Question

Given the following standard heat of reactions :
(i) heat of formation of water = $-68.3kcal$
(ii) heat of combustion of acetylene = $-310.6kcal$
(iii) heat of combustion of ethylene = $-337.2kcal$
Which of the following option(s) is / are correct for the hydrogenation of acetylene at constant volume $\left({27}^{\circ }C\right)$ ?

• A. Change in enthalpy of hydrogenation of acetylene is $-41.7\text{kcal}$
• B. Change in internal energy of hydrogenation of acetylene is $-41.7\text{kcal}$
• C. Change in enthalpy of hydrogenation of acetylene is $-41.1\text{kcal}$
• D. Change in internal energy of hydrogenation of acetylene is $-41.1\text{kcal}$

Answer 1

Answer: (A), (D)

The correct options are
A Change in enthalpy of hydrogenation of acetylene is $-41.7\text{kcal}$
D Change in internal energy of hydrogenation of acetylene is $-41.1\text{kcal}$

The given data can be written as follows
$\left(i\right)H_2\left(g\right)+\frac{1}{2}{O}_2\left(g\right)\to H_{2}O\left(l\right);\Delta H=-68.3kcal\left(ii\right)C_2{H}_2\left(g\right)+\frac{5}{2}{O}_2\left(g\right)\to H_{2}O\left(l\right)+2C{O}_2\left(g\right);\Delta H=-310.6kcal{C}_2{H}_4\left(g\right)+3O_2\to 2H_{2}O\left(l\right)+2C{O}_2\left(g\right);\Delta H=-337.2kcal$
The required thermochemical equation is obtained by substracting equation (iii) from the sum of the equations (i) and (ii)
$C_2{H}_2\left(g\right)+H_2\left(g\right)\to C_2{H}_4\left(g\right)$
$\Delta H=-68.3-310.6+337.2=-41.7kcal$

The change in internal energy of hydrogenation of acetylene at constant volume is given by :
$\Delta E=\Delta H-\Delta nRT=-41.7-\left(\right(-1)\times 2\times {10}^{-3}\times 300)=-41.7+0.6=-41.1kcal$