1. Tardigrade
  2. Question
  3. Chemistry
  4. Given the following standard heats of reactions (i) heat of formation of water =-68.3 kcal (ii) heat of combustion of acetylene =-310.6 kcal (iii) heat of combustion of ethylene =-337.2 kcal Calculate the heat of reaction for the hydrogenation of acetylene at constant volume (25° C ).

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Q. Given the following standard heats of reactions
(i) heat of formation of water $=-68.3 \,kcal$
(ii) heat of combustion of acetylene $=-310.6 \, kcal$
(iii) heat of combustion of ethylene $=-337.2 \, kcal$
Calculate the heat of reaction for the hydrogenation of acetylene at constant volume $\left(25^{\circ} C \right)$.

IIT JEEIIT JEE 1984Thermodynamics

Solution:

$C _{2} H _{2}+ H _{2} \longrightarrow C _{2} H _{4}$
$\Delta H^{\circ}=\Sigma \Delta H_{\text {comb }}^{\circ}$ (reactants) $-\Sigma \Delta H_{\text {comb }}^{\circ}$ (products)
$=-310.6-68.3-(-337.2)$
$=-41.7 \,kcal$