Question

Calculate the standard heat of formation of carbon disulphide(l). Given that standard heats of combustion of carbon(s), sulphur(s) and carbon disulphide are \[ - 393.3, - 239.7\,and\, - 1108.76kJmo{l^{ - 1}}\] respectively.

Answer 1

Hint: Heat of formation is also called as standard enthalpy of formation of compounds. We need to find the reactions for formation of carbon disulphide from the given elements. Combustion is the process wherein any component combines with oxygen.

Complete step by step answer:
The standard enthalpy change for the formation of one mole of a compound from its elements is called standard enthalpy or heat of formation. Its symbol is ${\Delta _f}{H^\Theta }$ .
When one mole of a substance undergoes combustion and all the reactants and products are in the standard state then the enthalpy change that occurs per mole is called as standard enthalpy of combustion. Combustion reactions are exothermic. Its symbol is ${\Delta _c}{H^\Theta }$ .
Now the equation for the formation of carbon disulphide is-
${C_{(s)}} + 2{S_{(s)}} \to C{S_2}(l)\,\,\, ------- (1)$
We have to find the standard heat of formation for carbon disuphide ${\Delta _f}{H^\Theta }$ .
Now from the given data we can write the equation for combustion reaction as follows-
${C_{(s)}} + {O_2}(g) \to C{O_2}(g)\,$ $\Delta H = - 393.3kJ\,\,\, -------- (2)$
$S(s) + {O_2}(g) \to S{O_2}(g)$ $\Delta H = - 293.72kJ\,\,\, --------- (3)$
$C{S_2}(l) + 3{O_2}(g) \to C{O_2}(g) + S{O_2}(g)$ $\Delta H = - 1108.76kJ\,\,\,\, ------- (4)$
Now we have to get the equation $1$ by using the above equation.
Multiply equation three by 2
$2S(s) + 2{O_2}(g) \to 2S{O_2}(g)\,\,\,\,\Delta H = - 587.44kJ\,\,\,\, --------- (5)$
Adding equation $2$ and $5$ and subtracting equation $4$ , we get-
$C(s) + 2S(s) + 3{O_2}(g) - C{S_2}(l) - 3{O_2}(g) \to C{O_2}(g) + 2S{O_2}(g) - C{O_2}(g) - 2S{O_2}(g) $
The final equation becomes-
${C_{(s)}} + 2{S_{(s)}} \to C{S_2}(l)\,\,$
So this is the required equation. Now we have to also add the enthalpy values accordingly.
${\Delta _f}{H^\Theta } = - 393.3 - 587.4 + 1108.76 = 128.02kJmo{l^{ - 1}}$
So the standard heat of formation of carbon disulphide is $128.02kJmo{l^{ - 1}}$.

Note: Enthalpy is a form of energy. In simple words it is the amount of internal energy that is present in any system.
Combustion is exothermic in nature. So enthalpy of combustion is always negative.
Enthalpy is a state function.
$1calorie = 4.184joules$ and $1kJ = 1000J$ .