The enthalpy of combustion of carbon disulfid | vedclass

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Question

(D) The combustion reaction of $CS_2$ is: $CS_2(l) + 3O_2(g) \rightarrow CO_2(g) + 2SO_2(g)$.The enthalpy of combustion is given by: $\Delta H_c = [\D

Vedclass · Accepted Answer

$-879.1 \, kJ \, mol^{-1}$

Vedclass · Answer

(D) The combustion reaction of $CS_2$ is: $CS_2(l) + 3O_2(g) ightarrow CO_2(g) + 2SO_2(g)$.The enthalpy of combustion is given by: $\Delta H_c = [\Delta H_f(CO_2) + 2 	imes \Delta H_f(SO_2)] - [\Delta H_f(CS_2) + 3 	imes \Delta H_f(O_2)]$.Given: $\Delta H_c = -110.2 \, kJ \, mol^{-1}$,$\Delta H_f(CO_2) = -394.5 \, kJ \, mol^{-1}$,$\Delta H_f(SO_2) = -297.4 \, kJ \, mol^{-1}$,and $\Delta H_f(O_2) = 0$.Substituting the values: $-110.2 = [-394.5 + 2 	imes (-297.4)] - [\Delta H_f(CS_2) + 0]$.$-110.2 = [-394.5 - 594.8] - \Delta H_f(CS_2)$.$-110.2 = -989.3 - \Delta H_f(CS_2)$.$\Delta H_f(CS_2) = -989.3 + 110.2 = -879.1 \, kJ \, mol^{-1}$.

The enthalpy of combustion of carbon disulfide $(CS_2)$ is $-110.2 \, kJ \, mol^{-1}$. If the enthalpies of formation of $SO_2$ and $CO_2$ are $-297.4 \, kJ \, mol^{-1}$ and $-394.5 \, kJ \, mol^{-1}$ respectively,the enthalpy of formation of carbon disulfide is:

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