Question

Calculate standard heats of formation of carbon-di-sulphide $\left(l\right)$. Given the standard heat of combustion of carbon $\left(s\right)$, sulphur $\left(s\right)$ and carbon-di-sulphide $\left(l\right)$ are: $-393.3,-293.72$ and $-1108.76kJ$ ${mol}^{-1}$ respectively (only magnitude in nearest integer in kJ).

Answer 1

$R_1:C\left(s\right)+S\left(s\right)\to C{S}_2\left(l\right);\Delta H_1=?$

Given that

$R_2:C\left(s\right)+O_2\left(g\right)\to C{O}_2\left(g\right);\Delta H_2=-393.3kJmo{l}^{-1}$

$R_3:S\left(s\right)+O_2\left(g\right)\to S{O}_2\left(g\right);\Delta H_3=-293.72kJmo{l}^{-1}$

$R_4:C{S}_2\left(l\right)+3O_2\left(g\right)\to C{O}_2\left(g\right)+2S{O}_2;\Delta H_4=-1108.76kJmo{l}^{-1}$

And

$R_1=R_2+2R_3-R_4$

$\Rightarrow \Delta H_1=\Delta H_2+2\Delta H_3-\Delta H_4=-393.3+2\times (-293.72)+1108.76=-128kJmo{l}^{-1}$

Answer = 128