $3 C _{2} H _{2}( g ) \rightarrow C _{6} H _{6}( l )$, is

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Solution:

$\Delta H =\sum \Delta H _{\text {Combustion }} ($ Reactant $)-\sum \Delta H _{\text {Combustion }}$(Product)
$=3 \times(-1300)-[-3268]$
$=-632 \,kJ \,mol ^{-1}$