Question

Calculate the standard heat of formation of carbon disulphide in liquid state. Given that the standard heat of combustion of carbon $(s)$ sulphur $(s)$ and carbon disulphide $(l)$ are $393.3, - 29a72$ and $-1108.76 kJ\,mol^{-1}$ respectively is:

• A. $+ 128.02\,kJ\,mol^{-1}$
• B. $-12802\,kJ\,mol^{-1}$
• C. $+ 12.802\,kJ\,mol^{-1}$
• D. $-128.02\, kJ\,mol^{-1}$

Answer 1

Answer: (A)

Use Hesss law to solve the problem.
Given :
$C(s)+O_{2}(g) \rightarrow C O_{2}(g) \Delta H_{1}=-393.3\, kJ$... (i) $S(s)+O_{2}(g) \rightarrow S O_{2}(g) \Delta H_{2}=-293.72\, kJ$...(ii)
$C S_{2}(l)+3 O_{2}(g) \rightarrow C O_{2}(g)+2 S O_{2}(g)$
$\Delta H_{3}=-1108.76\, kJ$ ... (iii)
Required equation $C(s)+2 S(s)-C S_{2}(g), \Delta H=?$
Multiply Eq. (ii) by 2 and add to Eq. (i)
$C(s)+2 S(s)-3 O_{2}(g) \rightarrow C O_{2}(g)+2 S O_{2}(g)$
$\Delta H_{4} =-392.3+2 \times-293.72=-979.14\, kJ$ ...(iv)
(iv) Subtract Eq. (iii) from Eq. (iv)
$C(s)+2 S(s) \rightarrow C S_{2}$
$\Delta H=-979.14-(-1108.76)=128.02\, kJ\, mol^{-1}$