Calculate standard enthalpies of formation of carbon-di-sulphide(l). Given the standard enthalpy of combustion of carbon(s), sulphur(s) & carbon-di-sulphide(l) are: $- 393, - 293$ and $- 1108 k J m o l^{- 1}$ respectively.

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Solution

Given:-
$C_{(s)} + {O_{2}}_{(g)} ⟶ {C O_{2}}_{(g)} Δ H = - 393 k J / m o l . . . . . (1)$
$S_{(s)} + {O_{2}}_{(g)} ⟶ {S O_{2}}_{(g)} Δ H = - 293 k J / m o l$
$2 \times [S_{(s)} + {O_{2}}_{(g)} ⟶ {S O_{2}}_{(g)} Δ H = - 293 k J / m o l]$
$2 S_{(s)} + 2 {O_{2}}_{(g)} ⟶ 2 {S O_{2}}_{(g)} Δ H = - 586 k J / m o l . . . . . (2)$
${C S_{2}}_{(l)} + 3 {O_{2}}_{(g)} ⟶ {C O_{2}}_{(g)} + 2 {S O_{2}}_{(g)} Δ H = - 1108 k J / m o l$
${C O_{2}}_{(g)} + 2 {S O_{2}}_{(g)} ⟶ {C S_{2}}_{(l)} + 3 {O_{2}}_{(g)} Δ H = 1108 k J / m o l . . . . . (3)$
Adding ${e q}^{n} (1), (2) & (3)$ , we have
$C_{(s)} + 2 S_{(s)} ⟶ {C S_{2}}_{(l)} Δ H = 129 k J / m o l$
Hence the standard enthalpy of formation of ${C S_{2}}_{(l)}$ is $129 k J / m o l$ .

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