Question

$\Delta {H^0}_f$ $298K$ of methanol is given by the chemical equation :
A. $C{H_4}_{(g)} + \dfrac{1}{2}{O_2}_{(g)} \to C{H_3}O{H_{(g)}}$
B. ${C_{(graphite)}} + \dfrac{1}{2}{O_2}_{(g)} + 2{H_2}_{(g)} \to C{H_3}O{H_{(l)}}$
C. ${C_{\left( {diamond} \right)}} + \dfrac{1}{2}{O_{2(g)}} + 2{H_{2\left( g \right)}} \to C{H_3}O{H_{(l)}}$
D. $C{O_{\left( g \right)}} + 2{H_{2\left( g \right)}} \to C{H_3}O{H_{\left( l \right)}}$

Answer 1

Hint: Here in this problem to know which equation is suitable for the formation of methanol we have to find out first the equation where all the constitutes are in their standard state .In the standard state, the heat of formation of constitutes is zero.

Complete answer:
The enthalpy change for chemical reaction is given by the equation ,$\Delta {H^\theta } = {H^\theta }_p - {H^\theta }_R$. Where $\Delta {H^\theta }$ is enthalpy change .The relation between enthalpy and internal energy is expressed by the equation ,$\Delta H = \Delta U + \Delta nRT$,where $\Delta U$ is change in internal energy ,$\Delta n$ sis equal to change in the number of the moles and $R =$gas constant.
In the given equation in ${C_{(graphite)}} + \dfrac{1}{2}{O_2}_{(g)} + 2{H_2}_{(g)} \to C{H_3}O{H_{(l)}}$ has all the constituents in their standard state so this equation is correct for the formation of methanol and we can easily calculate the heat of formation here because in rest of the option it is not possible to calculate heat of formation because either the constituents are not in their standard state or equation is not balanced. Graphite and diamond are the allotropes of carbon .The structure of graphite and diamond are simple hexagonal and face-centred cubic respectively. But graphite is the most stable form of carbon under standard conditions.

Note: Synthetically methanol can be prepared by heating by carbon monoxide and hydrogen gases under pressure in the presence of a catalyst and the reaction must contain all the constituents in their standard form. The reaction that takes place in the formation of methanol can be written as ;
${C_{(graphite)}} + \dfrac{1}{2}{O_2}_{(g)} \to C{O_{\left( g \right)}}......(i)$
$C{O_{\left( g \right)}} + 2{H_{2\left( g \right)}} \to C{H_3}O{H_{\left( l \right)}}......(ii)$
With the help of equations (i) and (ii),
${C_{(graphite)}} + \dfrac{1}{2}{O_2}_{(g)} + 2{H_2}_{(g)} \to C{H_3}O{H_{(l)}}$
Now we have the final equation which gives the heat of formation of methanol.