Question

Given below are some standard heat of reaction:

(a) Heat of formation of water $=-68.3kcal$

(b) Heat of combustion of acetylene $=-310.6kcal$

(c) Heat of combustion of ethylene $=-337.2kcal$.

The heat of reaction for the hydrogenation of acetylene at the constant volume at ${25}^{\circ }C$.

• A. $-41.104kcal$
• B. $-49.345kcal$
• C. $-62.203kcal$
• D. None of these

Answer 1

Answer: (A)

$-41.104kcal$

$H_2\left(g\right)+\frac{1}{2}{O}_2\to H_{2}O\left(l\right);$

$△H=-68.3Kcal$

$C_2{H}_2\left(g\right)+2\frac{1}{2}{O}_2\left(g\right)\to 2C{O}_2\left(g\right)+H_{2}O\left(1\right);$

$△H=-310.6Kcal$

$C_2{H}_4\left(g\right)+3O_2\left(g\right)\to 2C{O}_2\left(g\right)+2H_{2}O\left(1\right);$

$△H=-337.2Kcal$

$△E$ of the following equation is to be calculated:

$C_2{H}_2\left(g\right)+H_2\left(g\right)\to C_2{H}_4\left(g\right);△E=?$

Sum up the equations (i) and (ii) and subtract equation (iii)-

$\overline{C_2{H}_2\left(g\right)+H_2\left(g\right)\to C_2{H}_4\left(g\right);△H=-41.7Kcal}$

$\overline{}$

$△H$ is the heat of reaction at constant pressure. Heat at constant volume can be calculated by using the following equation:

$△H=△E+△n_{g}RT$

$△n_g$= moles of gaseous products - moles of gaseous reactants

$=1-2=-1$

$R=0.002Kcal/mole$

$T=273+25=298K$

$△E=-41.7-(-1\times 0.002\times 298)$

$=-41.104Kcal$