Question

The standard heat of formation of carbon disulphide (l), given that the standard heat of combustion of carbon (s), sulphur (s) and carbon disulphide (l) are $-393.3,-293.72$ and $-1108.76kJmo{l}^{-1}$ respectively is:

• A. $-128.02kJmo{l}^{-1}$
• B. $+12.802kJmo{l}^{-1}$
• C. $+128.02kJmo{l}^{-1}$
• D. $-12.802kJmo{l}^{-1}$

Answer 1

Answer: (C)

$+128.02kJmo{l}^{-1}$

Heat of formation of $C{S}_2$

$C\left(graphite\right)+2S⟶C{S}_2$

To get the above eq from the given reactions-

$C+O_2\to C{O}_2$ (i)

$S+O_2\to S{O}_2$(ii)

$C{S}_2+O_2\to C{O}_2+S{O}_2$ (iii)

Multiply the 2nd reaction with $2$ and reverse the third reaction.

From this we will get-

$\Delta H_f=-393.3-\left(2\right)\times 293.72+1108.76$

$\Delta H_f=+128.02\frac{KJ}{mol}$