Question

Compute the heat of formation of liquid methyl alcohol (in kJ mol-1) using the following data. The heat of vaporisation of liquid methyl alcohol is $38kJmo{l^{ - 1}}$
The heat of formation of gaseous atoms from the elements in their standard states:
$H = 218kJmo{l^{ - 1}}$,$C = 715kJmo{l^{ - 1}}$,$O = 249kJmo{l^{ - 1}}$
Average bond energies:
$C - H = 415kJmo{l^{ - 1}}$
$C - O = 356kJmo{l^{ - 1}}$
$O - H = 463kJmo{l^{ - 1}}$
A. $\Delta H = - 266kJmo{l^{ - 1}}$
B. $\Delta H = + 266kJmo{l^{ - 1}}$
C. $\Delta H = - 190kJmo{l^{ - 1}}$
D. None of these

Answer 1

We can say that heat of formation is the quantity of heat absorbed (or) released when one mole of a compound is produced from its fundamental elements, each substance being in its regular physical state such as gas, solid, (or) liquid. We have to know that the formation reactant is a process at constant temperature and constant pressure.

Complete step by step answer:
Given data contains,
The heat of formation of gaseous atoms from the elements in their standard states as,
$H = 218kJmo{l^{ - 1}}$,$C = 715kJmo{l^{ - 1}}$,$O = 249kJmo{l^{ - 1}}$
The average bond energies are:
$C - H = 415kJmo{l^{ - 1}}$
$C - O = 356kJmo{l^{ - 1}}$
$O - H = 463kJmo{l^{ - 1}}$
We can calculate the heat of formation using Hess’s law of summation of heat. We can calculate the heat absorbed (or) heat released in any chemical reaction by adding the known steps of formation (or) combustion for the step in that particular reaction.
We can write the thermochemical equation for the heat of vaporization of methyl alcohol as,
$C{H_3}O{H_{\left( l \right)}} \to C{H_3}O{H_{\left( g \right)}}$ $\Delta H = 38kJmo{l^{ - 1}}$..(1)
We can write the thermochemical equation for the hydrogen formation as,
$\dfrac{1}{2}{H_{2(g)}} \to {H_{\left( g \right)}}$ $\Delta H = 218kJmo{l^{ - 1}}$..(2)
The thermochemical equation for formation of carbon is written as,
$C\left( {graphite} \right) \to {C_{\left( g \right)}}$ $\Delta H = 715kJmo{l^{ - 1}}$..(3)
We can write the thermochemical equation for oxygen formation as,
$\dfrac{1}{2}{O_{2\left( g \right)}} \to {O_{\left( g \right)}}$ $\Delta H = 249kJmo{l^{ - 1}}$..(4)
We can write the equation for the average bond energy of $C - H$ as,
$C - {H_{\left( g \right)}} \to {C_{\left( g \right)}} + {H_{\left( g \right)}}$ $\Delta H = 415kJmo{l^{ - 1}}$..(5)
We can write the equation for the average bond energy of $C - O$ as,
$C - {O_{\left( g \right)}} \to {C_{\left( g \right)}} + {O_{\left( g \right)}}$ $\Delta H = 356kJmo{l^{ - 1}}$..(6)
We can write the equation for the average bond energy of $O - H$ as,
$O - {H_{\left( g \right)}} \to {H_{\left( g \right)}} + {O_{\left( g \right)}}$ $\Delta H = 463kJmo{l^{ - 1}}$..(7)
We can write the formation of methyl alcohol as,
${C_{\left( s \right)}} + 2{H_{2\left( g \right)}} + \dfrac{1}{2}{O_{2(g)}} \to C{H_3}O{H_{\left( g \right)}}$
We need the following energies:
$2{H_{2\left( g \right)}} \to 4{H_{\left( g \right)}}$
$C\left( {graphite} \right) \to {C_{\left( g \right)}}$
$\dfrac{1}{2}{O_{2\left( g \right)}} \to {O_{\left( g \right)}}$
We can write the total energy needed as,
Total energy $= \left( {4 \times 218} \right) + 715 + 249$
Total energy $= 1836kJmo{l^{ - 1}}$
So the total energy required is $1836kJmo{l^{ - 1}}$.
Using equation (5),(6) and (7), we can see the following bonds are formed. The number of $C - H$ bonds is three, the number of $C - O$ bonds is one and the number of $O - H$ bonds is one.
We can calculate the total energy liberated as,
Total energy liberated $= \left( {3 \times 415} \right) + 356 + 463$
Total energy liberated $= 2064kJmo{l^{ - 1}}$
Thus, the total energy liberated as $2064kJmo{l^{ - 1}}$.
The value $228$ is obtained from the total energy required and the total energy released.
We can calculate the heat of formation for methyl alcohol using the equations (1) and (2).
Heat of formation for methyl alcohol=$- 228 - 38$
Heat of formation for methyl alcohol=$- 266kJmo{l^{ - 1}}$
Thus, the heat of formation of methyl alcohol is $- 266kJmo{l^{ - 1}}$.

So, the correct answer is Option A.

Note: We can call heat of formation as standard heat of formation, standard enthalpy of formation. We have to know that standard enthalpy of formation for any element present in its standard state takes a value of zero. This is because there is no change taking place in their formation.