Question

The heat of formation of ${C_2}{H_5}OH(l)$ is $ - 66\text{kcal/mole}.$ The heat of combustion of the $C{H_3}OC{H_3}(g)$ is $ = - 348\text{kcal/mole}.$ . $\Delta {H_f}$ for ${H_2}O$ and $C{O_2}$ are $ - 68\text{kcal/mole}$ and $ - 94\text{kcal/mole}$ respectively. Then, $\Delta H$ for the isomerization reaction ${C_2}{H_5}OH(l) \to C{H_3}OC{H_3}\left( g \right)$ , and $\Delta E$ of the same reaction are at $T = {25^o}C.$
A.$\Delta H = 18\text{kcal/mole},\Delta E = 17.301\text{kcal/mole}$
B.$\Delta H = 22\text{kcal/mole},\Delta E = 21.408\text{kcal/mole}$
C.$\Delta H = 26\text{kcal/mole},\Delta E = 25.709\text{kcal/mole}$
D.$\Delta H = 30\text{kcal/mole},\Delta E = 28.522\text{kcal/mole}$

Answer 1

Hint: Heat of combustion is the heat liberated when one mole of a substance is burnt in the presence of oxygen at constant volume. The heat absorbed or evolved when one mole of a compound is formed from its constituent elements is called the heat of formation. Isomerization reaction is a chemical reaction where a compound is transformed into any of its isomeric forms. A negative value for the heat of formation $\Delta {H_f}$ indicates that the formation of a compound is exothermic and a positive value indicates heat of formation of a compound is endothermic.

Complete step by step answer:
First, let us write down the information given in the question.
The heat of formation of ${{{C}}_{{2}}}{{{H}}_{{5}}}{{OH(l) = - 66\text{kcal/mole}}}$
The heat of combustion of $C{H_3}OC{H_3}(g) = - 348\text{kcal/mole}$
$\Delta {H_f}$ for ${H_2}O = - 68\text{kcal/mole}$
$\Delta {H_f}$ for $C{O_2} = - 94\text{kcal/mole}$
We have to find $\Delta H$ and $\Delta E$ for the isomerization of ${C_2}{H_5}OH(l) \to C{H_3}OC{H_3}(g)$
Now let’s write down the reaction steps:
The reaction for the formation of ${C_2}{H_5}OH(l)$ is given below:
$2C + 3{H_2} + \dfrac{1}{2}{O_2} \to {C_2}{H_5}OH(l)$ , $\Delta H = - 66\text{kcal/mole}$ Reaction (I)
The combustion reaction $C{H_3}OC{H_3}$ is given below:
$C{H_3}OC{H_3} + 3{O_2} \to 2C{O_2} + 3{H_2}O$ , $\Delta H = - 348\text{kcal/mole}$ Reaction (II)
We know that the combustion reaction takes place in the presence of oxygen.
The reaction for the formation of ${H_2}O$ and $C{O_2}$ is given below:
${H_{{2_{(g)}}}} + \dfrac{1}{2}{O_{{2_{(g)}}}} \to {H_2}O$ , $\Delta {H_f} = - 68\text{kcal/mole}$ Reaction (III)
$C + {O_2} \to C{O_2}$ , $\Delta {H_f} = - 94\text{kcal/mole}$ Reaction (IV)
We know the isomerization reaction is ${C_2}{H_5}OH(l) \to C{H_3}OC{H_3}\left( g \right)$ Reaction (V)
To get the above isomerization reaction we have to follow certain steps as follows:
Here in the isomerization reaction ${C_2}{H_5}OH(l)$ is on the reactant side but in reaction (I) ${C_2}{H_5}OH(l)$ is on the product side. So we have to consider the reverse of the reaction (I) as shown here ${C_2}{H_5}O{H_{(l)}} \to 2C + 3{H_2} + \dfrac{1}{2}{O_2}$ Reaction (VI)
That is what we have $\Delta H = + 66\text{kcal/mole}$.
Similarly in the isomerization reaction (reaction (V)) $C{H_3}OC{H_3}(g)$ is on the product side so we have to reverse the reaction (II) that is to take $\Delta H = + 348\text{kcal/mole}$
Thus the reaction is $2C{O_2} + 3{H_2}O \to C{H_3}OC{H_3} + 3{O_2}$ Reaction (VII).
Next, we have to multiply the reaction (III) with and reaction (IV) with as shown below:
$({H_{{2_{(g)}}}} + \dfrac{1}{2}{O_{{2_{(g)}}}} \to {H_2}O) \times 3,\Delta H = - 68 \times 3 = - 204\text{kcal/mole}$ Reaction (VIII)
$(C + {O_2} \to C{O_2}) \times 2,\Delta H = - 94 \times 2 = - 188\text{kcal/mole}$ Reaction (IX)
Thus combining reaction (VI), (VII), (VIII) and reaction (IX) and canceling the common species on the product and reactant side we get the isomerization reaction (V) as follows
${C_2}{H_5}OH(l) \to C{H_3}OC{H_3}\left( g \right)$
Calculating $\Delta H$ for the above reaction as follows:
$\Delta H = + 66 + 348 - 204 - 188 = 22\text{kcal/mole}$
We know about the thermodynamic equation $\Delta H = \Delta E + \Delta {n_g}RT$ Equation (I)
${{\Delta }}{{{n}}_{{g}}}{{ = \text{The number of moles of gaseous products - number of moles of gaseous reactants}}}.$
Here in Reaction (V), there is only one gaseous product and there is no gaseous reactant. Thus for the isomerization reaction (reaction (V)), $\Delta {n_g} = 1 - 0 = 1$
Temperature $ = {25^o}C = 25 + 273 = 298K$
The value of the universal gas constant in $kcal$ is $1.987 \times {10^{ - 3}}kcal{K^{ - 1}}mo{l^{ - 1}}$
Substituting all the values in the above equation (I) we can calculate $\Delta E$ for the isomerization reaction (V) as shown below:
 $ \Delta H = 22\text{kcal/mole} $
$\Delta E = 21.408\text{kcal/mole} $
Therefore the answer is an option (B).

Note:
The equations for the thermodynamic quantity $\Delta H$ is:
$ \Rightarrow \Delta H = \Delta U + P\Delta V$
$ \Rightarrow \Delta H = \Delta E + P\Delta V$
$ \Rightarrow PV = nRT$
$ \Rightarrow P\Delta V = \Delta {n_g}RT$
$\therefore \Delta H = \Delta E + \Delta {n_g}RT$
The value of universal gas constant R should be in $kcalK^{ - 1}mol^{ - 1}$ since the unit of energy is in $kcal/mol$. The temperature should be in the unit kelvin.
${{\Delta }}{{{n}}_{{g}}}{{ = \text{The number of moles of gaseous products - number of moles of gaseous reactants}}}.$