QuestionJune 28, 2025

Calculate the enthalpy of formation of carbon disulfide given that the enthalpy of combustion of it is 110.2kJmol^-1 and those of sulfur and carbon are 297.4 kJ and 394.5kJ/g atoms respectively. C(s)+2S(s)arrow CS_(2)(l);Delta H=?

Calculate the enthalpy of formation of carbon disulfide given that the enthalpy of combustion of it is 110.2kJmol^-1 and those of sulfur and carbon are 297.4 kJ and 394.5kJ/g atoms respectively. C(s)+2S(s)arrow CS_(2)(l);Delta H=?
Calculate the enthalpy of formation of carbon disulfide given that the enthalpy of
combustion of it is 110.2kJmol^-1 and those of sulfur and carbon are 297.4 kJ and
394.5kJ/g atoms respectively.
C(s)+2S(s)arrow CS_(2)(l);Delta H=?

Solution
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Answer

\Delta H = -87.9 \text{ kJ/mol} Explanation 1. Identify the Reaction Enthalpies The enthalpy of combustion for CS_2 is given as 110.2 \text{ kJ/mol}. For sulfur, it is 297.4 \text{ kJ/mol}, and for carbon, it is 394.5 \text{ kJ/mol}. 2. Write Combustion Reactions 1. C(s) + O_2(g) \rightarrow CO_2(g); \Delta H = -394.5 \text{ kJ/mol} 2. S(s) + O_2(g) \rightarrow SO_2(g); \Delta H = -297.4 \text{ kJ/mol} 3. CS_2(l) + 3O_2(g) \rightarrow CO_2(g) + 2SO_2(g); \Delta H = -110.2 \text{ kJ/mol} 3. Apply Hess's Law Use Hess's Law to find \Delta H for C(s) + 2S(s) \rightarrow CS_2(l). Rearrange reactions: - Reverse reaction 3: CO_2(g) + 2SO_2(g) \rightarrow CS_2(l) + 3O_2(g); \Delta H = 110.2 \text{ kJ/mol} - Add reactions 1 and 2 twice (for two moles of sulfur): - C(s) + O_2(g) \rightarrow CO_2(g); \Delta H = -394.5 \text{ kJ/mol} - 2[S(s) + O_2(g) \rightarrow SO_2(g)]; \Delta H = 2(-297.4) \text{ kJ/mol} 4. Calculate Enthalpy of Formation Sum the enthalpies: \Delta H = 110.2 - 394.5 - 2(297.4)

Explanation

1. Identify the Reaction Enthalpies<br /> The enthalpy of combustion for $CS_2$ is given as $110.2 \text{ kJ/mol}$. For sulfur, it is $297.4 \text{ kJ/mol}$, and for carbon, it is $394.5 \text{ kJ/mol}$.<br /><br />2. Write Combustion Reactions<br /> 1. $C(s) + O_2(g) \rightarrow CO_2(g); \Delta H = -394.5 \text{ kJ/mol}$<br /> 2. $S(s) + O_2(g) \rightarrow SO_2(g); \Delta H = -297.4 \text{ kJ/mol}$<br /> 3. $CS_2(l) + 3O_2(g) \rightarrow CO_2(g) + 2SO_2(g); \Delta H = -110.2 \text{ kJ/mol}$<br /><br />3. Apply Hess's Law<br /> Use Hess's Law to find $\Delta H$ for $C(s) + 2S(s) \rightarrow CS_2(l)$.<br /> Rearrange reactions:<br />- Reverse reaction 3: $CO_2(g) + 2SO_2(g) \rightarrow CS_2(l) + 3O_2(g); \Delta H = 110.2 \text{ kJ/mol}$<br />- Add reactions 1 and 2 twice (for two moles of sulfur):<br /> - $C(s) + O_2(g) \rightarrow CO_2(g); \Delta H = -394.5 \text{ kJ/mol}$<br /> - $2[S(s) + O_2(g) \rightarrow SO_2(g)]; \Delta H = 2(-297.4) \text{ kJ/mol}$<br /><br />4. Calculate Enthalpy of Formation<br /> Sum the enthalpies:<br />$$\Delta H = 110.2 - 394.5 - 2(297.4)$$
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